APEXC IIVersion .

Gregory Hartman, Ph.D.Department of Applied Mathema cs

Virginia Military Ins tute

Contribu ng AuthorsTroy Siemers, Ph.D.

Department of Applied Mathema cs

Virginia Military Ins tute

Brian Heinold, Ph.D.Department of Mathema cs and Computer Science

Mount Saint Mary’s University

Dimplekumar Chalishajar, Ph.D.Department of Applied Mathema cs

Virginia Military Ins tute

EditorJennifer Bowen, Ph.D.

Department of Mathema cs and Computer Science

The College of Wooster

Copyright © Gregory HartmanLicensed to the public under Crea ve CommonsA ribu on-Noncommercial . Interna onal PublicLicense

Contents

Table of Contents iii

Preface v

Integra on. An deriva ves and Indefinite Integra on . . . . . . . . . . . .. The Definite Integral . . . . . . . . . . . . . . . . . . . . . . .. Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . .. The Fundamental Theorem of Calculus . . . . . . . . . . . . . .. Numerical Integra on . . . . . . . . . . . . . . . . . . . . . . .

Techniques of An differen a on. Subs tu on . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Integra on by Parts . . . . . . . . . . . . . . . . . . . . . . . .. Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . .. Trigonometric Subs tu on . . . . . . . . . . . . . . . . . . . .. Par al Frac on Decomposi on . . . . . . . . . . . . . . . . . .. Hyperbolic Func ons . . . . . . . . . . . . . . . . . . . . . . .. L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . .. Improper Integra on . . . . . . . . . . . . . . . . . . . . . . .

Applica ons of Integra on. Area Between Curves . . . . . . . . . . . . . . . . . . . . . . .. Volume by Cross-Sec onal Area; Disk and Washer Methods . . .. The Shell Method . . . . . . . . . . . . . . . . . . . . . . . . .. Arc Length and Surface Area . . . . . . . . . . . . . . . . . . .. Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Fluid Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Sequences and Series. Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . .. Integral and Comparison Tests . . . . . . . . . . . . . . . . . .. Ra o and Root Tests . . . . . . . . . . . . . . . . . . . . . . . .. Alterna ng Series and Absolute Convergence . . . . . . . . . .. Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . . .. Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . .

A Solu ons To Selected Problems A.

Index A.

PA Note on Using this Text

Thank you for reading this short preface. Allow us to share a few key pointsabout the text so that youmay be er understand what you will find beyond thispage.

This text is Part II of a three–text series on Calculus. The first part coversmaterial taught in many “Calc ” courses: limits, deriva ves, and the basics ofintegra on, found in Chapters through . . The second text covers materialo en taught in “Calc :” integra on and its applica ons, along with an introduc-on to sequences, series and Taylor Polynomials, found in Chapters through. The third text covers topics common in “Calc ” or “mul variable calc:” para-

metric equa ons, polar coordinates, vector–valued func ons, and func ons ofmore than one variable, found in Chapters through . All three are availableseparately for free at www.apexcalculus.com. These three texts are intendedto work together and make one cohesive text, APEX Calculus, which can also bedownloaded from the website.

Prin ng the en re text as one volumemakes for a large, heavy, cumbersomebook. One can certainly only print the pages they currently need, but someprefer to have a nice, bound copy of the text. Therefore this text has been splitinto these three manageable parts, each of which can be purchased for about$ at Amazon.com.

A result of this spli ng is that some mes a concept is said to be explored inan “earlier/later sec on,” though that sec on does not actually appear in thispar cular text. Also, the index makes reference to topics, and page numbers,that do not appear in this text. This is done inten onally to show the readerwhat topics are available for study. Downloading the .pdf of APEX Calculus willensure that you have all the content.

For Students: How to Read this Text

Mathema cs textbooks have a reputa on for being hard to read. High–levelmathema cal wri ng o en seeks to say much with few words, and this styleo en seeps into texts of lower–level topics. This book was wri en with the goalof being easier to read than many other calculus textbooks, without becomingtoo verbose.

Each chapter and sec on starts with an introduc on of the coming material,hopefully se ng the stage for “why you should care,” and endswith a look aheadto see how the just–learned material helps address future problems.

Please read the text; it is wri en to explain the concepts of Calculus. Thereare numerous examples to demonstrate the meaning of defini ons, the truthof theorems, and the applica on of mathema cal techniques. When you en-counter a sentence you don’t understand, read it again. If it s ll doesn’t makesense, read on anyway, as some mes confusing sentences are explained by latersentences.

You don’t have to read every equa on. The examples generally show “all”the steps needed to solve a problem. Some mes reading through each step ishelpful; some mes it is confusing. When the steps are illustra ng a new tech-nique, one probably should follow each step closely to learn the new technique.When the steps are showing the mathema cs needed to find a number to beused later, one can usually skip ahead and see how that number is being used,instead of ge ng bogged down in reading how the number was found.

Most proofs have been omi ed. In mathema cs, proving something is al-ways true is extremely important, and entails much more than tes ng to see ifit works twice. However, students o en are confused by the details of a proof,or become concerned that they should have been able to construct this proofon their own. To alleviate this poten al problem, we do not include the proofsto most theorems in the text. The interested reader is highly encouraged to findproofs online or from their instructor. In most cases, one is very capable of un-derstanding what a theorem means and how to apply it without knowing fullywhy it is true.

Interac ve, D Graphics

New to Version . was the addi on of interac ve, D graphics in the .pdfversion. Nearly all graphs of objects in space can be rotated, shi ed, and zoomedin/out so the reader can be er understand the object illustrated.

As of this wri ng, the only pdf viewers that support these D graphics areAdobe Reader & Acrobat (and only the versions for PC/Mac/Unix/Linux com-puters, not tablets or smartphones). To ac vate the interac ve mode, click onthe image. Once ac vated, one can click/drag to rotate the object and use thescroll wheel on a mouse to zoom in/out. (A great way to inves gate an imageis to first zoom in on the page of the pdf viewer so the graphic itself takes upmuch of the screen, then zoom inside the graphic itself.) A CTRL-click/drag pansthe object le /right or up/down. By right-clicking on the graph one can accessa menu of other op ons, such as changing the ligh ng scheme or perspec ve.One can also revert the graph back to its default view. If you wish to deac vatethe interac vity, one can right-click and choose the “Disable Content” op on.

Thanks

There are many people who deserve recogni on for the important role theyhave played in the development of this text. First, I thank Michelle for her sup-port and encouragement, even as this “project from work” occupied my meand a en on at home. Many thanks to Troy Siemers, whose most importantcontribu ons extend far beyond the sec ons he wrote or the figures hecoded in Asymptote for D interac on. He provided incredible support, adviceand encouragement for which I am very grateful. My thanks to Brian Heinoldand Dimplekumar Chalishajar for their contribu ons and to Jennifer Bowen forreading through somuchmaterial and providing great feedback early on. Thanksto Troy, Lee Dewald, Dan Joseph, Meagan Herald, Bill Lowe, John David, VondaWalsh, Geoff Cox, Jessica Liber ni and other faculty of VMI who have given menumerous sugges ons and correc ons based on their experience with teachingfrom the text. (Special thanks to Troy, Lee & Dan for their pa ence in teachingCalc III while I was s ll wri ng the Calc III material.) Thanks to Randy Cone forencouraging his tutors of VMI’s Open Math Lab to read through the text andcheck the solu ons, and thanks to the tutors for spending their me doing so.A very special thanks to Kris Brown and Paul Janiczek who took this opportu-nity far above & beyond what I expected, me culously checking every solu onand carefully reading every example. Their comments have been extraordinarilyhelpful. I am also thankful for the support provided by Wane Schneiter, who asmy Dean provided me with extra me to work on this project. Finally, a hugeheap of thanks is to be bestowed on the numerous people I do not know whotook the me to email me correc ons and sugges ons. I am blessed to have somany people give of their me to make this book be er.

APEX – Affordable Print and Electronic teXts

APEX is a consor um of authors who collaborate to produce high–quality,low–cost textbooks. The current textbook–wri ng paradigm is facing a poten-al revolu on as desktop publishing and electronic formats increase in popular-

ity. However, wri ng a good textbook is no easy task, as the me requirementsalone are substan al. It takes countless hours of work to produce text, writeexamples and exercises, edit and publish. Through collabora on, however, thecost to any individual can be lessened, allowing us to create texts that we freelydistribute electronically and sell in printed form for an incredibly low cost. Hav-ing said that, nothing is en rely free; someone always bears some cost. This text“cost” the authors of this book their me, and that was not enough. APEX Cal-culuswould not exist had not the Virginia Military Ins tute, through a generousJackson–Hope grant, given the lead author significant me away from teachingso he could focus on this text.

Each text is available as a free .pdf, protected by a Crea ve Commons At-tribu on - Noncommercial . copyright. That means you can give the .pdf toanyone you like, print it in any form you like, and even edit the original contentand redistribute it. If you do the la er, you must clearly reference this work andyou cannot sell your edited work for money.

We encourage others to adapt this work to fit their own needs. One mightadd sec ons that are “missing” or remove sec ons that your students won’tneed. The source files can be found at github.com/APEXCalculus.

You can learn more at www.vmi.edu/APEX.

Version .

Key changes from Version . to . :

• Numerous typographical and “small”mathema cal correc ons (again, thanksto all my close readers!).

• “Large”mathema cal correc ons and adjustments. Therewere a numberof places in Version . where a defini on/theorem was not correct asstated. See www.apexcalculus.com for more informa on.

• More useful numbering of Examples, Theorems, etc. “Defini on . . ”refers to the second defini on of Chapter , Sec on .

• The addi on of Sec on . : Triple Integra onwith Cylindrical and Spher-ical Coordinates

• The addi on of Chapter : Vector Analysis.

: IWe have spent considerable me considering the deriva ves of a func on andtheir applica ons. In the following chapters, we are going to star ng thinkingin “the other direc on.” That is, given a func on f(x), we are going to considerfunc ons F(x) such that F ′(x) = f(x). There are numerous reasons this willprove to be useful: these func ons will help us compute area, volume, mass,force, pressure, work, and much more.

. An deriva ves and Indefinite Integra onGiven a func on y = f(x), a differen al equa on is one that incorporates y, x,and the deriva ves of y. For instance, a simple differen al equa on is:

y ′ = x.

Solving a differen al equa on amounts to finding a func on y that sa sfiesthe given equa on. Take a moment and consider that equa on; can you find afunc on y such that y ′ = x?

Can you find another?And yet another?Hopefully one was able to come upwith at least one solu on: y = x . “Find-

ing another” may have seemed impossible un l one realizes that a func on likey = x + also has a deriva ve of x. Once that discovery is made, finding “yetanother” is not difficult; the func on y = x + , , also has a deriva-ve of x. The differen al equa on y ′ = x has many solu ons. This leads us

to some defini ons.

Defini on . . An deriva ves and Indefinite Integrals

Let a func on f(x) be given. An an deriva ve of f(x) is a func on F(x)such that F ′(x) = f(x).

The set of all an deriva ves of f(x) is the indefinite integral of f, denotedby ∫

f(x) dx.

Make a note about our defini on: we refer to an an deriva ve of f, as op-posed to the an deriva ve of f, since there is always an infinite number of them.

Chapter Integra on

We o en use upper-case le ers to denote an deriva ves.Knowing one an deriva ve of f allows us to find infinitely more, simply by

adding a constant. Not only does this give usmore an deriva ves, it gives us allof them.

Theorem . . An deriva ve Forms

Let F(x) and G(x) be an deriva ves of f(x) on an interval I. Then thereexists a constant C such that, on I,

G(x) = F(x) + C.

Given a func on f defined on an interval I and one of its an deriva ves F,we know all an deriva ves of f on I have the form F(x) + C for some constantC. Using Defini on . . , we can say that

∫

f(x) dx = F(x) + C.

Let’s analyze this indefinite integral nota on.

..

∫f(x) dx = F(x) + C

.

Integrand

.

Integra onsymbol

.

Differen alof x

.

Onean deriva ve

.

Constant ofintegra on

Figure . . : Understanding the indefinite integral nota on.

Figure . . shows the typical nota on of the indefinite integral. The inte-gra on symbol,

∫, is in reality an “elongated S,” represen ng “take the sum.”

We will later see how sums and an deriva ves are related.The func on we want to find an an deriva ve of is called the integrand. It

contains the differen al of the variable we are integra ngwith respect to. The∫

symbol and the differen al dx are not “bookends” with a func on sandwiched inbetween; rather, the symbol

∫means “find all an deriva ves of what follows,”

and the func on f(x) and dx are mul plied together; the dx does not “just sitthere.”

Let’s prac ce using this nota on.

Notes:

. An deriva ves and Indefinite Integra on

Example . . Evalua ng indefinite integralsEvaluate

∫

sin x dx.

S We are asked to find all func ons F(x) such that F ′(x) =sin x. Some thoughtwill leadus to one solu on: F(x) = − cos x, because d

dx (− cos x) =sin x.

The indefinite integral of sin x is thus− cos x, plus a constant of integra on.So: ∫

sin x dx = − cos x+ C.

A commonly asked ques on is “What happened to the dx?” The unenlight-ened response is “Don’t worry about it. It just goes away.” A full understandingincludes the following.

This process of an differen a on is really solving a differen al ques on. Theintegral ∫

sin x dx

presents us with a differen al, dy = sin x dx. It is asking: “What is y?” We foundlots of solu ons, all of the form y = − cos x+ C.

Le ng dy = sin x dx, rewrite∫

sin x dx as∫

dy.

This is asking: “What func ons have a differen al of the form dy?” The answeris “Func ons of the form y+ C, where C is a constant.” What is y? We have lotsof choices, all differing by a constant; the simplest choice is y = − cos x.

Understanding all of this is more important later as we try to find an deriva-ves of more complicated func ons. In this sec on, we will simply explore the

rules of indefinite integra on, and one can succeed for now with answering“What happened to the dx?” with “It went away.”

Let’s prac ce once more before sta ng integra on rules.

Example . . Evalua ng indefinite integralsEvaluate

∫

( x + x+ ) dx.

S We seek a func on F(x) whose deriva ve is x + x + .When taking deriva ves, we can consider func ons term–by–term, so we canlikely do that here.

What func ons have a deriva ve of x ? Some thought will lead us to acubic, specifically x + C , where C is a constant.

Notes:

Chapter Integra on

What func ons have a deriva ve of x? Here the x term is raised to the firstpower, so we likely seek a quadra c. Some thought should lead us to x + C ,where C is a constant.

Finally, what func ons have a deriva ve of ? Func ons of the form x+C ,where C is a constant.

Our answer appears to be

∫

( x + x+ ) dx = x + C + x + C + x+ C .

We do not need three separate constants of integra on; combine them as oneconstant, giving the final answer of

∫

( x + x+ ) dx = x + x + x+ C.

It is easy to verify our answer; take the deriva ve of x + x + x + C andsee we indeed get x + x+ .

This final step of “verifying our answer” is important both prac cally andtheore cally. In general, taking deriva ves is easier than finding an deriva vesso checking our work is easy and vital as we learn.

We also see that taking the deriva ve of our answer returns the func on inthe integrand. Thus we can say that:

ddx

(∫

f(x) dx)

= f(x).

Differen a on “undoes” the work done by an differen a on.

Theorem . . gave a list of the deriva ves of common func ons we hadlearned at that point. We restate part of that list here to stress the rela onshipbetween deriva ves and an deriva ves. This list will also be useful as a glossaryof common an deriva ves as we learn.

Notes:

. An deriva ves and Indefinite Integra on

Theorem . . Deriva ves and An deriva ves

Common Differen a on Rules

. ddx

(cf(x)

)= c · f ′(x)

. ddx

(f(x)± g(x)

)=

f ′(x)± g′(x)

. ddx

(C)=

. ddx

(x)=

. ddx

(xn)= n · xn−

. ddx

(sin x

)= cos x

. ddx

(cos x

)= − sin x

. ddx

(tan x

)= sec x

. ddx

(csc x

)= − csc x cot x

. ddx

(sec x

)= sec x tan x

. ddx

(cot x

)= − csc x

. ddx

(ex)= ex

. ddx

(ax)= ln a · ax

. ddx

(ln x)= x

Common Indefinite Integral Rules

.∫c · f(x) dx = c ·

∫f(x) dx

.∫ (

f(x)± g(x))dx =

∫f(x) dx±

∫g(x) dx

.∫

dx = C

.∫

dx =∫dx = x+ C

.∫xn dx = n+ xn+ + C (n ̸= − )

.∫cos x dx = sin x+ C

.∫sin x dx = − cos x+ C

.∫sec x dx = tan x+ C

.∫csc x cot x dx = − csc x+ C

.∫sec x tan x dx = sec x+ C

.∫csc x dx = − cot x+ C

.∫ex dx = ex + C

.∫ax dx = ln a · ax + C

.∫

x dx = ln |x|+ C

We highlight a few important points from Theorem . . :

• Rule # states∫c · f(x) dx = c ·

∫f(x) dx. This is the Constant Mul ple

Rule: we can temporarily ignore constants when finding an deriva ves,just as we did when compu ng deriva ves (i.e., d

dx

(x)is just as easy to

compute as ddx

(x)). An example:

∫

cos x dx = ·∫

cos x dx = · (sin x+ C) = sin x+ C.

In the last step we can consider the constant as also being mul plied by

Notes:

Chapter Integra on

, but “ mes a constant” is s ll a constant, so we just write “C ”.

• Rule # is the Sum/Difference Rule: we can split integrals apart when theintegrand contains terms that are added/subtracted, as we did in Example. . . So:

∫

( x + x+ ) dx =∫

x dx+∫

x dx+∫

dx

=

∫

x dx+∫

x dx+∫

dx

= · x + · x + x+ C

= x + x + x+ C

In prac ce we generally do not write out all these steps, but we demon-strate them here for completeness.

• Rule # is the Power Rule of indefinite integra on. There are two impor-tant things to keep in mind:

. No ce the restric on that n ̸= − . This is important:∫

x dx ̸=“ x + C”; rather, see Rule # .

. We are presen ng an differen a on as the “inverse opera on” ofdifferen a on. Here is a useful quote to remember:

“Inverse opera ons do the opposite things in the oppositeorder.”

When taking a deriva ve using the Power Rule, we first mul ply bythe power, then second subtract from the power. To find the an-deriva ve, do the opposite things in the opposite order: first add

one to the power, then second divide by the power.

• Note that Rule # incorporates the absolute value of x. The exercises willwork the reader through why this is the case; for now, know the absolutevalue is important and cannot be ignored.

Ini al Value Problems

In Sec on . we saw that the deriva ve of a posi on func on gave a velocityfunc on, and the deriva ve of a velocity func on describes accelera on. Wecan now go “the other way:” the an deriva ve of an accelera on func on givesa velocity func on, etc. While there is just one deriva ve of a given func on,there are infinitely many an deriva ves. Therefore we cannot ask “What is thevelocity of an object whose accelera on is− /s ?”, since there is more thanone answer.

Notes:

. An deriva ves and Indefinite Integra on

We can find the answer if we provide more informa on with the ques on,as done in the following example. O en the addi onal informa on comes in theform of an ini al value, a value of the func on that one knows beforehand.

Example . . Solving ini al value problemsThe accelera on due to gravity of a falling object is − /s . At me t = ,a falling object had a velocity of − /s. Find the equa on of the object’svelocity.

S We want to know a velocity func on, v(t). We know twothings:

• The accelera on, i.e., v ′(t) = − , and

• the velocity at a specific me, i.e., v( ) = − .

Using the first piece of informa on, we know that v(t) is an an deriva ve ofv ′(t) = − . So we begin by finding the indefinite integral of− :

∫

(− ) dt = − t+ C = v(t).

Now we use the fact that v( ) = − to find C:

v(t) = − t+ Cv( ) = −

− ( ) + C = −C =

Thus v(t) = − t+ . We can use this equa on to understand the mo onof the object: when t = , the object had a velocity of v( ) = /s. Since thevelocity is posi ve, the object was moving upward.

When did the object begin moving down? Immediately a er v(t) = :

− t+ = ⇒ t = ≈ . s.

Recognize that we are able to determine quite a bit about the path of the objectknowing just its accelera on and its velocity at a single point in me.

Example . . Solving ini al value problemsFind f(t), given that f ′′(t) = cos t, f ′( ) = and f( ) = .

S We start by finding f ′(t), which is an an deriva ve of f ′′(t):∫

f ′′(t) dt =∫

cos t dt = sin t+ C = f ′(t).

Notes:

Chapter Integra on

So f ′(t) = sin t + C for the correct value of C. We are given that f ′( ) = ,so:

f ′( ) = ⇒ sin + C = ⇒ C = .

Using the ini al value, we have found f ′(t) = sin t+ .We now find f(t) by integra ng again.

f(t) =∫

f ′(t) dt =∫

(sin t+ ) dt = − cos t+ t+ C.

We are given that f( ) = , so

− cos + ( ) + C =

− + C =

C =

Thus f(t) = − cos t+ t+ .

This sec on introduced an deriva ves and the indefinite integral. We foundthey are needed when finding a func on given informa on about its deriva-ve(s). For instance, we found a velocity func on given an accelera on func-on.In the next sec on, we will see how posi on and velocity are unexpectedly

related by the areas of certain regions on a graph of the velocity func on. Then,in Sec on . , wewill see howareas and an deriva ves are closely ed together.This connec on is incredibly important, as indicated by the nameof the theoremthat describes it: The Fundamental Theorem of Calculus.

Notes:

Exercises .Terms and Concepts. Define the term “an deriva ve” in your own words.

. Is it more accurate to refer to “the” an deriva ve of f(x) or“an” an deriva ve of f(x)?

. Use your own words to define the indefinite integral off(x).

. Fill in the blanks: “Inverse opera ons do thethings in the order.”

. What is an “ini al value problem”?

. The deriva ve of a posi on func on is a func-on.

. The an deriva ve of an accelera on func on is afunc on.

. If F(x) is an an deriva ve of f(x), and G(x) is an an deriva-ve of g(x), give an an deriva ve of f(x) + g(x).

ProblemsIn Exercises – , evaluate the given indefinite integral.

.∫

x dx

.∫

x dx

.∫

( x − ) dx

.∫

dt

.∫

ds

.∫

tdt

.∫

tdt

.∫

√xdx

.∫

sec θ dθ

.∫

sin θ dθ

.∫

(sec x tan x+ csc x cot x) dx

.∫

eθ dθ

.∫

t dt

.∫ t

dt

.∫

( t+ ) dt

.∫

(t + )(t − t) dt

.∫

x x dx

.∫

eπ dx

.∫

a dx

. This problem inves gates why Theorem . . states that∫

xdx = ln |x|+ C.

(a) What is the domain of y = ln x?(b) Find d

dx

(

ln x)

.(c) What is the domain of y = ln(−x)?(d) Find d

dx

(

ln(−x))

.(e) You should find that /x has two types of an deriva-

ves, depending on whether x > or x < . Inone expression, give a formula for

∫

xdx that takes

these different domains into account, and explainyour answer.

In Exercises – , find f(x) described by the given ini alvalue problem.

. f ′(x) = sin x and f( ) =

. f ′(x) = ex and f( ) =

. f ′(x) = x − x and f(− ) =

. f ′(x) = sec x and f(π/ ) =

. f ′(x) = x and f( ) =

. f ′′(x) = and f ′( ) = , f( ) =

. f ′′(x) = x and f ′( ) = − , f( ) =

. f ′′(x) = ex and f ′( ) = , f( ) =

. f ′′(θ) = sin θ and f ′(π) = , f(π) =

. f ′′(x) = x + x − cos x and f ′( ) = , f( ) =

. f ′′(x) = and f ′( ) = , f( ) =

Review

. Use informa on gained from the first and second deriva-ves to sketch f(x) =

ex +.

. Given y = x ex cos x, find dy.

..... 5. 10.

5

.t (s)

.

y ( /s)

Figure . . : The area under a constantvelocity func on corresponds to distancetraveled.

.....

5

..

5

.−

.

5

.

t (s)

.

y ( /s)

Figure . . : The total displacement is thearea above the t–axis minus the area be-low the t–axis.

. The Definite Integral

. The Definite IntegralWe start with an easy problem. An object travels in a straight line at a constantvelocity of /s for seconds. How far away from its star ng point is the ob-ject?

We approach this problemwith the familiar “Distance= Rate× Time” equa-on. In this case, Distance = /s× s= feet.It is interes ng to note that this solu on of feet can be represented graph-

ically. Consider Figure . . , where the constant velocity of /s is graphed onthe axes. Shading the area under the line from t = to t = gives a rectanglewith an area of square units; when one considers the units of the axes, wecan say this area represents .

Now consider a slightly harder situa on (and not par cularly realis c): anobject travels in a straight line with a constant velocity of /s for seconds,then instantly reverses course at a rate of /s for seconds. (Since the objectis traveling in the opposite direc on when reversing course, we say the velocityis a constant− /s.) How far away from the star ng point is the object – whatis its displacement?

Here we use “Distance= Rate × Time + Rate × Time ,” which is

Distance = · + (− ) · = .

Hence the object is feet from its star ng loca on.We can again depict this situa on graphically. In Figure . . we have the

veloci es graphed as straight lines on [ , ] and [ , ], respec vely. The dis-placement of the object is

“Area above the t–axis − Area below the t–axis,”

which is easy to calculate as − = feet.Now consider a more difficult problem.

Example . . Finding posi on using velocityThe velocity of an object moving straight up/down under the accelera on ofgravity is given as v(t) = − t+ , where me t is given in seconds and velocityis in /s. When t = , the object had a height of .

. What was the ini al velocity of the object?

. What was the maximum height of the object?

. What was the height of the object at me t = ?

S It is straigh orward to find the ini al velocity; at me t = ,v( ) = − · + = /s.

Notes:

........−5.

5

.

t (s)

.

y ( /s)

Figure . . : A graph of v(t) = − t +; the shaded areas help determine dis-

placement.

Chapter Integra on

To answer ques ons about the height of the object, we need to find theobject’s posi on func on s(t). This is an ini al value problem, which we studiedin the previous sec on. We are told the ini al height is , i.e., s( ) = . Weknow s ′(t) = v(t) = − t+ . To find s, we find the indefinite integral of v(t):

∫

v(t) dt =∫

(− t+ ) dt = − t + t+ C = s(t).

Since s( ) = , we conclude that C = and s(t) = − t + t.To find the maximum height of the object, we need to find the maximum of

s. Recalling our work finding extreme values, we find the cri cal points of s byse ng its deriva ve equal to and solving for t:

s ′(t) = − t+ = ⇒ t = / = . s.

(No ce how we ended up just finding when the velocity was /s!) The firstderiva ve test shows this is a maximum, so the maximum height of the objectis found at

s( . ) = − ( . ) + ( . ) = .

The height at me t = is now straigh orward to compute: it is s( ) = .

While we have answered all three ques ons, let’s look at them again graph-ically, using the concepts of area that we explored earlier.

Figure . . shows a graph of v(t) on axes from t = to t = . It is againstraigh orward to find v( ). How can we use the graph to find the maximumheight of the object?

Recall how in our previous work that the displacement of the object (in thiscase, its height) was found as the area under the velocity curve, as shaded in thefigure. Moreover, the area between the curve and the t–axis that is below thet–axis counted as “nega ve” area. That is, it represents the object coming backtoward its star ng posi on. So to find the maximum distance from the star ngpoint – the maximum height – we find the area under the velocity line that isabove the t–axis, i.e., from t = to t = . . This region is a triangle; its area is

Area = Base× Height = × . s× /s = ,

which matches our previous calcula on of the maximum height.Finally, to find the height of the object at me t = we calculate the total

“signed area” (where some area is nega ve) under the velocity func on fromt = to t = . This signed area is equal to s( ), the displacement (i.e., signeddistance) from the star ng posi on at t = to the posi on at me t = . Thatis,

Displacement = Area above the t–axis− Area below t–axis.

Notes:

. The Definite Integral

The regions are triangles, and we find

Displacement = ( . s)( /s)− (. s)( /s) = .

This also matches our previous calcula on of the height at t = .No ce howweanswered each ques on in this example in twoways. Our first

methodwas tomanipulate equa ons using our understanding of an deriva vesand deriva ves. Our second method was geometric: we answered ques onslooking at a graph and finding the areas of certain regions of this graph.

The above example does not prove a rela onship between area under a ve-locity func on and displacement, but it does imply a rela onship exists. Sec on. will fully establish fact that the area under a velocity func on is displace-

ment.Given a graph of a func on y = f(x), we will find that there is great use in

compu ng the area between the curve y = f(x) and the x-axis. Because of this,we need to define some terms.

Defini on . . The Definite Integral, Total Signed Area

Let y = f(x) be defined on a closed interval [a, b]. The total signed areafrom x = a to x = b under f is:(area under f and above the x–axis on [a, b])− (area above f and under

the x–axis on [a, b]).

The definite integral of f on [a, b] is the total signed area of f on [a, b],denoted

∫ b

af(x) dx,

where a and b are the bounds of integra on.

By our defini on, the definite integral gives the “signed area under f.” Weusually drop the word “signed” when talking about the definite integral, andsimply say the definite integral gives “the area under f ” or, more commonly,“the area under the curve.”

The previous sec on introduced the indefinite integral, which related to an-deriva ves. We have now defined the definite integral, which relates to areas

under a func on. The two are very much related, as we’ll see when we learnthe Fundamental Theorem of Calculus in Sec on . . Recall that earlier we saidthat the “

∫” symbol was an “elongated S” that represented finding a “sum.” In

the context of the definite integral, this nota on makes a bit more sense, as weare adding up areas under the func on f.

Notes:

..........−..

x

.

y

Figure . . : A graph of f(x) in Example. . .

..........−..

x

.

y

Figure . . : A graph of f in Example. . . (Yes, it looks just like the graph of

f in Figure . . , just with a different y-scale.)

Chapter Integra on

We prac ce using this nota on.

Example . . Evalua ng definite integralsConsider the func on f given in Figure . . .

Find:

.∫

f(x) dx

.∫

f(x) dx

.∫

f(x) dx

.∫

f(x) dx

.∫

f(x) dx

S

.∫

f(x) dx is the area under f on the interval [ , ]. This region is a triangle,so the area is

∫f(x) dx = ( )( ) = . .

.∫

f(x) dx represents the area of the triangle found under the x–axis on[ , ]. The area is ( )( ) = ; since it is found under the x–axis, this is“nega ve area.” Therefore

∫f(x) dx = − .

.∫

f(x) dx is the total signed area under fon [ , ]. This is . +(− ) = . .

.∫

f(x) dx is the area under f on [ , ]. This is sketched in Figure . . .Again, the region is a triangle, with height mes that of the height of theoriginal triangle. Thus the area is

∫f(x) dx = ( )( ) = . .

.∫

f(x) dx is the area under f on the “interval” [ , ]. This describes a linesegment, not a region; it has no width. Therefore the area is .

This example illustrates some of the proper es of the definite integral, givenhere.

Notes:

. The Definite Integral

Theorem . . Proper es of the Definite Integral

Let f and g be defined on a closed interval I that contains the values a, band c, and let k be a constant. The following hold:

.∫ a

af(x) dx =

.∫ b

af(x) dx+

∫ c

bf(x) dx =

∫ c

af(x) dx

.∫ b

af(x) dx = −

∫ a

bf(x) dx

.∫ b

a

(f(x)± g(x)

)dx =

∫ b

af(x) dx±

∫ b

ag(x) dx

.∫ b

ak · f(x) dx = k ·

∫ b

af(x) dx

We give a brief jus fica on of Theorem . . here.

. As demonstrated in Example . . , there is no “area under the curve”when the region has no width; hence this definite integral is .

. This states that total area is the sum of the areas of subregions. It is easilyconsidered when we let a < b < c. We can break the interval [a, c] intotwo subintervals, [a, b] and [b, c]. The total area over [a, c] is the area over[a, b] plus the area over [b, c].It is important to note that this s ll holds true even if a < b < c is nottrue. We discuss this in the next point.

. This property can be viewed a merely a conven on to make other proper-esworkwell. (Later wewill see how this property has a jus fica on all its

own, not necessarily in support of other proper es.) Suppose b < a < c.The discussion from the previous point clearly jus fies

∫ a

bf(x) dx+

∫ c

af(x) dx =

∫ c

bf(x) dx. ( . )

However, we s ll claim that, as originally stated,∫ b

af(x) dx+

∫ c

bf(x) dx =

∫ c

af(x) dx. ( . )

Notes:

.....

a

.

b

.

c

.

x

.

y

Figure . . : A graph of a func on in Ex-ample . . .

Chapter Integra on

How do Equa ons ( . ) and ( . ) relate? Start with Equa on ( . ):∫ a

bf(x) dx+

∫ c

af(x) dx =

∫ c

bf(x) dx

∫ c

af(x) dx = −

∫ a

bf(x) dx+

∫ c

bf(x) dx

Property ( ) jus fies changing the sign and switching the bounds of inte-

gra on on the −∫ a

bf(x) dx term; when this is done, Equa ons ( . ) and

( . ) are equivalent.The conclusion is this: by adop ng the conven on of Property ( ), Prop-erty ( ) holds no ma er the order of a, b and c. Again, in the next sec onwe will see another jus fica on for this property.

, . Each of these may be non–intui ve. Property ( ) states that when onescales a func on by, for instance, , the area of the enclosed region alsois scaled by a factor of . Both Proper es ( ) and ( ) can be proved usinggeometry. The details are not complicated but are not discussed here.

Example . . Evalua ng definite integrals using Theorem . . .Consider the graph of a func on f(x) shown in Figure . . . Answer the follow-ing:

. Which value is greater:∫ b

af(x) dx or

∫ c

bf(x) dx?

. Is∫ c

af(x) dx greater or less than ?

. Which value is greater:∫ b

af(x) dx or

∫ b

cf(x) dx?

S

.∫ ba f(x) dx has a posi ve value (since the area is above the x–axis) whereas∫ cb f(x) dx has a nega ve value. Hence

∫ ba f(x) dx is bigger.

.∫ ca f(x) dx is the total signed area under f between x = a and x = c. Sincethe region below the x–axis looks to be larger than the region above, weconclude that the definite integral has a value less than .

. Note how the second integral has the bounds “reversed.” Therefore∫ bc f(x)dx

represents a posi ve number, greater than the area described by the firstdefinite integral. Hence

∫ bc f(x) dx is greater.

Notes:

.....(− ,−8).

(5, 6)

.

R

.

R

.

−

..

5

. −.

−5

.

5

..

x

.

y

(a)

.....−3. 3.

5

.....x

.

y

(b)

Figure . . : A graph of f(x) = x − in(a) and f(x) =

√− x in (b), from Exam-

ple . . .

.....

11

.

11

.

38

.

−5

.

5

.

10

.

15

.

a

.

b

.

c

.

t (s)

.

y ( /s)

Figure . . : A graph of a velocity in Ex-ample . . .

. The Definite Integral

The area defini on of the definite integral allows us to use geometry to com-pute the definite integral of some simple func ons.

Example . . Evalua ng definite integrals using geometryEvaluate the following definite integrals:

.

∫

−( x− ) dx .

∫

−

√

− x dx.

S

. It is useful to sketch the func on in the integrand, as shown in Figure. . (a). We see we need to compute the areas of two regions, which

we have labeled R and R . Both are triangles, so the area computa on isstraigh orward:

R : ( )( ) = R : ( ) = .

Region R lies under the x–axis, hence it is counted as nega ve area (wecan think of the triangle’s height as being “− ”), so

∫

−( x− ) dx = − + = − .

. Recognize that the integrand of this definite integral describes a half circle,as sketched in Figure . . (b), with radius . Thus the area is:

∫

−

√

− x dx = πr = π.

Example . . Understanding mo on given velocityConsider the graph of a velocity func on of an object moving in a straight line,given in Figure . . , where the numbers in the given regions gives the area ofthat region. Assume that the definite integral of a velocity func on gives dis-placement. Find the maximum speed of the object and its maximum displace-ment from its star ng posi on.

S Since the graph gives velocity, finding the maximum speedis simple: it looks to be /s.

At me t = , the displacement is ; the object is at its star ng posi on. Atme t = a, the object has moved backward feet. Between mes t = a and

Notes:

........

5

..x

.

y

Figure . . : What is the area below y =x on [ , ]? The region is not a usual ge-ometric shape.

Chapter Integra on

t = b, the object moves forward feet, bringing it into a posi on feet for-ward of its star ng posi on. From t = b to t = c the object is moving backwardsagain, hence its maximum displacement is feet from its star ng posi on.

In our examples, we have either found the areas of regions that have nicegeometric shapes (such as rectangles, triangles and circles) or the areas weregiven to us. Consider Figure . . , where a region below y = x is shaded. Whatis its area? The func on y = x is rela vely simple, yet the shape it defines hasan area that is not simple to find geometrically.

In the next sec on we will explore how to find the areas of such regions.

Notes:

Exercises .Terms and Concepts

. What is “total signed area”?

. What is “displacement”?

. What is∫

sin x dx?

. Give a single definite integral that has the same value as∫

( x+ ) dx+∫

( x+ ) dx.

Problems

In Exercises – , a graph of a func on f(x) is given. Usingthe geometry of the graph, evaluate the definite integrals.

.

.....

y = −2x + 4

.

2

.

4

. −4.

−2

.

2

.

4

.

x

.

y

(a)∫

(− x+ ) dx

(b)∫

(− x+ ) dx

(c)∫

(− x+ ) dx

(d)∫

(− x+ ) dx

(e)∫

(− x+ ) dx

(f)∫

(− x+ ) dx

.

.....

y = f(x)

......−.

−

...

x

.

y

(a)∫

f(x) dx

(b)∫

f(x) dx

(c)∫

f(x) dx

(d)∫

f(x) dx

(e)∫

f(x) dx

(f)∫

− f(x) dx

.

.....

y = f(x)

.......x

.

y

(a)∫

f(x) dx

(b)∫

f(x) dx

(c)∫

f(x) dx

(d)∫

x dx

(e)∫

( x− ) dx

(f)∫

( x− ) dx

.

.....

y = x −

.....−

....

x

.

y

(a)∫

(x− ) dx

(b)∫

(x− ) dx

(c)∫

(x− ) dx

(d)∫

(x− ) dx

(e)∫

(x− ) dx

(f)∫

(

(x− ) +)

dx

.

.....

f(x) =√

− (x − )

........x

.

y

(a)∫

f(x) dx

(b)∫

f(x) dx

(c)∫

f(x) dx

(d)∫

f(x) dx

.

f(x) =

5x

y

(a)∫

f(x) dx

(b)∫

f(x) dx

(c)∫

f(x) dx

(d)∫ b

af(x) dx, where

≤ a ≤ b ≤

In Exercises – , a graph of a func on f(x) is given; thenumbers inside the shaded regions give the area of that re-gion. Evaluate the definite integrals using this area informa-on.

.

.....

y = f(x)

.

59

......−.

−5

.

5

.

x

.

y

(a)∫

f(x) dx

(b)∫

f(x) dx

(c)∫

f(x) dx

(d)∫

− f(x) dx

.

.....

f(x) = sin(πx/ )

.

/π

.

/π

.....

−

..

x

.

y

(a)∫

f(x) dx

(b)∫

f(x) dx

(c)∫

f(x) dx

(d)∫

f(x) dx

.

f(x) = x −

− −

−

x

y

(a)∫ −

−f(x) dx

(b)∫

f(x) dx

(c)∫

−f(x) dx

(d)∫

f(x) dx

.

.....

f(x) = x

. /. 7/....... x.

y

(a)∫

x dx

(b)∫

(x + ) dx

(c)∫

(x− ) dx

(d)∫

(

(x− ) +)

dx

In Exercises – , a graph of the velocity func on of an ob-ject moving in a straight line is given. Answer the ques onsbased on that graph.

.

........− ...

t (s)

.

y ( /s)

(a) What is the object’s maximum velocity?(b) What is the object’s maximum displacement?(c) What is the object’s total displacement on [ , ]?

.

.............t (s)

.

y ( /s)

(a) What is the object’s maximum velocity?(b) What is the object’s maximum displacement?(c) What is the object’s total displacement on [ , ]?

. An object is thrown straight up with a velocity, in /s, givenby v(t) = − t + , where t is in seconds, from a heightof feet.

(a) What is the object’s maximum velocity?(b) What is the object’s maximum displacement?(c) When does the maximum displacement occur?(d) When will the object reach a height of ? (Hint: find

when the displacement is− .)

. An object is thrown straight up with a velocity, in /s, givenby v(t) = − t + , where t is in seconds, from a heightof feet.

(a) What is the object’s ini al velocity?(b) When is the object’s displacement ?(c) How long does it take for the object to return to its

ini al height?(d) When will the object reach a height of feet?

In Exercises – , let

•∫

f(x) dx = ,

•∫

f(x) dx = ,

•∫

g(x) dx = − , and

•∫

g(x) dx = .

Use these values to evaluate the given definite integrals.

.∫

(

f(x) + g(x))

dx

.∫

(

f(x)− g(x))

dx

.∫

(

f(x) + g(x))

dx

. Find nonzero values for a and b such that∫

(

af(x) + bg(x))

dx =

In Exercises – , let

•∫

s(t) dt = ,

•∫

s(t) dt = ,

•∫

r(t) dt = − , and

•∫

r(t) dt = .

Use these values to evaluate the given definite integrals.

.∫

(

s(t) + r(t))

dt

.∫

(

s(t)− r(t))

dt

.∫

(

πs(t)− r(t))

dt

. Find nonzero values for a and b such that∫

(

ar(t) + bs(t))

dt =

ReviewIn Exercises – , evaluate the given indefinite integral.

.∫

(

x − x + x−)

dx

.∫

(

sin x− cos x+ sec x)

dx

.∫

(√t+

t+ t) dt

.∫(

x− csc x cot x

)

dx

.............

x

.

y

Figure . . : A graph of f(x) = x − x .What is the area of the shaded region?

.....

RHR

.

MPR

.

LHR

.

other

.........

x

.

y

Figure . . : Approxima ng∫

( x −x ) dx using rectangles. The heights of therectangles are determined using differentrules.

Chapter Integra on

. Riemann SumsIn the previous sec on we defined the definite integral of a func on on [a, b] tobe the signed area between the curve and the x–axis. Some areas were simpleto compute; we ended the sec on with a region whose area was not simple tocompute. In this sec on we develop a technique to find such areas.

A fundamental calculus technique is to first answer a given problem with anapproxima on, then refine that approxima on to make it be er, then use limitsin the refining process to find the exact answer. That is what we will do here.

Consider the region given in Figure . . , which is the area under y = x−xon [ , ]. What is the signed area of this region – i.e., what is

∫( x− x ) dx?

We start by approxima ng. We can surround the region with a rectanglewith height and width of and find the area is approximately square units.This is obviously an over–approxima on; we are including area in the rectanglethat is not under the parabola.

We have an approxima on of the area, using one rectangle. How can werefine our approxima on tomake it be er? The key to this sec on is this answer:use more rectangles.

Let’s use rectangles with an equal width of . This par ons the interval[ , ] into subintervals, [ , ], [ , ], [ , ] and [ , ]. On each subinterval wewill draw a rectangle.

There are three common ways to determine the height of these rectangles:the Le Hand Rule, the Right Hand Rule, and theMidpoint Rule. The Le HandRule says to evaluate the func on at the le –hand endpoint of the subintervaland make the rectangle that height. In Figure . . , the rectangle drawn on theinterval [ , ] has height determined by the Le Hand Rule; it has a height off( ). (The rectangle is labeled “LHR.”)

The Right Hand Rule says the opposite: on each subinterval, evaluate thefunc on at the right endpoint and make the rectangle that height. In the figure,the rectangle drawn on [ , ] is drawn using f( ) as its height; this rectangle islabeled “RHR.”.

The Midpoint Rule says that on each subinterval, evaluate the func on atthe midpoint and make the rectangle that height. The rectangle drawn on [ , ]was made using the Midpoint Rule, with a height of f( . ). That rectangle islabeled “MPR.”

These are the three most common rules for determining the heights of ap-proxima ng rectangles, but one is not forced to use one of these threemethods.The rectangle on [ , ] has a height of approximately f( . ), very close to theMidpoint Rule. It was chosen so that the area of the rectangle is exactly the areaof the region under f on [ , ]. (Later you’ll be able to figure how to do this, too.)

The following example will approximate the value of∫

( x − x ) dx usingthese rules.

Notes:

.................

x

.

y

(a)

.................

x

.

y

(b)

.....

.7

.

.7

.

.7

.

.7

.........

x

.

y

(c)

Figure . . : Approxima ng∫

( x −x ) dx in Example . . . In (a), the LeHand Rule is used; in (b), the Right HandRule is used; in (c), the Midpoint Rule isused.

. Riemann Sums

Example . . Using the Le Hand, Right Hand and Midpoint RulesApproximate the value of

∫( x − x ) dx using the Le Hand Rule, the Right

Hand Rule, and the Midpoint Rule, using equally spaced subintervals.

S We break the interval [ , ] into four subintervals as before.In Figure . . (a) we see rectangles drawn on f(x) = x − x using the LeHand Rule. (The areas of the rectangles are given in each figure.)

Note how in the first subinterval, [ , ], the rectangle has height f( ) = .We add up the areas of each rectangle (height× width) for our Le Hand Ruleapproxima on:

f( ) · + f( ) · + f( ) · + f( ) · =

+ + + = .

Figure . . (b) shows rectangles drawn under f using the Right Hand Rule;note how the [ , ] subinterval has a rectangle of height .

In this example, these rectangle seem to be the mirror image of those foundin part (a) of the Figure. This is because of the symmetry of our shaded region.Our approxima on gives the same answer as before, though calculated a differ-ent way:

f( ) · + f( ) · + f( ) · + f( ) · =

+ + + = .

Figure . . (c) shows rectangles drawn under f using the Midpoint Rule.This gives an approxima on of

∫( x− x ) dx as:

f( . ) · + f( . ) · + f( . ) · + f( . ) · =

. + . + . + . = .

Our three methods provide two approxima ons of∫

( x− x ) dx: and .

Summa on Nota on

It is hard to tell at this moment which is a be er approxima on: or ?We can con nue to refine our approxima on by using more rectangles. Thenota on can become unwieldy, though, as we add up longer and longer lists ofnumbers. We introduce summa on nota on to ameliorate this problem.

Notes:

Chapter Integra on

Suppose we wish to add up a list of numbers a , a , a , …, a . Instead ofwri ng

a + a + a + a + a + a + a + a + a ,

we use summa on nota on and write

..

9∑

i=1

ai.

.i=indexof summa on

. lowerbound

.

upperbound

.

summand

Figure . . : Understanding summa on nota on.

The upper case sigma represents the term “sum.” The index of summa onin this example is i; any symbol can be used. By conven on, the index takes ononly the integer values between (and including) the lower and upper bounds.

Let’s prac ce using this nota on.

Example . . Using summa on nota onLet the numbers {ai} be defined as ai = i − for integers i, where i ≥ . Soa = , a = , a = , etc. (The output is the posi ve odd integers). Evaluatethe following summa ons:

.∑

i=

ai .∑

i=

( ai − ) .∑

i=

(ai)

S

.∑

i=

ai = a + a + a + a + a + a

= + + + + +

= .

. Note the star ng value is different than :

∑

i=

( ai − ) = ( a − ) + ( a − ) + ( a − ) + ( a − ) + ( a − )

= + + + +

= .

Notes:

. Riemann Sums

.

∑

i=

(ai) = (a ) + (a ) + (a ) + (a )

= + + +

= .

It might seem odd to stress a new, concise way of wri ng summa ons onlyto write each term out as we add them up. It is. The following theorem givessome of the proper es of summa ons that allow us to work with them withoutwri ng individual terms. Examples will follow.

Theorem . . Proper es of Summa ons

.n∑

i=

c = c · n, where c is a constant.

.n∑

i=m

(ai ± bi) =n∑

i=m

ai ±n∑

i=m

bi

.n∑

i=m

c · ai = c ·n∑

i=m

ai

.j∑

i=m

ai +n∑

i=j+

ai =n∑

i=m

ai

.n∑

i=

i =n(n+ )

.n∑

i=

i =n(n+ )( n+ )

.n∑

i=

i =

(n(n+ )

)

Example . . Evalua ng summa ons using Theorem . .Revisit Example . . and, using Theorem . . , evaluate

∑

i=

ai =∑

i=

( i− ).

Notes:

.......

x

.

x

.

x9

.

x

.

x 7

Figure . . : Dividing [ , ] intoequally spaced subintervals.

Chapter Integra on

S

∑

i=

( i− ) =∑

i=

i−∑

i=

( )

=

(∑

i=

i

)

−

=( + ) −

= − =

We obtained the same answer without wri ng out all six terms. When dealingwith small sizes of n, it may be faster to write the terms out by hand. However,Theorem . . is incredibly important when dealing with large sums as we’llsoon see.

Riemann Sums

Consider again∫

( x − x ) dx. We will approximate this definite integralusing equally spaced subintervals and the Right Hand Rule in Example . . .Before doing so, it will pay to do some careful prepara on.

Figure . . shows a number line of [ , ] divided, or par oned, intoequally spaced subintervals. Wedenote as x ; wehavemarked the values of x ,x , x and x . We couldmark themall, but the figurewould get crowded. Whileit is easy to figure that x = . , in general, we want a method of determiningthe value of xi without consul ng the figure. Consider:

..

xi = x1 + (i− 1)∆x

. star ngvalue

.

number ofsubintervals

between x1 and xi

. subintervalsize

So x = x + ( / ) = . .If we had par oned [ , ] into equally spaced subintervals, each subin-

terval would have length∆x = / = . . We could compute x as

x = x + ( / ) = . .

(That was far faster than crea ng a sketch first.)

Notes:

. Riemann Sums

Given any subdivision of [ , ], the first subinterval is [x , x ]; the second is[x , x ]; the i th subinterval is [xi, xi+ ].

When using the Le Hand Rule, the height of the i th rectangle will be f(xi).Whenusing theRightHandRule, the height of the i th rectanglewill be f(xi+ ).

Whenusing theMidpoint Rule, the height of the i th rectanglewill be f(xi + xi+

)

.

Thus approxima ng∫

( x− x ) dx with equally spaced subintervals canbe expressed as follows, where∆x = / = / :

Le Hand Rule:∑

i=

f(xi)∆x

Right Hand Rule:∑

i=

f(xi+ )∆x

Midpoint Rule:∑

i=

f(xi + xi+

)

∆x

Weuse these formulas in the next two examples. The following example letsus prac ce using the Right Hand Rule and the summa on formulas introducedin Theorem . . .

Example . . Approxima ng definite integrals using sumsApproximate

∫( x−x ) dx using the Right Hand Rule and summa on formulas

with and equally spaced intervals.

S Using the formula derived before, using equally spacedintervals and the Right Hand Rule, we can approximate the definite integral as

∑

i=

f(xi+ )∆x.

We have∆x = / = . . Since xi = + (i− )∆x, we have

xi+ = +((i+ )−

)∆x

= i∆x

Notes:

.............

x

.

y

Figure . . : Approxima ng∫

( x −x ) dx with the Right Hand Rule andevenly spaced subintervals.

Chapter Integra on

Using the summa on formulas, consider:∫

( x− x ) dx ≈∑

i=

f(xi+ )∆x

=∑

i=

f(i∆x)∆x

=∑

i=

(i∆x− (i∆x)

)∆x

=∑

i=

( i∆x − i ∆x )

= ( ∆x )∑

i=

i−∆x∑

i=

i ( . )

= ( ∆x )· −∆x

( )( )(∆x = . )

= .

Wewere able to sum up the areas of rectangles with very li le computa on.In Figure . . the func on and the rectangles are graphed. While somerectangles over–approximate the area, other under–approximate the area (byabout the same amount). Thus our approximate area of . is likely a fairlygood approxima on.

No ce Equa on ( . ); by changing the ’s to , ’s (and appropriatelychanging the value of ∆x), we can use that equa on to sum up rectan-gles! We do so here, skipping from the original summand to the equivalent ofEqua on ( . ) to save space. Note that∆x = / = . .

∫

( x− x ) dx ≈∑

i=

f(xi+ )∆x

= ( ∆x )∑

i=

i−∆x∑

i=

i

= ( ∆x )· −∆x

( )( )

= .

Using many, many rectangles, we have a likely good approxima on of∫

( x− x )∆x. That is,∫

( x− x ) dx ≈ . .

Notes:

. Riemann Sums

Before the above example, we statedwhat the summa ons for the Le Hand,Right Hand and Midpoint Rules looked like. Each had the same basic structure,which was:

. each rectangle has the same width, which we referred to as∆x, and

. each rectangle’s height is determined by evalua ng f at a par cular pointin each subinterval. For instance, the Le Hand Rule states that each rect-angle’s height is determined by evalua ng f at the le hand endpoint ofthe subinterval the rectangle lives on.

One could par on an interval [a, b]with subintervals that do not have the samesize. We refer to the length of the i th subinterval as∆xi. Also, one could deter-mine each rectangle’s height by evalua ng f at any point ci in the i th subinterval.Thus the height of the i th subinterval would be f(ci), and the area of the i th rect-angle would be f(ci)∆xi. These ideas are formally defined below.

Defini on . . Par on

A par on ∆x of a closed interval [a, b] is a set of numbers x , x , . . .xn+ where

a = x < x < . . . < xn < xn+ = b.

The length of the i th subinterval, [xi, xi+ ], is ∆xi = xi+ − xi. If [a, b] ispar oned into subintervals of equal length, we let ∆x represent thelength of each subinterval.

The size of the par on, denoted ||∆x||, is the length of the largestsubinterval of the par on.

Summa ons of rectangleswith area f(ci)∆xi are named a ermathema cianGeorg Friedrich Bernhard Riemann, as given in the following defini on.

Defini on . . Riemann Sum

Let f be defined on a closed interval [a, b], let∆x be a par on of [a, b]and let ci denote any value in the i th subinterval.The sum

n∑

i=

f(ci)∆xi

is a Riemann sum of f on [a, b].

Notes:

.............

x

.

y

Figure . . : An example of a general Rie-mann sum to approximate

∫

( x−x ) dx.

Chapter Integra on

Figure . . shows the approxima ng rectangles of a Riemann sumof∫

( x−x ) dx. While the rectangles in this example do not approximate well the shadedarea, they demonstrate that the subinterval widths may vary and the heights ofthe rectangles can be determined without following a par cular rule.

“Usually” Riemann sums are calculated using one of the three methods wehave introduced. The uniformity of construc on makes computa ons easier.Beforeworking another example, let’s summarize someofwhatwehave learnedin a convenient way.

Key Idea . . Riemann Sum Concepts

Consider∫ b

af(x) dx ≈

n∑

i=

f(ci)∆xi.

. When the n subintervals have equal length,∆xi = ∆x =b− an

.

. The i th term of an equally spaced par on is xi = a + (i − )∆x.(Thus x = a and xn+ = b.)

. The Le Hand Rule summa on is:n∑

i=

f(xi)∆x.

. The Right Hand Rule summa on is:n∑

i=

f(xi+ )∆x.

. The Midpoint Rule summa on is:n∑

i=

f(xi + xi+

)

∆x.

Let’s do another example.

Example . . Approxima ng definite integrals with sumsApproximate

∫

− ( x + ) dx using the Midpoint Rule and equally spacedintervals.

S Following Key Idea . . , we have

∆x =− (− )

= / and xi = (− ) + ( / )(i− ) = i/ − / .

Notes:

.....

−

.

−

.....

7

. −8.

x

.

y

Figure . . : Approxima ng∫

− ( x +) dx using the Midpoint Rule and

evenly spaced subintervals in Example. . .

. Riemann Sums

As we are using the Midpoint Rule, we will also need xi+ andxi + xi+ . Since

xi = i/ − / , xi+ = (i+ )/ − / = i/ − . This gives

xi + xi+=

(i/ − / ) + (i/ − )=

i− /= i/ − / .

We now construct the Riemann sum and compute its value using summa onformulas.

∫

−( x+ ) dx ≈

∑

i=

f(xi + xi+

)

∆x

=∑

i=

f(i/ − / )∆x

=∑

i=

((i/ − / ) +

)∆x

= ∆x∑

i=

[( )

i−]

= ∆x

(∑

i=

(i)−∑

i=

( ))

=

(

· ( ) − ·)

= = .

Note the graph of f(x) = x + in Figure . . . The regions whose area iscomputed by the definite integral are triangles, meaning we can find the exactanswer without summa on techniques. We find that the exact answer is indeed

. . One of the strengths of the Midpoint Rule is that o en each rectangleincludes area that should not be counted, but misses other area that should.When the par on size is small, these two amounts are about equal and theseerrors almost “cancel each other out.” In this example, since our func on is aline, these errors are exactly equal and they do cancel each other out, giving usthe exact answer.

Note too thatwhen the func on is nega ve, the rectangles have a “nega ve”height. When we compute the area of the rectangle, we use f(ci)∆x; when f isnega ve, the area is counted as nega ve.

No ce in the previous example that while we used equally spaced inter-vals, the number “ ” didn’t play a big role in the calcula ons un l the very end.

Notes:

Chapter Integra on

Mathema cians love to abstract ideas; let’s approximate the area of another re-gion using n subintervals, wherewe do not specify a value of n un l the very end.

Example . . Approxima ngdefinite integralswith a formula, using sumsRevisit

∫( x−x )dx yet again. Approximate this definite integral using theRight

Hand Rule with n equally spaced subintervals.

S Using Key Idea . . , we know ∆x = −n = /n. We also

find xi = +∆x(i − ) = (i − )/n. The Right Hand Rule uses xi+ , which isxi+ = i/n.

We construct the Right Hand Rule Riemann sum as follows. Be sure to fol-low each step carefully. If you get stuck, and do not understand how one lineproceeds to the next, you may skip to the result and consider how this resultis used. You should come back, though, and work through each step for fullunderstanding.∫

( x− x ) dx ≈n∑

i=

f(xi+ )∆x

=

n∑

i=

f(

in

)

∆x

=

n∑

i=

[

in−(

in

) ]

∆x

=

n∑

i=

(∆xn

)

i−n∑

i=

(∆xn

)

i

=

(∆xn

) n∑

i=

i−(

∆xn

) n∑

i=

i

=

(∆xn

)

· n(n+ ) −(

∆xn

)n(n+ )( n+ ) (

recall∆x = /n

)

=(n+ )

n− (n+ )( n+ )

n(now simplify)

=

(

−n

)

The result is an amazing, easy to use formula. To approximate the definiteintegral with equally spaced subintervals and the Right Hand Rule, set n =and compute

∫

( x− x ) dx ≈(

−)

= . .

Notes:

. Riemann Sums

Recall how earlier we approximated the definite integral with subintervals;with n = , the formula gives , our answer as before.

It is noweasy to approximate the integralwith , , subintervals! Hand-held calculators will round off the answer a bit prematurely giving an answer of

. . (The actual answer is . .)We now take an important leap. Up to this point, our mathema cs has been

limited to geometry and algebra (finding areas and manipula ng expressions).Now we apply calculus. For any finite n, we know that

∫

( x− x ) dx ≈(

−n

)

.

Both common sense and high–level mathema cs tell us that as n gets large, theapproxima on gets be er. In fact, if we take the limit as n → ∞, we get theexact area described by

∫( x− x ) dx. That is,

∫

( x− x ) dx = limn→∞

(

−n

)

= ( − )

= = .

This is a fantas c result. By considering n equally–spaced subintervals, we ob-tained a formula for an approxima on of the definite integral that involved ourvariable n. As n grows large – without bound – the error shrinks to zero and weobtain the exact area.

This sec on started with a fundamental calculus technique: make an ap-proxima on, refine the approxima on to make it be er, then use limits in therefining process to get an exact answer. That is precisely what we just did.

Let’s prac ce this again.

Example . . Approxima ngdefinite integralswith a formula, using sumsFind a formula that approximates

∫

− x dx using the Right Hand Rule and nequally spaced subintervals, then take the limit as n → ∞ to find the exactarea.

S Following Key Idea . . , we have ∆x = −(− )n = /n.

We have xi = (− ) + (i − )∆x; as the Right Hand Rule uses xi+ , we havexi+ = (− ) + i∆x.

The Riemann sum corresponding to the Right Hand Rule is (followed by sim-

Notes:

.....−

........ x.

y

Figure . . : Approxima ng∫

− x dx us-ing the Right Hand Rule and evenlyspaced subintervals.

Chapter Integra on

plifica ons):∫

−x dx ≈

n∑

i=

f(xi+ )∆x

=n∑

i=

f(− + i∆x)∆x

=

n∑

i=

(− + i∆x) ∆x

=

n∑

i=

((i∆x) − (i∆x) + i∆x−

)∆x (now distribute∆x)

=

n∑

i=

(i ∆x − i ∆x + i∆x −∆x

)(now split up summa on)

= ∆xn∑

i=

i − ∆xn∑

i=

i + ∆xn∑

i=

i−n∑

i=

∆x

= ∆x(n(n+ )

)

− ∆xn(n+ )( n+ )

+ ∆xn(n+ ) − n∆x

(use∆x = /n)

=n

· n (n+ ) −n

· n(n+ )( n+ )+

nn(n+ ) −

(now do a sizable amount of algebra to simplify)

= +n

+n

Once again, we have found a compact formula for approxima ng the definiteintegral with n equally spaced subintervals and the Right Hand Rule. Usingsubintervals, we have an approxima on of . (these rectangles are shownin Figure . . ). Using n = gives an approxima on of . .

Now find the exact answer using a limit:∫

−x dx = lim

n→∞

(

+n

+n

)

= .

Limits of Riemann Sums

We have used limits to evaluate given definite integrals. Will this alwayswork? We will show, given not–very–restric ve condi ons, that yes, it will al-ways work.

Notes:

. Riemann Sums

The previous two examples demonstrated how an expression such as

n∑

i=

f(xi+ )∆x

can be rewri en as an expression explicitly involving n, such as / ( − /n ).Viewed in this manner, we can think of the summa on as a func on of n.

An n value is given (where n is a posi ve integer), and the sum of areas of nequally spaced rectangles is returned, using the Le Hand, Right Hand, or Mid-point Rules.

Given a definite integral∫ ba f(x) dx, let:

• SL(n) =n∑

i=

f(xi)∆x, the sum of equally spaced rectangles formed using

the Le Hand Rule,

• SR(n) =n∑

i=

f(xi+ )∆x, the sum of equally spaced rectangles formed us-

ing the Right Hand Rule, and

• SM(n) =

n∑

i=

f(xi + xi+

)

∆x, the sum of equally spaced rectangles

formed using the Midpoint Rule.

Recall the defini on of a limit as n → ∞: limn→∞

SL(n) = K if, given any ε > ,there exists N > such that

|SL(n)− K| < ε when n ≥ N.

The following theorem states that we can use any of our three rules to findthe exact value of a definite integral

∫ ba f(x) dx. It also goes two steps further.

The theorem states that the height of each rectangle doesn’t have to be deter-mined following a specific rule, but could be f(ci), where ci is any point in the i thsubinterval, as discussed before Riemann Sums were defined in Defini on . . .

The theorem goes on to state that the rectangles do not need to be of thesame width. Using the nota on of Defini on . . , let ∆xi denote the lengthof the i th subinterval in a par on of [a, b] and let ||∆x|| represent the lengthof the largest subinterval in the par on: that is, ||∆x|| is the largest of all the∆xi’s. If ||∆x|| is small, then [a, b] must be par oned into many subintervals,since all subintervals must have small lengths. “Taking the limit as ||∆x|| goesto zero” implies that the number n of subintervals in the par on is growing to

Notes:

Chapter Integra on

infinity, as the largest subinterval length is becoming arbitrarily small. We theninterpret the expression

lim||∆x||→

n∑

i=

f(ci)∆xi

as “the limit of the sum of the areas of rectangles, where the width of eachrectangle can be different but ge ng small, and the height of each rectangle isnot necessarily determined by a par cular rule.” The theorem states that thisRiemann Sum also gives the value of the definite integral of f over [a, b].

Theorem . . Definite Integrals and the Limit of Riemann Sums

Let f be con nuous on the closed interval [a, b] and let SL(n), SR(n),SM(n),∆x,∆xi and ci be defined as before. Then:

. limn→∞

SL(n) = limn→∞

SR(n) = limn→∞

SM(n) = limn→∞

n∑

i=

f(ci)∆x,

. limn→∞

n∑

i=

f(ci)∆x =∫ b

af(x) dx, and

. lim∥∆x∥→

n∑

i=

f(ci)∆xi =∫ b

af(x) dx.

We summarize what we have learned over the past few sec ons here.

• Knowing the “area under the curve” can be useful. One commonexample:the area under a velocity curve is displacement.

• We have defined the definite integral,∫ ba f(x) dx, to be the signed area

under f on the interval [a, b].

• While we can approximate a definite integral manyways, we have focusedon using rectangles whose heights can be determined using the Le HandRule, the Right Hand Rule and the Midpoint Rule.

• Sums of rectangles of this type are called Riemann sums.

• The exact value of the definite integral can be computed using the limit ofa Riemann sum. We generally use one of the above methods as it makesthe algebra simpler.

Notes:

. Riemann Sums

We first learned of deriva ves through limits then learned rules that madethe process simpler. We knowof away to evaluate a definite integral using limits;in the next sec onwewill see how the Fundamental Theorem of Calculusmakesthe process simpler. The key feature of this theorem is its connec on betweenthe indefinite integral and the definite integral.

Notes:

Exercises .Terms and Concepts. A fundamental calculus technique is to use to re-fine approxima ons to get an exact answer.

. What is the upper bound in the summa on∑

i=

( i− )?

. This sec on approximates definite integrals using what ge-ometric shape?

. T/F: A sum using the Right Hand Rule is an example of aRiemann Sum.

ProblemsIn Exercises – , write out each term of the summa on andcompute the sum.

.∑

i=

i

.∑

i=−

( i− )

.∑

i=−

sin(πi/ )

.∑

i=

.∑

i=i

.∑

i=

(− )ii

.∑

i=

(

i−

i+

)

.∑

i=

(− )i cos(πi)

In Exercises – , write each sum in summa on nota on.

. + + + +

. − + + + + + + + +

. + + +

. − e+ e − e + e

In Exercises – , evaluate the summa on using Theorem. . .

.∑

i=

.∑

i=

i

.∑

i=

( i − i)

.∑

i=

( i − )

.∑

i=

(− i + i − i+ )

.∑

i=

(i − i + i+ )

. + + + . . .+ +

. + + + . . .+ +

Theorem . . statesn∑

i=

ai =k∑

i=

ai +n∑

i=k+

ai , so

n∑

i=k+

ai =n∑

i=

ai −k∑

i=

ai .

Use this fact, alongwith other parts of Theorem . . , to eval-uate the summa ons given in Exercises – .

.∑

i=

i

.∑

i=

i

.∑

i=

.∑

i=

i

In Exercises – , a definite integral∫ b

af(x) dx is given.

(a) Graph f(x) on [a, b].(b) Add to the sketch rectangles using the provided rule.

(c) Approximate∫ b

af(x) dx by summing the areas of the

rectangles.

.∫

−x dx, with rectangles using the Le Hand Rule.

.∫

( − x ) dx, with rectangles using the Midpoint Rule.

.∫ π

sin x dx, with rectangles using the Right Hand Rule.

.∫

x dx, with rectangles using the Le Hand Rule.

.∫

ln x dx, with rectangles using the Midpoint Rule.

.∫

xdx, with rectangles using the Right Hand Rule.

In Exercises – , a definite integral∫ b

af(x) dx is given. As demonstrated in Examples . .

and . . , do the following.

(a) Find a formula to approximate∫ b

af(x) dx using n

subintervals and the provided rule.(b) Evaluate the formula using n = , and , .(c) Find the limit of the formula, as n → ∞, to find the

exact value of∫ b

af(x) dx.

.∫

x dx, using the Right Hand Rule.

.∫

−x dx, using the Le Hand Rule.

.∫

−( x− ) dx, using the Midpoint Rule.

.∫

( x − ) dx, using the Le Hand Rule.

.∫

−( − x) dx, using the Right Hand Rule.

.∫

(x − x ) dx, using the Right Hand Rule.

ReviewIn Exercises – , find an an deriva ve of the given func-on.

. f(x) = sec x

. f(x) =x

. g(t) = t − t +

. g(t) = · t

. g(t) = cos t+ sin t

. f(x) = √x

.....a

.x

.b

.

t

.

y

Figure . . : The area of the shaded re-gion is F(x) =

∫ xa f(t) dt.

Chapter Integra on

. The Fundamental Theorem of Calculus

Let f(t)be a con nuous func ondefinedon [a, b]. The definite integral∫ ba f(x)dx

is the “area under f ” on [a, b]. We can turn this concept into a func on by le ngthe upper (or lower) bound vary.

Let F(x) =∫ xa f(t) dt. It computes the area under f on [a, x] as illustrated

in Figure . . . We can study this func on using our knowledge of the definiteintegral. For instance, F(a) = since

∫ aa f(t) dt = .

We can also apply calculus ideas to F(x); in par cular, we can compute itsderiva ve. While thismay seem like an innocuous thing to do, it has far–reachingimplica ons, as demonstrated by the fact that the result is given as an importanttheorem.

Theorem . . The Fundamental Theorem of Calculus, Part

Let f be con nuous on [a, b] and let F(x) =∫ xa f(t) dt. Then F is a differ-

en able func on on (a, b), and

F ′(x) = f(x).

Ini ally this seems simple, as demonstrated in the following example.

Example . . Using the Fundamental Theorem of Calculus, PartLet F(x) =

∫ x

−(t + sin t) dt. What is F ′(x)?

S Using the Fundamental Theoremof Calculus, wehave F ′(x) =x + sin x.

This simple example reveals something incredible: F(x) is an an deriva veof x + sin x! Therefore, F(x) = x − cos x + C for some value of C. (We canfind C, but generally we do not care. We know that F(− ) = , which allows usto compute C. In this case, C = cos(− ) + .)

We have done more than found a complicated way of compu ng an an-deriva ve. Consider a func on f defined on an open interval containing a, b

and c. Suppose we want to compute∫ ba f(t) dt. First, let F(x) =

∫ xc f(t) dt. Using

Notes:

. The Fundamental Theorem of Calculus

the proper es of the definite integral found in Theorem . . , we know∫ b

af(t) dt =

∫ c

af(t) dt+

∫ b

cf(t) dt

= −∫ a

cf(t) dt+

∫ b

cf(t) dt

= −F(a) + F(b)= F(b)− F(a).

We now see how indefinite integrals and definite integrals are related: we canevaluate a definite integral using an deriva ves! This is the second part of theFundamental Theorem of Calculus.

Theorem . . The Fundamental Theorem of Calculus, Part

Let f be con nuous on [a, b] and let F be any an deriva ve of f. Then∫ b

af(x) dx = F(b)− F(a).

Example . . Using the Fundamental Theorem of Calculus, PartWe spent a great deal of me in the previous sec on studying

∫( x − x ) dx.

Using the Fundamental Theorem of Calculus, evaluate this definite integral.

S We need an an deriva ve of f(x) = x− x . All an deriva-ves of f have the form F(x) = x − x + C; for simplicity, choose C = .The Fundamental Theorem of Calculus states

∫

( x− x ) dx = F( )− F( ) =(( ) −

)−(

−)= − = / .

This is the same answer we obtained using limits in the previous sec on, justwith much less work.

Nota on: A special nota on is o en used in the process of evalua ng definiteintegrals using the Fundamental Theorem of Calculus. Instead of explicitly writ-ing F(b)− F(a), the nota on F(x)

∣∣∣

b

ais used. Thus the solu on to Example . .

would be wri en as:∫

( x− x ) dx =(

x − x)∣∣∣∣

=(( ) −

)−(

−)= / .

Notes:

Chapter Integra on

The Constant C: Any an deriva ve F(x) can be chosen when using the Funda-mental Theorem of Calculus to evaluate a definite integral, meaning any valueof C can be picked. The constant always cancels out of the expression whenevalua ng F(b) − F(a), so it does not ma er what value is picked. This beingthe case, we might as well let C = .

Example . . Using the Fundamental Theorem of Calculus, PartEvaluate the following definite integrals.

.

∫

−x dx .

∫ π

sin x dx .

∫

et dt .

∫ √u du .

∫

dx

S

.∫

−x dx = x

∣∣∣∣−

=

( )

−(

(− )

)

= .

.∫ π

sin x dx = − cos x∣∣∣

π

= − cos π −(− cos

)= + = .

(This is interes ng; it says that the area under one “hump” of a sine curveis .)

.∫

et dt = et∣∣∣ = e − e = e − ≈ . .

.∫ √

u du =

∫

u du = u∣∣∣∣

=(

−)

=(

−)= .

.∫

dx = x∣∣∣ = ( )− = ( − ) = .

This integral is interes ng; the integrand is a constant func on, hence weare finding the area of a rectangle with width ( − ) = and height .No ce how the evalua on of the definite integral led to ( ) = .

In general, if c is a constant, then∫ ba c dx = c(b− a).

Understanding Mo on with the Fundamental Theorem ofCalculus

We established, star ng with Key Idea . . , that the deriva ve of a posi onfunc on is a velocity func on, and the deriva ve of a velocity func on is an ac-celera on func on. Now consider definite integrals of velocity and accelera on

func ons. Specifically, if v(t) is a velocity func on, what does∫ b

av(t) dtmean?

Notes:

. The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that∫ b

av(t) dt = V(b)− V(a),

where V(t) is any an deriva ve of v(t). Since v(t) is a velocity func on, V(t)must be a posi on func on, and V(b)− V(a)measures a change in posi on, ordisplacement.

Example . . Finding displacementA ball is thrown straight up with velocity given by v(t) = − t + /s, where

t is measured in seconds. Find, and interpret,∫

v(t) dt.

S Using the Fundamental Theorem of Calculus, we have∫

v(t) dt =∫

(− t+ ) dt

= − t + t∣∣∣

= .

Thus if a ball is thrown straight up into the air with velocity v(t) = − t + ,the height of the ball, second later, will be feet above the ini al height. (Notethat the ball has traveled much farther. It has gone up to its peak and is fallingdown, but the difference between its height at t = and t = is .)

Integra ng a rate of change func on gives total change. Velocity is the rateof posi on change; integra ng velocity gives the total change of posi on, i.e.,displacement.

Integra ng a speed func on gives a similar, though different, result. Speedis also the rate of posi on change, but does not account for direc on. So inte-gra ng a speed func on gives total change of posi on, without the possibilityof “nega ve posi on change.” Hence the integral of a speed func on gives dis-tance traveled.

As accelera on is the rate of velocity change, integra ng an accelera onfunc on gives total change in velocity. We do not have a simple term for thisanalogous to displacement. If a(t) = miles/h and t is measured in hours,then ∫

a(t) dt =

means the velocity has increased by m/h from t = to t = .

Notes:

Chapter Integra on

The Fundamental Theorem of Calculus and the Chain Rule

Part of the Fundamental Theoremof Calculus (FTC) states that given F(x) =∫ x

af(t) dt, F ′(x) = f(x). Using other nota on,

ddx(F(x)

)= f(x). While we have

just prac ced evalua ng definite integrals, some mes finding an deriva ves isimpossible and we need to rely on other techniques to approximate the valueof a definite integral. Func ons wri en as F(x) =

∫ xa f(t) dt are useful in such

situa ons.It may be of further use to compose such a func on with another. As an

example, we may compose F(x) with g(x) to get

F(g(x)

)=

∫ g(x)

af(t) dt.

What is the deriva ve of such a func on? The Chain Rule can be employed tostate

ddx

(

F(g(x)

))

= F ′(g(x)

)g ′(x) = f

(g(x)

)g ′(x).

An example will help us understand this.

Example . . The FTC, Part , and the Chain Rule

Find the deriva ve of F(x) =∫ x

ln t dt.

S We can view F(x) as being the func on G(x) =

∫ xln t dt

composed with g(x) = x ; that is, F(x) = G(g(x)

). The Fundamental Theorem

of Calculus states that G ′(x) = ln x. The Chain Rule gives us

F ′(x) = G ′(g(x))g ′(x)

= ln(g(x))g ′(x)= ln(x ) x= x ln x

Normally, the steps defining G(x) and g(x) are skipped.

Prac ce this once more.

Example . . The FTC, Part , and the Chain Rule

Find the deriva ve of F(x) =∫

cos xt dt.

Notes:

.....

f(x)

.

g(x)

.a

.b

.

x

.

y

.....

f(x)

.

g(x)

.a

.b

.

x

.

y

Figure . . : Finding the area bounded bytwo func ons on an interval; it is foundby subtrac ng the area under g from thearea under f.

.....

y = x + x −

.y = x −.

−

.

−

........

x

.

y

Figure . . : Sketching the region en-closed by y = x + x− and y = x−in Example . . .

. The Fundamental Theorem of Calculus

S Note that F(x) = −∫ cos x

t dt. Viewed this way, the deriva-

ve of F is straigh orward:

F ′(x) = sin x cos x.

Area Between Curves

Consider con nuous func ons f(x) and g(x) defined on [a, b], where f(x) ≥g(x) for all x in [a, b], as demonstrated in Figure . . . What is the area of theshaded region bounded by the two curves over [a, b]?

The area can be found by recognizing that this area is “the area under f −the area under g.” Using mathema cal nota on, the area is

∫ b

af(x) dx−

∫ b

ag(x) dx.

Proper es of the definite integral allow us to simplify this expression to

∫ b

a

(f(x)− g(x)

)dx.

Theorem . . Area Between Curves

Let f(x) and g(x) be con nuous func ons defined on [a, b] where f(x) ≥g(x) for all x in [a, b]. The area of the region bounded by the curvesy = f(x), y = g(x) and the lines x = a and x = b is

∫ b

a

(f(x)− g(x)

)dx.

Example . . Finding area between curvesFind the area of the region enclosed by y = x + x− and y = x− .

S It will help to sketch these two func ons, as done in Figure. . . The region whose area we seek is completely bounded by these two

func ons; they seem to intersect at x = − and x = . To check, set x +x− =

Notes:

.........

x

.

y

Figure . . : A graph of a func on f to in-troduce the Mean Value Theorem.

.........

x

.

y

(a)

.........

x

.

y

(b)

.........

x

.

y

(c)

Figure . . : Differently sized rectan-gles give upper and lower bounds on∫

f(x) dx; the last rectangle matches thearea exactly.

Chapter Integra on

x− and solve for x:

x + x− = x−(x + x− )− ( x− ) =

x − x− =

(x− )(x+ ) =

x = − , .

Following Theorem . . , the area is

∫

−

(x− − (x + x− )

)dx =

∫

−(−x + x+ ) dx

=

(

− x + x + x)∣∣∣∣−

= − ( ) + + −(

+ −)

= = .

The Mean Value Theorem and Average Value

Consider the graph of a func on f in Figure . . and the area defined by∫

f(x) dx. Three rectangles are drawn in Figure . . ; in (a), the height of therectangle is greater than f on [ , ], hence the area of this rectangle is is greaterthan

∫f(x) dx.

In (b), the height of the rectangle is smaller than f on [ , ], hence the areaof this rectangle is less than

∫f(x) dx.

Finally, in (c) the height of the rectangle is such that the area of the rectangleis exactly that of

∫f(x) dx. Since rectangles that are “too big”, as in (a), and

rectangles that are “too li le,” as in (b), give areas greater/lesser than∫

f(x) dx,it makes sense that there is a rectangle, whose top intersects f(x) somewhereon [ , ], whose area is exactly that of the definite integral.

We state this idea formally in a theorem.

Notes:

...

.....

c

.

π

.

sin .69

.

x

.

y

Figure . . : A graph of y = sin x on[ , π] and the rectangle guaranteed bythe Mean Value Theorem.

.....

y = f(x)

.

a

.b

.

c

.

f(c)

.

x

.

y

(a)

.....

y = f(x) − f(c)

.

a

.b

.

c

.

f(c)

.

x

.

y

(b)

Figure . . : In (a), a graph of y =f(x) and the rectangle guaranteed by theMean Value Theorem. In (b), y = f(x) isshi ed down by f(c); the resul ng “areaunder the curve” is .

. The Fundamental Theorem of Calculus

Theorem . . The Mean Value Theorem of Integra on

Let f be con nuous on [a, b]. There exists a value c in [a, b] such that∫ b

af(x) dx = f(c)(b− a).

This is an existen al statement; c exists, but we do not provide a method offinding it. Theorem . . is directly connected to the Mean Value Theorem ofDifferen a on, given as Theorem . . ; we leave it to the reader to see how.

We demonstrate the principles involved in this version of the Mean ValueTheorem in the following example.

Example . . Using the Mean Value TheoremConsider

∫ π sin x dx. Find a value c guaranteed by the Mean Value Theorem.

S We first need to evaluate∫ π sin x dx. (This was previously

done in Example . . .)∫ π

sin x dx = − cos x∣∣∣

π

= .

Thus we seek a value c in [ , π] such that π sin c = .

π sin c = ⇒ sin c = /π ⇒ c = arcsin( /π) ≈ . .

In Figure . . sin x is sketched along with a rectangle with height sin( . ).The area of the rectangle is the same as the area under sin x on [ , π].

Let f be a func on on [a, b]with c such that f(c)(b−a) =∫ ba f(x) dx. Consider

∫ ba

(f(x)− f(c)

)dx:∫ b

a

(f(x)− f(c)

)dx =

∫ b

af(x)−

∫ b

af(c) dx

= f(c)(b− a)− f(c)(b− a)= .

When f(x) is shi ed by −f(c), the amount of area under f above the x–axis on[a, b] is the same as the amount of area below the x–axis above f; see Figure. . for an illustra on of this. In this sense, we can say that f(c) is the average

value of f on [a, b].

Notes:

Chapter Integra on

The value f(c) is the average value in another sense. First, recognize that theMean Value Theorem can be rewri en as

f(c) =b− a

∫ b

af(x) dx,

for some value of c in [a, b]. Next, par on the interval [a, b] into n equallyspaced subintervals, a = x < x < . . . < xn+ = b and choose any ci in[xi, xi+ ]. The average of the numbers f(c ), f(c ), …, f(cn) is:

n

(

f(c ) + f(c ) + . . .+ f(cn))

=n

n∑

i=

f(ci).

Mul ply this last expression by in the form of (b−a)(b−a) :

n

n∑

i=

f(ci) =n∑

i=

f(ci)n

=

n∑

i=

f(ci)n(b− a)(b− a)

=b− a

n∑

i=

f(ci)b− an

=b− a

n∑

i=

f(ci)∆x (where∆x = (b − a)/n)

Now take the limit as n → ∞:

limn→∞ b− a

n∑

i=

f(ci)∆x =b− a

∫ b

af(x) dx = f(c).

This tells us this: when we evaluate f at n (somewhat) equally spaced points in[a, b], the average value of these samples is f(c) as n → ∞.

This leads us to a defini on.

Defini on . . The Average Value of f on [a, b]

Let f be con nuous on [a, b]. The average value of f on [a, b] is f(c),where c is a value in [a, b] guaranteed by the Mean Value Theorem. I.e.,

Average Value of f on [a, b] =b− a

∫ b

af(x) dx.

Notes:

. The Fundamental Theorem of Calculus

An applica on of this defini on is given in the following example.

Example . . Finding the average value of a func onAn object moves back and forth along a straight line with a velocity given byv(t) = (t − ) on [ , ], where t is measured in seconds and v(t) is measuredin /s.

What is the average velocity of the object?

S By our defini on, the average velocity is:

−

∫

(t− ) dt =∫(t − t+

)dt =

(

t − t + t)∣∣∣∣

= /s.

We can understand the above example through a simpler situa on. Supposeyou drove miles in hours. What was your average speed? The answer issimple: displacement/ me = miles/ hours = mph.

What was the displacement of the object in Example . . ? We calculatethis by integra ng its velocity func on:

∫(t − ) dt = . Its final posi on

was feet from its ini al posi on a er seconds: its average velocity was /s.

This sec on has laid the groundwork for a lot of great mathema cs to fol-low. The most important lesson is this: definite integrals can be evaluated usingan deriva ves. Since the previous sec on established that definite integrals arethe limit of Riemann sums, we can later create Riemann sums to approximatevalues other than “area under the curve,” convert the sums to definite integrals,then evaluate these using the Fundamental Theorem of Calculus. This will allowus to compute the work done by a variable force, the volume of certain solids,the arc length of curves, and more.

The downside is this: generally speaking, compu ng an deriva ves is muchmore difficult than compu ng deriva ves. The next chapter is devoted to tech-niques of finding an deriva ves so that a wide variety of definite integrals canbe evaluated. Before that, the next sec on explores techniques of approximat-ing the value of definite integrals beyond using the Le Hand, Right Hand andMidpoint Rules. These techniques are invaluable when an deriva ves cannotbe computed, or when the actual func on f is unknown and all we know is thevalue of f at certain x-values.

Notes:

Exercises .Terms and Concepts. How are definite and indefinite integrals related?

. What constant of integra on is most commonly used whenevalua ng definite integrals?

. T/F: If f is a con nuous func on, then F(x) =∫ x

af(t) dt is

also a con nuous func on.

. The definite integral can be used to find “the area under acurve.” Give two other uses for definite integrals.

ProblemsIn Exercises – , evaluate the definite integral.

.∫

( x − x+ ) dx

.∫

(x− ) dx

.∫

−(x − x ) dx

.∫ π

π/

cos x dx

.∫ π/

sec x dx

.∫ e

xdx

.∫

−

x dx

.∫ −

−( − x ) dx

.∫ π

( cos x− sin x) dx

.∫

ex dx

.∫ √

t dt

.∫

√tdt

.∫ √

x dx

.∫

xdx

.∫

xdx

.∫

xdx

.∫

x dx

.∫

x dx

.∫

x dx

.∫

x dx

.∫

−dx

.∫ −

−dx

.∫

−dx

.∫ π/

π/

csc x cot x dx

. Explain why:

(a)∫

−xn dx = , when n is a posi ve, odd integer, and

(b)∫

−xn dx =

∫

xn dx when n is a posi ve, even

integer.

. Explain why∫ a+ π

asin t dt = for all values of a.

In Exercises – , find a value c guaranteed by the MeanValue Theorem.

.∫

x dx

.∫

−x dx

.∫

ex dx

.∫ √

x dx

In Exercises – , find the average value of the func on onthe given interval.

. f(x) = sin x on [ , π/ ]

. y = sin x on [ , π]

. y = x on [ , ]

. y = x on [ , ]

. y = x on [ , ]

. g(t) = /t on [ , e]

In Exercises – , a velocity func on of an object movingalong a straight line is given. Find the displacement of theobject over the given me interval.

. v(t) = − t+ /s on [ , ]

. v(t) = − t+ /s on [ , ]

. v(t) = /s on [ , ].

. v(t) = tmph on [− , ]

. v(t) = cos t /s on [ , π/ ]

. v(t) =√t /s on [ , ]

In Exercises – , an accelera on func on of an objectmoving along a straight line is given. Find the change of theobject’s velocity over the given me interval.

. a(t) = − /s on [ , ]

. a(t) = /s on [ , ]

. a(t) = t /s on [ , ]

. a(t) = cos t /s on [ , π]

In Exercises – , sketch the given func ons and find thearea of the enclosed region.

. y = x, y = x, and x = .

. y = −x+ , y = x+ , x = and x = − .

. y = x − x+ , y = x− .

. y = x + x− , y = x + x+ .

In Exercises – , find F ′(x).

. F(x) =∫ x +x

tdt

. F(x) =∫

xt dt

. F(x) =∫ x

x(t+ ) dt

. F(x) =∫ ex

ln xsin t dt

.....

y = e−x

. .5..

.5

..x

.

y

(a)

.....

y = sin(x3)

.

−1

.

1

.−0.5

.

0.5

.

1

.

x

.

y

(b)

.....

y =

sin xx

.5

.10

.15

.

0.5

.

1

.

x

.

y

(c)

Figure . . : Graphically represen ngthree definite integrals that cannot beevaluated using an deriva ves.

Chapter Integra on

. Numerical Integra onThe Fundamental Theorem of Calculus gives a concrete technique for findingthe exact value of a definite integral. That technique is based on compu ng an-deriva ves. Despite the power of this theorem, there are s ll situa ons where

we must approximate the value of the definite integral instead of finding its ex-act value. The first situa on we explore is where we cannot compute the an-deriva ve of the integrand. The second case is when we actually do not know

the func on in the integrand, but only its valuewhen evaluated at certain points.

An elementary func on is any func on that is a combina on of polynomial,nth root, ra onal, exponen al, logarithmic and trigonometric func ons. We cancompute the deriva ve of any elementary func on, but there aremany elemen-tary func ons of which we cannot compute an an deriva ve. For example, thefollowing func ons do not have an deriva ves that we can express with ele-mentary func ons:

e−x , sin(x ) andsin xx

.

The simplest way to refer to the an deriva ves of e−x is to simply write∫e−x dx.This sec on outlines three common methods of approxima ng the value of

definite integrals. We describe each as a systema c method of approxima ngarea under a curve. By approxima ng this area accurately, we find an accurateapproxima on of the corresponding definite integral.

We will apply the methods we learn in this sec on to the following definiteintegrals:

∫

e−x dx,∫ π

− π

sin(x ) dx, and∫ π

.

sin(x)x

dx,

as pictured in Figure . . .

The Le and Right Hand Rule Methods

In Sec on . we addressed the problem of evalua ng definite integrals byapproxima ng the area under the curve using rectangles. We revisit those ideashere before introducing other methods of approxima ng definite integrals.

We start with a review of nota on. Let f be a con nuous func on on the

interval [a, b]. We wish to approximate∫ b

af(x) dx. We par on [a, b] into n

equally spaced subintervals, each of length∆x =b− an

. The endpoints of these

Notes:

.....

y = e−x

. .. .. .. .8..

.

..x

.

y

(a)

.....

y = e−x

. .. .. .. .8..

.

..x

.

y

(b)

Figure . . : Approxima ng∫

e−x dx inExample . . .

. Numerical Integra on

subintervals are labeled as

x = a, x = a+∆x, x = a+ ∆x, . . . , xi = a+ (i− )∆x, . . . , xn+ = b.

Key Idea . . states that to use the Le Hand Rule we use the summa onn∑

i=

f(xi)∆x and to use the Right Hand Rule we usen∑

i=

f(xi+ )∆x. We review

the use of these rules in the context of examples.

Example . . Approxima ng definite integrals with rectangles

Approximate∫

e−x dx using the Le and Right Hand Rules with equally

spaced subintervals.

S We begin by par oning the interval [ , ] into equallyspaced intervals. We have∆x = − = / = . , so

x = , x = . , x = . , x = . , x = . , and x = .

Using the Le Hand Rule, we have:

n∑

i=

f(xi)∆x =(f(x ) + f(x ) + f(x ) + f(x ) + f(x )

)∆x

=(f( ) + f( . ) + f( . ) + f( . ) + f( . )

)∆x

≈(

+ . + . + . + . )( . )

≈ . .

Using the Right Hand Rule, we have:

n∑

i=

f(xi+ )∆x =(f(x ) + f(x ) + f(x ) + f(x ) + f(x )

)∆x

=(f( . ) + f( . ) + f( . ) + f( . ) + f( )

)∆x

≈(. + . + . + . + . )( . )

≈ . .

Figure . . shows the rectangles used in each method to approximate thedefinite integral. These graphs show that in this par cular case, the Le HandRule is an over approxima on and the Right Hand Rule is an under approxima-on. To get a be er approxima on, we could use more rectangles, as we did in

Notes:

xi Exact Approx. sin(xi )x −π/ − . − .x − π/ − . − .x −π/ − . − .x −π/ − .x π/ . .x π/ . .x π/ . .x π/ . .x π/ . .x π/ . .x π/ . − .

Figure . . : Table of values used toapproximate

∫π

− π sin(x ) dx in Example. . .

.....

y = sin(x3)

.

−1

.

1

.−0.5

.

0.5

.

1

.

x

.

y

(a)

.....

y = sin(x3)

.

−1

.

1

.−0.5

.

0.5

.

1

.

x

.

y

(b)

Figure . . : Approxima ng∫

π

− π sin(x ) dx in Example . . .

Chapter Integra on

Sec on . . We could also average the Le and Right Hand Rule results together,giving

. + .= . .

The actual answer, accurate to places a er the decimal, is . , showingour average is a good approxima on.

Example . . Approxima ng definite integrals with rectangles

Approximate∫ π

− π

sin(x ) dx using the Le and Right Hand Rules with equally

spaced subintervals.

S We begin by finding∆x:

b− an

=π/ − (−π/ )

=π ≈ . .

It is useful to write out the endpoints of the subintervals in a table; in Figure. . , we give the exact values of the endpoints, their decimal approxima ons,

and decimal approxima ons of sin(x ) evaluated at these points.Once this table is created, it is straigh orward to approximate the definite

integral using the Le and Right Hand Rules. (Note: the table itself is easy tocreate, especially with a standard spreadsheet program on a computer. The lasttwo columns are all that are needed.) The Le Hand Rule sums the first valuesof sin(xi ) and mul plies the sum by ∆x; the Right Hand Rule sums the lastvalues of sin(xi ) and mul plies by∆x. Therefore we have:

Le Hand Rule:∫ π

− π

sin(x ) dx ≈ ( . )( . ) = . .

Right Hand Rule:∫ π

− π

sin(x ) dx ≈ ( . )( . ) = . .

Average of the Le and Right Hand Rules: . .The actual answer, accurate to places a er the decimal, is . . Our ap-

proxima ons were once again fairly good. The rectangles used in each approx-ima on are shown in Figure . . . It is clear from the graphs that using morerectangles (and hence, narrower rectangles) should result in a more accurateapproxima on.

The Trapezoidal Rule

In Example . . we approximated the value of∫

e−x dxwith rectangles

of equal width. Figure . . shows the rectangles used in the Le and Right

Notes:

.....

y = e−x

. .. .. .. .8..

.

..x

.

y

Figure . . : Approxima ng∫

e−x dxusing trapezoids of equal widths.

..a.

b

.h

.

Area = a+b2 h

Figure . . : The area of a trapezoid.

xi e−xi

. .

. .

. .

. ..

Figure . . : A table of values of e−x .

. Numerical Integra on

Hand Rules. These graphs clearly show that rectangles do not match the shapeof the graph all that well, and that accurate approxima ons will only come byusing lots of rectangles.

Instead of using rectangles to approximate the area, we can instead usetrapezoids. In Figure . . , we show the region under f(x) = e−x on [ , ]approximated with trapezoids of equal width; the top “corners” of each trape-zoid lies on the graph of f(x). It is clear from this figure that these trapezoidsmore accurately approximate the area under f and hence should give a be erapproxima on of

∫e−x dx. (In fact, these trapezoids seem to give a great ap-

proxima on of the area!)The formula for the area of a trapezoid is given in Figure . . . We approxi-

mate∫

e−x dx with these trapezoids in the following example.

Example . . Approxima ng definite integrals using trapezoids

Use trapezoids of equal width to approximate∫

e−x dx.

S To compute the areas of the trapezoids in Figure . . , itwill again be useful to create a table of values as shown in Figure . . .

The le most trapezoid has legs of length and . and a height of . .Thus, by our formula, the area of the le most trapezoid is:

+ .( . ) = . .

Moving right, the next trapezoid has legs of length . and . and a heightof . . Thus its area is:

. + .( . ) = . .

The sum of the areas of all trapezoids is:

+ .( . ) +

. + .( . ) +

. + .( . )+

. + .( . ) +

. + .( . ) = . .

We approximate∫

e−x dx ≈ . .

There are many things to observe in this example. Note how each term inthe final summa onwasmul plied by both / and by∆x = . . We can factorthese coefficients out, leaving a more concise summa on as:

( . )[

( + . )+( . + . )+( . + . )+( . + . )+( . + . )]

.

Notes:

Chapter Integra on

Now no ce that all numbers except for the first and the last are added twice.Therefore we can write the summa on even more concisely as

. [+ ( . + . + . + . ) + .

]

.

This is the heart of the Trapezoidal Rule, wherein a definite integral∫ b

af(x)dx

is approximated by using trapezoids of equal widths to approximate the corre-sponding area under f. Using n equally spaced subintervals with endpoints x ,x , . . ., xn+ , we again have∆x =

b− an

. Thus:

∫ b

af(x) dx ≈

n∑

i=

f(xi) + f(xi+ )∆x

=∆x n∑

i=

(f(xi) + f(xi+ )

)

=∆x[

f(x ) +n∑

i=

f(xi) + f(xn+ )]

.

Example . . Using the Trapezoidal Rule

Revisit Example . . and approximate∫ π

− π

sin(x ) dx using the Trapezoidal Rule

and equally spaced subintervals.

S Werefer back to Figure . . for the table of values of sin(x ).Recall that∆x = π/ ≈ . . Thus we have:∫ π

− πsin(x ) dx ≈ . [

− . +(

− . + (− . ) + . . .+ .)

+ (− . )]

= . .

No ce how “quickly” the Trapezoidal Rule can be implemented once the ta-ble of values is created. This is true for all the methods explored in this sec on;the real work is crea ng a table of xi and f(xi) values. Once this is completed, ap-proxima ng the definite integral is not difficult. Again, using technology is wise.Spreadsheets can make quick work of these computa ons and make using lotsof subintervals easy.

Also no ce the approxima ons the Trapezoidal Rule gives. It is the averageof the approxima ons given by the Le and Right Hand Rules! This effec vely

Notes:

........... x.

y

Figure . . : A graph of a func on f anda parabola that approximates it well on[ , ].

. Numerical Integra on

renders the Le and Right Hand Rules obsolete. They are useful when first learn-ing about definite integrals, but if a real approxima on is needed, one is gener-ally be er off using the Trapezoidal Rule instead of either the Le or Right HandRule.

How can we improve on the Trapezoidal Rule, apart from using more andmore trapezoids? The answer is clear once we look back and consider what wehave really done so far. The Le Hand Rule is not really about using rectangles toapproximate area. Instead, it approximates a func on f with constant func onson small subintervals and then computes the definite integral of these constantfunc ons. The Trapezoidal Rule is really approxima ng a func on fwith a linearfunc on on a small subinterval, then computes the definite integral of this linearfunc on. In both of these cases the definite integrals are easy to compute ingeometric terms.

So we have a progression: we start by approxima ng fwith a constant func-on and then with a linear func on. What is next? A quadra c func on. By

approxima ng the curve of a func on with lots of parabolas, we generally getan even be er approxima on of the definite integral. We call this process Simp-son’s Rule, named a er Thomas Simpson ( - ), even though others hadused this rule as much as years prior.

Simpson’s Rule

Given one point, we can create a constant func on that goes through thatpoint. Given two points, we can create a linear func on that goes through thosepoints. Given three points, we can create a quadra c func on that goes throughthose three points (given that no two have the same x–value).

Consider three points (x , y ), (x , y ) and (x , y )whose x–values are equallyspaced and x < x < x . Let fbe the quadra c func on that goes through thesethree points. It is not hard to show that

∫ x

xf(x) dx =

x − x (y + y + y

). ( . )

Consider Figure . . . A func on f goes through the points shown and theparabola g that also goes through those points is graphed with a dashed line.Using our equa on from above, we know exactly that

∫

g(x) dx =− (

+ ( ) +)= .

Since g is a good approxima on for f on [ , ], we can state that∫

f(x) dx ≈ .

Notes:

xi e−xi

. .. .. .

.

(a)

.....

y = e−x

. . 5. .5. .75..

.5

..x

.

y

(b)

Figure . . : A table of values to approxi-mate

∫

e−x dx, alongwith a graph of thefunc on.

xi sin(xi )− . − .− . − .− . − .− .

. .

. .

. .

. .. .. .. − .

Figure . . : Table of values used toapproximate

∫π

− π sin(x ) dx in Example. . .

Chapter Integra on

No ce how the interval [ , ]was split into two subintervals as we neededpoints. Because of this, whenever we use Simpson’s Rule, we need to break theinterval into an even number of subintervals.

In general, to approximate∫ b

af(x) dx using Simpson’s Rule, subdivide [a, b]

into n subintervals, where n is even and each subinterval has width∆x = (b−a)/n. We approximate fwith n/ parabolic curves, using Equa on ( . ) to com-pute the area under these parabolas. Adding up these areas gives the formula:∫ b

af(x)dx ≈ ∆x[

f(x )+ f(x )+ f(x )+ f(x )+. . .+ f(xn− )+ f(xn)+f(xn+ )]

.

Note how the coefficients of the terms in the summa on have the pa ern , ,, , , , . . ., , , .

Let’s demonstrate Simpson’s Rule with a concrete example.

Example . . Using Simpson’s Rule

Approximate∫

e−x dxusing Simpson’s Rule and equally spaced subintervals.

S We begin bymaking a table of values as we have in the past,as shown in Figure . . (a). Simpson’s Rule states that∫

e−x dx ≈ . [

+ ( . ) + ( . ) + ( . ) + .]

= . .

Recall in Example . . we stated that the correct answer, accurate toplaces a er the decimal, was . . Our approxima on with Simpson’s Rule,with subintervals, is be er than our approxima on with the Trapezoidal Ruleusing !

Figure . . (b) shows f(x) = e−x along with its approxima ng parabolas,demonstra ng how good our approxima on is. The approxima ng curves arenearly indis nguishable from the actual func on.

Example . . Using Simpson’s Rule

Approximate∫ π

− π

sin(x ) dx using Simpson’s Rule and equally spaced inter-

vals.

S Figure . . shows the table of values that we used in thepast for this problem, shown here again for convenience. Again, ∆x = (π/ +π/ )/ ≈ . .

Notes:

.....

y = sin(x3)

.

−1

.

1

.−0.5

.

0.5

.

1

.

x

.

y

Figure . . : Approxima ng∫

π

− π sin(x ) dx in Example . . withSimpson’s Rule and equally spacedintervals.

. Numerical Integra on

Simpson’s Rule states that∫ π

− πsin(x ) dx ≈ . [

(− . ) + (− . ) + (− . ) + . . .

. . .+ ( . ) + ( . ) + (− . )]

= .

Recall that the actual value, accurate to decimal places, is . . Our ap-proxima on iswithin one / th of the correct value. The graph in Figure . .shows how closely the parabolas match the shape of the graph.

Summary and Error Analysis

We summarize the key concepts of this sec on thus far in the following KeyIdea.

Key Idea . . Numerical Integra on

Let f be a con nuous func on on [a, b], let n be a posi ve integer, and let∆x =b− an

.Set x = a, x = a+∆x, . . ., xi = a+ (i− )∆x, xn+ = b.

Consider∫ b

af(x) dx.

Le Hand Rule:∫ b

af(x) dx ≈ ∆x

[

f(x ) + f(x ) + . . .+ f(xn)].

Right Hand Rule:∫ b

af(x) dx ≈ ∆x

[

f(x ) + f(x ) + . . .+ f(xn+ )].

Trapezoidal Rule:∫ b

af(x) dx ≈ ∆x[

f(x ) + f(x ) + f(x ) + . . .+ f(xn) + f(xn+ )].

Simpson’s Rule:∫ b

af(x) dx ≈ ∆x[

f(x ) + f(x ) + f(x ) + . . .+ f(xn) + f(xn+ )](n even).

In our examples, we approximated the value of a definite integral using agiven method then compared it to the “right” answer. This should have raisedseveral ques ons in the reader’s mind, such as:

. How was the “right” answer computed?

. If the right answer can be found, what is the point of approxima ng?

. If there is value to approxima ng, how are we supposed to know if theapproxima on is any good?

Notes:

Chapter Integra on

These are good ques ons, and their answers are educa onal. In the exam-ples, the right answer was never computed. Rather, an approxima on accurateto a certain number of places a er the decimal was given. In Example . . , wedo not know the exact answer, but we know it starts with . . These moreaccurate approxima ons were computed using numerical integra on but withmore precision (i.e., more subintervals and the help of a computer).

Since the exact answer cannot be found, approxima on s ll has its place.How are we to tell if the approxima on is any good?

“Trial and error” provides one way. Using technology, make an approxima-on with, say, , , and subintervals. This likely will not take much me

at all, and a trend should emerge. If a trend does not emerge, try using yet moresubintervals. Keep in mind that trial and error is never foolproof; you mightstumble upon a problem in which a trend will not emerge.

A second method is to use Error Analysis. While the details are beyond thescope of this text, there are some formulas that give bounds for how good yourapproxima on will be. For instance, the formula might state that the approx-ima on is within . of the correct answer. If the approxima on is . , thenone knows that the correct answer is between . and . . By using lots ofsubintervals, one can get an approxima on as accurate as one likes. Theorem. . states what these bounds are.

Theorem . . Error Bounds in the Trapezoidal Rule andSimpson’s Rule

. Let ET be the error in approxima ng∫ b

af(x) dx using the Trape-

zoidal Rule with n subintervals.If f has a con nuous nd deriva ve on [a, b] and M is any upperbound of

∣∣f ′′(x)

∣∣ on [a, b], then

ET ≤(b− a)

nM.

. Let ES be the error in approxima ng∫ b

af(x) dx using Simpson’s

Rule with n subintervals.If f has a con nuous th deriva ve on [a, b] and M is any upperbound of

∣∣f ( )

∣∣ on [a, b], then

ES ≤(b− a)

nM.

Notes:

.....

y = e−x (4x − )

.

.5

..−.

−

.

x

.

y

Figure . . : Graphing f ′′(x) in Example. . to help establish error bounds.

. Numerical Integra on

There are some key things to note about this theorem.

. The larger the interval, the larger the error. This should make sense intu-i vely.

. The error shrinks as more subintervals are used (i.e., as n gets larger).

. The error in Simpson’s Rule has a term rela ng to the th deriva ve of f.Consider a cubic polynomial: it’s th deriva ve is . Therefore, the error inapproxima ng the definite integral of a cubic polynomial with Simpson’sRule is – Simpson’s Rule computes the exact answer!

We revisit Examples . . and . . and compute the error bounds usingTheorem . . in the following example.

Example . . Compu ng error bounds

Find the error bounds when approxima ng∫

e−x dx using the Trapezoidal

Rule and subintervals, and using Simpson’s Rule with subintervals.

STrapezoidal Rule with n = :

We start by compu ng the nd deriva ve of f(x) = e−x :

f ′′(x) = e−x ( x − ).

Figure . . shows a graph of f ′′(x) on [ , ]. It is clear that the largest value off ′′, in absolute value, is . Thus we letM = and apply the error formula fromTheorem . . .

ET =( − )

· · = . .

Our error es ma on formula states that our approxima on of . foundin Example . . is within . of the correct answer, hence we know that

. − . = . ≤∫

e−x dx ≤ . = . + . .

We had earlier computed the exact answer, correct to decimal places, to be. , affirming the validity of Theorem . . .

Simpson’s Rule with n = :We start by compu ng the th deriva ve of f(x) = e−x :

f ( )(x) = e−x ( x − x + ).

Notes:

.....

y = e−x ( x − 8x + )

.

.

..−

...

x

.

y

Figure . . : Graphing f ( )(x) in Exam-ple . . to help establish error bounds.

Time Speed(mph)

Figure . . : Speed data collected atsecond intervals for Example . . .

Chapter Integra on

Figure . . shows a graph of f ( )(x) on [ , ]. It is clear that the largest valueof f ( ), in absolute value, is . Thus we letM = and apply the error formulafrom Theorem . . .

Es =( − )

· · = . .

Our error es ma on formula states that our approxima onof . foundin Example . . is within . of the correct answer, hence we know that

. − . = . ≤∫

e−x dx ≤ . = . + . .

Once again we affirm the validity of Theorem . . .

At the beginning of this sec on we men oned two main situa ons wherenumerical integra on was desirable. We have considered the case where anan deriva ve of the integrand cannot be computed. We now inves gate thesitua on where the integrand is not known. This is, in fact, the most widelyused applica on of Numerical Integra on methods. “Most of the me” we ob-serve behavior but do not know “the” func on that describes it. We insteadcollect data about the behavior and make approxima ons based on this data.We demonstrate this in an example.

Example . . Approxima ng distance traveledOne of the authors drove his daughter home from school while she recordedtheir speed every seconds. The data is given in Figure . . . Approximatethe distance they traveled.

S Recall that by integra ng a speed func on we get distancetraveled. We have informa on about v(t); we will use Simpson’s Rule to approx-

imate∫ b

av(t) dt.

Themost difficult aspect of this problem is conver ng the given data into theform we need it to be in. The speed is measured in miles per hour, whereas theme is measured in second increments.We need to compute∆x = (b − a)/n. Clearly, n = . What are a and b?

Since we start at me t = , we have that a = . The final recorded me camea er periods of seconds, which is minutes or / of an hour. Thus wehave

∆x =b− an

=/ −

= ;∆x

= .

Notes:

. Numerical Integra on

Thus the distance traveled is approximately:∫ .

v(t) dt ≈[

f(x ) + f(x ) + f(x ) + · · ·+ f(xn) + f(xn+ )]

=[

+ · + · + · · ·+ · + · +]

≈ . miles.

We approximate the author drove . miles. (Because we are sure the readerwants to know, the author’s odometer recorded the distance as about .miles.)

We started this chapter learning about an deriva ves and indefinite inte-grals. We then seemed to change focus by looking at areas between the graphof a func on and the x-axis. We defined these areas as the definite integral ofthe func on, using a nota on very similar to the nota on of the indefinite inte-gral. The Fundamental Theorem of Calculus ed these two seemingly separateconcepts together: we can find areas under a curve, i.e., we can evaluate a def-inite integral, using an deriva ves.

We ended the chapter by no ng that an deriva ves are some mes morethan difficult to find: they are impossible. Therefore we developed numericaltechniques that gave us good approxima ons of definite integrals.

We used the definite integral to compute areas, and also to compute dis-placements and distances traveled. There is far more we can do than that. InChapter we’ll see more applica ons of the definite integral. Before that, inChapter we’ll learn advanced techniques of integra on, analogous to learningrules like the Product, Quo ent and Chain Rules of differen a on.

Notes:

Exercises .Terms and Concepts

. T/F: Simpson’s Rule is a method of approxima ng an-deriva ves.

. What are the two basic situa ons where approxima ng thevalue of a definite integral is necessary?

. Why are the Le and Right Hand Rules rarely used?

. Simpson’s Rule is based on approxima ng por ons of afunc on with what type of func on?

ProblemsIn Exercises – , a definite integral is given.

(a) Approximate the definite integral with the TrapezoidalRule and n = .

(b) Approximate the definite integral with Simpson’s Ruleand n = .

(c) Find the exact value of the integral.

.∫

−x dx

.∫

x dx

.∫ π

sin x dx

.∫ √

x dx

.∫

(x + x − x+ ) dx

.∫

x dx

.∫ π

cos x dx

.∫

−

√− x dx

In Exercises – , approximate the definite integral withthe Trapezoidal Rule and Simpson’s Rule, with n = .

.∫

cos(

x)

dx

.∫

−ex dx

.∫ √

x + dx

.∫ π

x sin x dx

.∫ π/ √

cos x dx

.∫

ln x dx

.∫

− sin x+dx

.∫

sin x+dx

In Exercises – , find n such that the error in approximat-ing the given definite integral is less than . when using:

(a) the Trapezoidal Rule

(b) Simpson’s Rule

.∫ π

sin x dx

.∫

√xdx

.∫ π

cos(

x)

dx

.∫

x dx

In Exercises – , a region is given. Find the area of theregion using Simpson’s Rule:

(a) where the measurements are in cen meters, taken incm increments, and

(b) where the measurements are in hundreds of yards,taken in yd increments.

. ..

.

.

.

.

.9

. .. .1

.. .

..

.

.

.

.

.

.

.

: TA

The previous chapter introduced the an deriva ve and connected it to signedareas under a curve through the Fundamental Theorem of Calculus. The nextchapter explores more applica ons of definite integrals than just area. As eval-ua ng definite integrals will become important, we will want to find an deriva-ves of a variety of func ons.This chapter is devoted to exploring techniques of an differen a on. While

not every func on has an an deriva ve in terms of elementary func ons (aconcept introduced in the sec on on Numerical Integra on), we can s ll findan deriva ves of a wide variety of func ons.

. Subs tu onWe mo vate this sec on with an example. Let f(x) = (x + x − ) . We cancompute f ′(x) using the Chain Rule. It is:

f ′(x) = (x + x− ) · ( x+ ) = ( x+ )(x + x− ) .

Now consider this: What is∫( x+ )(x + x− ) dx? We have the answer

in front of us;∫

( x+ )(x + x− ) dx = (x + x− ) + C.

How would we have evaluated this indefinite integral without star ng with f(x)as we did?

This sec on explores integra on by subs tu on. It allows us to “undo theChain Rule.” Subs tu on allows us to evaluate the above integral without know-ing the original func on first.

The underlying principle is to rewrite a “complicated” integral of the form∫f(x) dx as a not–so–complicated integral

∫h(u) du. We’ll formally establish

later how this is done. First, consider again our introductory indefinite integral,∫( x + )(x + x − ) dx. Arguably the most “complicated” part of the

integrand is (x + x − ) . We wish to make this simpler; we do so through asubs tu on. Let u = x + x− . Thus

(x + x− ) = u .

Chapter Techniques of An differen a on

We have established u as a func on of x, so now consider the differen al of u:

du = ( x+ )dx.

Keep inmind that ( x+ ) and dx aremul plied; the dx is not “just si ng there.”Return to the original integral and do some subs tu ons through algebra:∫

( x+ )(x + x− ) dx =∫

( x+ )(x + x− ) dx

=

∫

(x + x−︸ ︷︷ ︸

u

) ( x+ ) dx︸ ︷︷ ︸

du

=

∫

u du

= u + C (replace u with x + x − )

= (x + x− ) + C

One might well look at this and think “I (sort of) followed how that worked,but I could never come up with that on my own,” but the process is learnable.This sec on contains numerous examples through which the reader will gainunderstanding and mathema cal maturity enabling them to regard subs tu onas a natural tool when evalua ng integrals.

We stated before that integra on by subs tu on “undoes” the Chain Rule.Specifically, let F(x) and g(x) be differen able func ons and consider the deriva-ve of their composi on:

ddx

(

F(g(x)

))

= F ′(g(x))g ′(x).

Thus ∫

F ′(g(x))g ′(x) dx = F(g(x)) + C.

Integra on by subs tu on works by recognizing the “inside” func on g(x) andreplacing it with a variable. By se ng u = g(x), we can rewrite the deriva veas

ddx

(

F(u))

= F ′(u)u ′.

Since du = g ′(x)dx, we can rewrite the above integral as∫

F ′(g(x))g ′(x) dx =∫

F ′(u)du = F(u) + C = F(g(x)) + C.

This concept is important so we restate it in the context of a theorem.

Notes:

. Subs tu on

Theorem . . Integra on by Subs tu on

Let F and g be differen able func ons, where the range of g is an intervalI contained in the domain of F. Then

∫

F ′(g(x))g ′(x) dx = F(g(x)) + C.

If u = g(x), then du = g ′(x)dx and∫

F ′(g(x))g ′(x) dx =∫

F ′(u) du = F(u) + C = F(g(x)) + C.

The point of subs tu on is to make the integra on step easy. Indeed, thestep

∫F ′(u) du = F(u)+C looks easy, as the an deriva ve of the deriva ve of F

is just F, plus a constant. The “work” involved is making the proper subs tu on.There is not a step–by–step process that one can memorize; rather, experiencewill be one’s guide. To gain experience, we now embark on many examples.

Example . . Integra ng by subs tu onEvaluate

∫

x sin(x + ) dx.

S Knowing that subs tu on is related to the Chain Rule, wechoose to let u be the “inside” func on of sin(x + ). (This is not always a goodchoice, but it is o en the best place to start.)

Let u = x + , hence du = x dx. The integrand has an x dx term, butnot a x dx term. (Recall that mul plica on is commuta ve, so the x does notphysically have to be next to dx for there to be an x dx term.) We can divide bothsides of the du expression by :

du = x dx ⇒ du = x dx.

We can now subs tute.∫

x sin(x + ) dx =∫

sin(x +︸ ︷︷ ︸

u

) x dx︸︷︷︸

du

=

∫

sin u du

Notes:

Chapter Techniques of An differen a on

= − cos u+ C (now replace u with x + )

= − cos(x + ) + C.

Thus∫x sin(x + ) dx = − cos(x + ) + C. We can check our work by eval-

ua ng the deriva ve of the right hand side.

Example . . Integra ng by subs tu onEvaluate

∫

cos( x) dx.

S Again let u replace the “inside” func on. Le ng u = x, wehave du = dx. Since our integrand does not have a dx term, we can dividethe previous equa on by to obtain du = dx. We can now subs tute.

∫

cos( x) dx =∫

cos( x︸︷︷︸

u

) dx︸︷︷︸

du

=

∫

cos u du

= sin u+ C

= sin( x) + C.

We can again check our work through differen a on.

The previous example exhibited a common, and simple, type of subs tu on.The “inside” func on was a linear func on (in this case, y = x). When theinside func on is linear, the resul ng integra on is very predictable, outlinedhere.

Key Idea . . Subs tu on With A Linear Func on

Consider∫F ′(ax + b) dx, where a ̸= and b are constants. Le ng

u = ax+ b gives du = a · dx, leading to the result∫

F ′(ax+ b) dx =aF(ax+ b) + C.

Thus∫sin( x− ) dx = − cos( x− ) + C. Our next example can use Key

Idea . . , but we will only employ it a er going through all of the steps.

Notes:

. Subs tu on

Example . . Integra ng by subs tu ng a linear func onEvaluate

∫

− x+dx.

S View the integrand as the composi on of func ons f(g(x)),where f(x) = /x and g(x) = − x+ . Employing our understanding of subs -tu on, we let u = − x+ , the inside func on. Thus du = − dx. The integrandlacks a − ; hence divide the previous equa on by − to obtain −du/ = dx.We can now evaluate the integral through subs tu on.

∫

− x+dx =

∫

udu−

=− ∫

duu

=−

ln |u|+ C

= − ln | − x+ |+ C.

Using Key Idea . . is faster, recognizing that u is linear and a = − . One maywant to con nue wri ng out all the steps un l they are comfortable with thispar cular shortcut.

Not all integrals that benefit from subs tu on have a clear “inside” func on.Several of the following examples will demonstrate ways in which this occurs.

Example . . Integra ng by subs tu onEvaluate

∫

sin x cos x dx.

S There is not a composi onof func onhere to exploit; rather,just a product of func ons. Do not be afraid to experiment; when given an inte-gral to evaluate, it is o en beneficial to think “If I let u be this, then dumust bethat …” and see if this helps simplify the integral at all.

In this example, let’s set u = sin x. Then du = cos x dx, which we have aspart of the integrand! The subs tu on becomes very straigh orward:

∫

sin x cos x dx =∫

u du

= u + C

= sin x+ C.

Notes:

Chapter Techniques of An differen a on

One would do well to ask “What would happen if we let u = cos x?” The resultis just as easy to find, yet looks very different. The challenge to the reader is toevaluate the integral le ng u = cos x and discover why the answer is the same,yet looks different.

Our examples so far have required “basic subs tu on.” The next exampledemonstrates how subs tu ons can be made that o en strike the new learneras being “nonstandard.”

Example . . Integra ng by subs tu onEvaluate

∫

x√x+ dx.

S Recognizing the composi on of func ons, set u = x + .Then du = dx, giving what seems ini ally to be a simple subs tu on. But at thisstage, we have: ∫

x√x+ dx =

∫

x√u du.

We cannot evaluate an integral that has both an x and an u in it. We need toconvert the x to an expression involving just u.

Since we set u = x+ , we can also state that u− = x. Thus we can replacex in the integrand with u− . It will also be helpful to rewrite

√u as u .

∫

x√x+ dx =

∫

(u− )u du

=

∫(u − u

)du

= u − u + C

= (x+ ) − (x+ ) + C.

Checking your work is always a good idea. In this par cular case, some algebrawill be needed to make one’s answer match the integrand in the original prob-lem.

Example . . Integra ng by subs tu onEvaluate

∫

x ln xdx.

S This is another example where there does not seem to bean obvious composi on of func ons. The line of thinking used in Example . .is useful here: choose something for u and consider what this implies du must

Notes:

. Subs tu on

be. If u can be chosen such that du also appears in the integrand, then we havechosen well.

Choosing u = /xmakes du = − /x dx; that does not seem helpful. How-ever, se ng u = ln xmakes du = /x dx, which is part of the integrand. Thus:

∫

x ln xdx =

∫

ln x︸︷︷︸

u

xdx

︸︷︷︸

du

=

∫

udu

= ln |u|+ C= ln | ln x|+ C.

The final answer is interes ng; the natural log of the natural log. Take the deriva-ve to confirm this answer is indeed correct.

Integrals Involving Trigonometric Func ons

Sec on . delves deeper into integrals of a variety of trigonometric func-ons; here we use subs tu on to establish a founda on that wewill build upon.

Thenext three exampleswill help fill in somemissing pieces of our an deriva-ve knowledge. We know the an deriva ves of the sine and cosine func ons;

what about the other standard func ons tangent, cotangent, secant and cose-cant? We discover these next.

Example . . Integra on by subs tu on: an deriva ves of tan xEvaluate

∫

tan x dx.

S The previous paragraph established that we did not knowthe an deriva ves of tangent, hence we must assume that we have learnedsomething in this sec on that can help us evaluate this indefinite integral.

Rewrite tan x as sin x/ cos x. While the presence of a composi on of func-ons may not be immediately obvious, recognize that cos x is “inside” the /x

func on. Therefore, we see if se ng u = cos x returns usable results. We have

Notes:

Chapter Techniques of An differen a on

that du = − sin x dx, hence−du = sin x dx. We can integrate:

∫

tan x dx =∫

sin xcos x

dx

=

∫

cos x︸︷︷︸

u

sin x dx︸ ︷︷ ︸

−du

=

∫ −u

du

= − ln |u|+ C= − ln | cos x|+ C.

Some texts prefer to bring the− inside the logarithm as a power of cos x, as in:

− ln | cos x|+ C = ln |(cos x)− |+ C

= ln∣∣∣∣cos x

∣∣∣∣+ C

= ln | sec x|+ C.

Thus the result they give is∫tan x dx = ln | sec x| + C. These two answers are

equivalent.

Example . . Integra ng by subs tu on: an deriva ves of sec xEvaluate

∫

sec x dx.

S This example employs a wonderful trick: mul ply the inte-grand by “ ” so that we see how to integrate more clearly. In this case, we write“ ” as

=sec x+ tan xsec x+ tan x

.

This may seem like it came out of le field, but it works beau fully. Consider:

∫

sec x dx =∫

sec x · sec x+ tan xsec x+ tan x

dx

=

∫sec x+ sec x tan x

sec x+ tan xdx.

Notes:

. Subs tu on

Now let u = sec x + tan x; this means du = (sec x tan x + sec x) dx, which isour numerator. Thus:

=

∫duu

= ln |u|+ C= ln | sec x+ tan x|+ C.

We can use similar techniques to those used in Examples . . and . .to find an deriva ves of cot x and csc x (which the reader can explore in theexercises.) We summarize our results here.

Theorem . . An deriva ves of Trigonometric Func ons

.∫

sin x dx = − cos x+ C

.∫

cos x dx = sin x+ C

.∫

tan x dx = − ln | cos x|+C

.∫

csc x dx = − ln | csc x+ cot x|+ C

.∫

sec x dx = ln | sec x+ tan x|+ C

.∫

cot x dx = ln | sin x|+ C

We explore one more common trigonometric integral.

Example . . Integra on by subs tu on: powers of cos x and sin xEvaluate

∫

cos x dx.

S We have a composi on of func ons as cos x =(cos x

).

However, se ng u = cos xmeans du = − sin x dx, which we do not have in theintegral. Another technique is needed.

The process we’ll employ is to use a Power Reducing formula for cos x (per-haps consult the back of this text for this formula), which states

cos x =+ cos( x)

.

The right hand side of this equa on is not difficult to integrate. We have:∫

cos x dx =∫

+ cos( x)dx

=

∫ (

+ cos( x))

dx.

Notes:

Chapter Techniques of An differen a on

Now use Key Idea . . :

= x+sin( x)

+ C

= x+sin( x)

+ C.

We’ll make significant use of this power–reducing technique in future sec ons.

Simplifying the Integrand

It is common to be reluctant to manipulate the integrand of an integral; atfirst, our grasp of integra on is tenuous and one may think that working withthe integrand will improperly change the results. Integra on by subs tu onworks using a different logic: as long as equality is maintained, the integrand canbe manipulated so that its form is easier to deal with. The next two examplesdemonstrate common ways in which using algebra first makes the integra oneasier to perform.

Example . . Integra on by subs tu on: simplifying first

Evaluate∫

x + x + x+x + x+

dx.

S One may try to start by se ng u equal to either the numer-ator or denominator; in each instance, the result is not workable.

When dealing with ra onal func ons (i.e., quo ents made up of polynomialfunc ons), it is an almost universal rule that everything works be er when thedegree of the numerator is less than the degree of the denominator. Hence weuse polynomial division.

We skip the specifics of the steps, but note that when x + x+ is dividedinto x + x + x+ , it goes in x+ mes with a remainder of x+ . Thus

x + x + x+x + x+

= x+ +x+

x + x+.

Integra ng x + is simple. The frac on can be integrated by se ng u = x +x+ , giving du = ( x+ ) dx. This is very similar to the numerator. Note that

Notes:

. Subs tu on

du/ = (x+ ) dx and then consider the following:∫

x + x + x+x + x+

dx =∫ (

x+ +x+

x + x+

)

dx

=

∫

(x+ ) dx+∫

(x+ )

x + x+dx

= x + x+ C +

∫

udu

= x + x+ C + ln |u|+ C

= x + x+ ln |x + x+ |+ C.

In some ways, we “lucked out” in that a er dividing, subs tu on was able to bedone. In later sec ons we’ll develop techniques for handling ra onal func onswhere subs tu on is not directly feasible.

Example . . Integra on by alternate methods

Evaluate∫

x + x+√x

dx with, and without, subs tu on.

S We already know how to integrate this par cular example.Rewrite

√x as x and simplify the frac on:

x + x+x /

= x + x + x− .

We can now integrate using the Power Rule:∫

x + x+x /

dx =∫ (

x + x + x−)

dx

= x + x + x + C

This is a perfectly fine approach. We demonstrate how this can also be solvedusing subs tu on as its implementa on is rather clever.

Let u =√x = x ; therefore

du = x− dx = √xdx ⇒ du = √

xdx.

This gives us∫

x + x+√x

dx =∫

(x + x+ ) · du. What are we to do

with the other x terms? Since u = x , u = x, etc. We can then replace x and

Notes:

Chapter Techniques of An differen a on

x with appropriate powers of u. We thus have∫

x + x+√x

dx =∫

(x + x+ ) · du

=

∫

(u + u + ) du

= u + u + u+ C

= x + x + x + C,

which is obviously the same answer we obtained before. In this situa on, sub-s tu on is arguably more work than our other method. The fantas c thing isthat it works. It demonstrates how flexible integra on is.

Subs tu on and Inverse Trigonometric Func ons

When studying deriva ves of inverse func ons, we learned that

ddx(tan− x

)=

+ x.

Applying the Chain Rule to this is not difficult; for instance,

ddx(tan− x

)=

+ x.

Wenow explore how Subs tu on can be used to “undo” certain deriva ves thatare the result of the Chain Rule applied to Inverse Trigonometric func ons. Webegin with an example.

Example . . Integra ngby subs tu on: inverse trigonometric func onsEvaluate

∫

+ xdx.

S The integrand looks similar to the deriva ve of the arctan-gent func on. Note:

+ x=

( + x )

=( +

( x ) )

=+( x )

.

Notes:

. Subs tu on

Thus ∫

+ xdx =

∫

+( x )

dx.

This can be integrated using Subs tu on. Set u = x/ , hence du = dx/ ordx = du. Thus

∫

+ xdx =

∫

+( x )

dx

=

∫

+ udu

= tan− u+ C

= tan−( x)

+ C

Example . . demonstrates a general technique that can be applied toother integrands that result in inverse trigonometric func ons. The results aresummarized here.

Theorem . . Integrals Involving Inverse Trigonometric Func ons

Let a > .

.∫

a + xdx =

atan−

( xa

)

+ C

.∫

√a − x

dx = sin−( xa

)

+ C

.∫

x√x − a

dx =asec−

( |x|a

)

+ C

Let’s prac ce using Theorem . . .

Example . . Integra ngby subs tu on: inverse trigonometric func onsEvaluate the given indefinite integrals.

.

∫

+ xdx, .

∫

x√

x −dx .

∫

√− x

dx.

Notes:

Chapter Techniques of An differen a on

S Each can be answered using a straigh orward applica on ofTheorem . . .

.∫

+ xdx = tan−

x+ C, as a = .

.∫

x√

x −dx = sec− x+ C, as a = .

.∫

√− x

= sin−x√ + C, as a =

√.

Most applica ons of Theorem . . are not as straigh orward. The nextexamples show some common integrals that can s ll be approached with thistheorem.

Example . . Integra ng by subs tu on: comple ng the squareEvaluate

∫

x − x+dx.

S Ini ally, this integral seems to have nothing in commonwiththe integrals in Theorem . . . As it lacks a square root, it almost certainly is notrelated to arcsine or arcsecant. It is, however, related to the arctangent func on.

We see this by comple ng the square in the denominator. We give a briefreminder of the process here.

Start with a quadra c with a leading coefficient of . It will have the form ofx +bx+c. Take / of b, square it, and add/subtract it back into the expression.I.e.,

x + bx+ c = x + bx+b

︸ ︷︷ ︸

(x+b/ )

−b+ c

=

(

x+b)

+ c− b

In our example, we take half of − and square it, ge ng . We add/subtract itinto the denominator as follows:

x − x+=

x − x+︸ ︷︷ ︸

(x− )

− +

=(x− ) +

Notes:

. Subs tu on

We can now integrate this using the arctangent rule. Technically, we need tosubs tute first with u = x− , but we can employ Key Idea . . instead. Thuswe have

∫

x − x+dx =

∫

(x− ) +dx = tan−

x−+ C.

Example . . Integrals requiring mul ple methodsEvaluate

∫ − x√− x

dx.

S This integral requires two different methods to evaluate it.We get to those methods by spli ng up the integral:

∫ − x√− x

dx =∫

√− x

dx−∫

x√− x

dx.

The first integral is handled using a straigh orward applica on of Theorem . . ;the second integral is handled by subs tu on, with u = −x . We handle eachseparately.∫

√− x

dx = sin−x+ C.

∫x√− x

dx: Set u = − x , so du = − xdx and xdx = −du/ . We

have∫

x√− x

dx =∫ −du/√

u

= −∫

√udu

= −√u+ C

= −√

− x + C.

Combining these together, we have∫ − x√

− xdx = sin−

x+√

− x + C.

Subs tu on and Definite Integra on

This sec on has focused on evalua ng indefinite integrals as we are learninga new technique for finding an deriva ves. However, much of the me integra-on is used in the context of a definite integral. Definite integrals that require

subs tu on can be calculated using the following workflow:

Notes:

Chapter Techniques of An differen a on

. Start with a definite integral∫ b

af(x) dx that requires subs tu on.

. Ignore the bounds; use subs tu on to evaluate∫

f(x) dx and find an an-

deriva ve F(x).

. Evaluate F(x) at the bounds; that is, evaluate F(x)∣∣∣

b

a= F(b)− F(a).

This workflow works fine, but subs tu on offers an alterna ve that is powerfuland amazing (and a li le me saving).

At its heart, (using the nota on of Theorem . . ) subs tu on converts inte-grals of the form

∫F ′(g(x))g ′(x) dx into an integral of the form

∫F ′(u) du with

the subs tu on of u = g(x). The following theorem states how the bounds ofa definite integral can be changed as the subs tu on is performed.

Theorem . . Subs tu on with Definite Integrals

Let F and g be differen able func ons, where the range of g is an intervalI that is contained in the domain of F. Then

∫ b

aF ′(g(x)

)g ′(x) dx =

∫ g(b)

g(a)F ′(u) du.

In effect, Theorem . . states that once you convert to integra ng with re-spect to u, you do not need to switch back to evalua ng with respect to x. A fewexamples will help one understand.

Example . . Definite integrals and subs tu on: changing the bounds

Evaluate∫

cos( x− ) dx using Theorem . . .

S Observing the composi on of func ons, let u = x − ,hence du = dx. As dx does not appear in the integrand, divide the la erequa on by to get du/ = dx.

By se ng u = x− , we are implicitly sta ng that g(x) = x− . Theorem. . states that the new lower bound is g( ) = − ; the new upper bound is

Notes:

.....

y = cos( x − )

.

−

...... −.

− .

.

.

..

x

.

y

(a)

.....

y = cos(u)

.

−

...... −.

− .

.

.

..

u

.

y

(b)

Figure . . : Graphing the areas de-fined by the definite integrals of Example. . .

.....

y = sin x cos x

.. − .5.

.5

..

π

.

x

.

y

(a)

.....

y = u

.. − .5.

.5

..

π

.

u

.

y

(b)

Figure . . : Graphing the areas de-fined by the definite integrals of Example. . .

. Subs tu on

g( ) = . We now evaluate the definite integral:∫

cos( x− ) dx =∫

−cos u

du

= sin u∣∣∣−

=(sin − sin(− )

)≈ − . .

No ce how once we converted the integral to be in terms of u, we never wentback to using x.

The graphs in Figure . . tell more of the story. In (a) the area defined bythe original integrand is shaded, whereas in (b) the area defined by the new in-tegrand is shaded. In this par cular situa on, the areas look very similar; thenew region is “shorter” but “wider,” giving the same area.

Example . . Definite integrals and subs tu on: changing the bounds

Evaluate∫ π/

sin x cos x dx using Theorem . . .

S Wesaw the corresponding indefinite integral in Example . . .In that example we set u = sin x but stated that we could have let u = cos x.For variety, we do the la er here.

Let u = g(x) = cos x, giving du = − sin x dx and hence sin x dx = −du. Thenew upper bound is g(π/ ) = ; the new lower bound is g( ) = . Note howthe lower bound is actually larger than the upper bound now. We have

∫ π/

sin x cos x dx =∫

−u du (switch bounds & change sign)

=

∫

u du

= u∣∣∣ = / .

In Figure . . we have again graphed the two regions defined by our definiteintegrals. Unlike the previous example, they bear no resemblance to each other.However, Theorem . . guarantees that they have the same area.

Integra on by subs tu on is a powerful and useful integra on technique.The next sec on introduces another technique, called Integra on by Parts. Assubs tu on “undoes” the Chain Rule, integra on by parts “undoes” the ProductRule. Together, these two techniques provide a strong founda ononwhichmostother integra on techniques are based.

Notes:

Exercises .Terms and Concepts

. Subs tu on “undoes” what deriva ve rule?

. T/F: One can use algebra to rewrite the integrand of an in-tegral to make it easier to evaluate.

ProblemsIn Exercises – , evaluate the indefinite integral to developan understanding of Subs tu on.

.∫

x(

x −)

dx

.∫

( x− )(

x − x+)

dx

.∫

x(

x +)

dx

.∫

( x+ )(

x + x−)

dx

.∫

x+dx

.∫

√x+

dx

.∫

x√x+

dx

.∫

x − x√x

dx

.∫

e√

x√xdx

.∫

x√x +

dx

.∫

x +

xdx

.∫

ln(x)x

dx

In Exercises – , use Subs tu on to evaluate the indefi-nite integral involving trigonometric func ons.

.∫

sin (x) cos(x)dx

.∫

cos (x) sin(x)dx

.∫

cos( − x)dx

.∫

sec ( − x)dx

.∫

sec( x)dx

.∫

tan (x) sec (x)dx

.∫

x cos(

x)

dx

.∫

tan (x)dx

.∫

cot x dx. Do not just refer to Theorem . . for the an-swer; jus fy it through Subs tu on.

.∫

csc x dx. Do not just refer to Theorem . . for the an-swer; jus fy it through Subs tu on.

In Exercises – , use Subs tu on to evaluate the indefi-nite integral involving exponen al func ons.

.∫

e x− dx

.∫

ex x dx

.∫

ex − x+ (x− )dx

.∫

ex +ex

dx

.∫

ex

ex +dx

.∫

ex − e−x

e x dx

.∫

xdx

.∫

xdx

In Exercises – , use Subs tu on to evaluate the indefi-nite integral involving logarithmic func ons.

.∫

ln xx

dx

.∫

(

ln x)

xdx

.∫ ln

(

x)

xdx

.∫

x ln (x )dx

In Exercises – , use Subs tu on to evaluate the indefi-nite integral involving ra onal func ons.

.∫

x + x+x

dx

.∫

x + x + x+x

dx

.∫

x −x+

dx

.∫

x + x−x− dx

.∫

x − x+x+

dx

.∫

x + x+x + x + x

dx

In Exercises – , use Subs tu on to evaluate the indefi-nite integral involving inverse trigonometric func ons.

.∫

x +dx

.∫

√− x

dx

.∫

√− x

dx

.∫

x√x −

dx

.∫

√x − x

dx

.∫

x√− x

dx

.∫

x − x+dx

.∫

√−x + x+

dx

.∫

√−x + x+

dx

.∫

x + x+dx

In Exercises – , evaluate the indefinite integral.

.∫

x(x + )

dx

.∫

(

x + x) (

x + x +)

dx

.∫

x√− x

dx

.∫

x csc(

x +)

dx

.∫

sin(x)√

cos(x)dx

.∫

sin(

x+)

dx

.∫

x− dx

.∫

x+dx

.∫

x + x + x−x + x+

dx

.∫

x+x + x+

dx

.∫

( x+ )

x + x+dx

.∫ −x + x − x−

x − x+dx

.∫

xx +

dx

.∫

x +dx

.∫

x√

x −dx

.∫

√− x

dx

.∫

x−x − x+

dx

.∫ − x

x + x+dx

.∫

x + x−x − x+

dx

.∫

xx +

dx

.∫

x − xx + x+

dx

.∫

sin(x)cos (x) +

dx

.∫

cos(x)sin (x) +

dx

.∫

cos(x)− sin (x)

dx

.∫

x−√x − x−

dx

.∫

x−√x − x+

dx

In Exercises – , evaluate the definite integral.

.∫

x− dx

.∫

x√x− dx

.∫ π/

−π/

sin x cos x dx

.∫

x( − x ) dx

.∫ −

−(x+ )ex + x+ dx

.∫

− + xdx

.∫

x − x+dx

.∫

√

√− x

dx

. Integra on by Parts

. Integra on by PartsHere’s a simple integral that we can’t yet evaluate:

∫

x cos x dx.

It’s a simple ma er to take the deriva ve of the integrand using the ProductRule, but there is no Product Rule for integrals. However, this sec on introducesIntegra on by Parts, a method of integra on that is based on the Product Rulefor deriva ves. It will enable us to evaluate this integral.

The Product Rule says that ifu and v are func ons of x, then (uv)′ = u ′v+uv ′.For simplicity, we’ve wri en u for u(x) and v for v(x). Suppose we integrate bothsides with respect to x. This gives

∫

(uv)′ dx =∫

(u ′v+ uv ′) dx.

By the Fundamental Theoremof Calculus, the le side integrates to uv. The rightside can be broken up into two integrals, and we have

uv =∫

u ′v dx+∫

uv ′ dx.

Solving for the second integral we have∫

uv ′ dx = uv−∫

u ′v dx.

Using differen al nota on, we can write du = u ′(x)dx and dv = v ′(x)dx andthe expression above can be wri en as follows:

∫

u dv = uv−∫

v du.

This is the Integra on by Parts formula. For reference purposes, we state this ina theorem.

Theorem . . Integra on by Parts

Let u and v be differen able func ons of x on an interval I containing aand b. Then ∫

u dv = uv−∫

v du,

and∫ x=b

x=au dv = uv

∣∣∣

b

a−∫ x=b

x=av du.

Notes:

Chapter Techniques of An differen a on

Let’s try an example to understand our new technique.

Example . . Integra ng using Integra on by PartsEvaluate

∫

x cos x dx.

S The key to Integra on by Parts is to iden fy part of the in-tegrand as “u” and part as “dv.” Regular prac ce will help one make good iden-fica ons, and later we will introduce some principles that help. For now, let

u = x and dv = cos x dx.It is generally useful to make a small table of these values as done below.

Right now we only know u and dv as shown on the le of Figure . . ; on theright we fill in the rest of what we need. If u = x, then du = dx. Sincedv = cos x dx, v is an an deriva ve of cos x. We choose v = sin x.

u = x v = ?du = ? dv = cos x dx

⇒ u = x v = sin xdu = dx dv = cos x dx

Figure . . : Se ng up Integra on by Parts.

Now subs tute all of this into the Integra on by Parts formula, giving∫

x cos x dx = x sin x−∫

sin x dx.

We can then integrate sin x to get− cos x+ C and overall our answer is∫

x cos x dx = x sin x+ cos x+ C.

Note how the an deriva ve contains a product, x sin x. This product is whatmakes Integra on by Parts necessary.

The example above demonstrates how Integra on by Parts works in general.We try to iden fy u and dv in the integral we are given, and the key is that weusually want to choose u and dv so that du is simpler than u and v is hopefullynot too much more complicated than dv. This will mean that the integral on theright side of the Integra on by Parts formula,

∫v du will be simpler to integrate

than the original integral∫u dv.

In the example above, we chose u = x and dv = cos x dx. Then du = dxwassimpler than u and v = sin x is no more complicated than dv. Therefore, insteadof integra ng x cos x dx, we could integrate sin x dx, which we knew how to do.

A useful mnemonic for helping to determine u is “LIATE,” where

L = Logarithmic, I = Inverse Trig., A = Algebraic (polynomials),T = Trigonometric, and E = Exponen al.

Notes:

. Integra on by Parts

If the integrand contains both a logarithmic and an algebraic term, in generalle ng u be the logarithmic term works best, as indicated by L coming before Ain LIATE.

We now consider another example.

Example . . Integra ng using Integra on by PartsEvaluate

∫

xex dx.

S The integrand contains anAlgebraic term (x) and an Exponen alterm (ex). Our mnemonic suggests le ng u be the algebraic term, so we chooseu = x and dv = ex dx. Then du = dx and v = ex as indicated by the tables below.

u = x v = ?du = ? dv = ex dx

⇒ u = x v = ex

du = dx dv = ex dx

Figure . . : Se ng up Integra on by Parts.

We see du is simpler than u, while there is no change in going from dv to v.This is good. The Integra on by Parts formula gives

∫

xex dx = xex −∫

ex dx.

The integral on the right is simple; our final answer is∫

xex dx = xex − ex + C.

Note again how the an deriva ves contain a product term.

Example . . Integra ng using Integra on by PartsEvaluate

∫

x cos x dx.

S Themnemonic suggests le ngu = x insteadof the trigono-metric func on, hence dv = cos x dx. Then du = x dx and v = sin x as shownbelow.

u = x v = ?du = ? dv = cos x dx

⇒ u = x v = sin xdu = x dx dv = cos x dx

Figure . . : Se ng up Integra on by Parts.

Notes:

Chapter Techniques of An differen a on

The Integra on by Parts formula gives∫

x cos x dx = x sin x−∫

x sin x dx.

At this point, the integral on the right is indeed simpler than the one we startedwith, but to evaluate it, we need to do Integra on by Parts again. Here wechoose u = x and dv = sin x and fill in the rest below.

u = x v = ?du = ? dv = sin x dx

⇒ u = x v = − cos xdu = dx dv = sin x dx

Figure . . : Se ng up Integra on by Parts (again).

∫

x cos x dx = x sin x−(

− x cos x−∫

− cos x dx)

.

The integral all the way on the right is now something we can evaluate. It eval-uates to − sin x. Then going through and simplifying, being careful to keep allthe signs straight, our answer is

∫

x cos x dx = x sin x+ x cos x− sin x+ C.

Example . . Integra ng using Integra on by PartsEvaluate

∫

ex cos x dx.

S This is a classic problem. Our mnemonic suggests le ng ube the trigonometric func on instead of the exponen al. In this par cular ex-ample, one can let u be either cos x or ex; to demonstrate that we do not haveto follow LIATE, we choose u = ex and hence dv = cos x dx. Then du = ex dxand v = sin x as shown below.

u = ex v = ?du = ? dv = cos x dx

⇒ u = ex v = sin xdu = ex dx dv = cos x dx

Figure . . : Se ng up Integra on by Parts.

No ce that du is no simpler than u, going against our general rule (but bearwith us). The Integra on by Parts formula yields

∫

ex cos x dx = ex sin x−∫

ex sin x dx.

Notes:

. Integra on by Parts

The integral on the right is not much different than the one we started with, soit seems like we have go en nowhere. Let’s keep working and apply Integra onby Parts to the new integral, using u = ex and dv = sin x dx. This leads us to thefollowing:

u = ex v = ?du = ? dv = sin x dx

⇒ u = ex v = − cos xdu = ex dx dv = sin x dx

Figure . . : Se ng up Integra on by Parts (again).

The Integra on by Parts formula then gives:∫

ex cos x dx = ex sin x−(

−ex cos x−∫

−ex cos x dx)

= ex sin x+ ex cos x−∫

ex cos x dx.

It seems we are back right where we started, as the right hand side contains∫ex cos x dx. But this is actually a good thing.

Add∫

ex cos x dx to both sides. This gives

∫

ex cos x dx = ex sin x+ ex cos x

Now divide both sides by :∫

ex cos x dx =(ex sin x+ ex cos x

).

Simplifying a li le and adding the constant of integra on, our answer is thus∫

ex cos x dx = ex (sin x+ cos x) + C.

Example . . Integra ng using Integra on by Parts: an deriva ve of ln xEvaluate

∫

ln x dx.

S Onemay have no ced that we have rules for integra ng thefamiliar trigonometric func ons and ex, but we have not yet given a rule forintegra ng ln x. That is because ln x can’t easily be integrated with any of therules we have learned up to this point. But we can find its an deriva ve by a

Notes:

Chapter Techniques of An differen a on

clever applica on of Integra on by Parts. Set u = ln x and dv = dx. This is agood, sneaky trick to learn as it can help in other situa ons. This determinesdu = ( /x) dx and v = x as shown below.

u = ln x v = ?du = ? dv = dx

⇒ u = ln x v = xdu = /x dx dv = dx

Figure . . : Se ng up Integra on by Parts.

Pu ng this all together in the Integra on by Parts formula, things work outvery nicely: ∫

ln x dx = x ln x−∫

xxdx.

The new integral simplifies to∫

dx, which is about as simple as things get. Itsintegral is x+ C and our answer is

∫

ln x dx = x ln x− x+ C.

Example . . Integra ng using Int. by Parts: an deriva ve of arctan xEvaluate

∫

arctan x dx.

S The same sneaky trick we used above works here. Let u =arctan x and dv = dx. Then du = /( + x ) dx and v = x. The Integra on byParts formula gives

∫

arctan x dx = x arctan x−∫

x+ x

dx.

The integral on the right can be solved by subs tu on. Taking u = + x , weget du = x dx. The integral then becomes

∫

arctan x dx = x arctan x−∫

udu.

The integral on the right evaluates to ln |u| + C, which becomes ln( + x ) + C(we can drop the absolute values as + x is always pos ve). Therefore, theanswer is ∫

arctan x dx = x arctan x− ln( + x ) + C.

Notes:

. Integra on by Parts

Subs tu on Before Integra on

When taking deriva ves, it was common to employ mul ple rules (such asusing both theQuo ent and the Chain Rules). It should then come as no surprisethat some integrals are best evaluated by combining integra on techniques. Inpar cular, here we illustrate making an “unusual” subs tu on first before usingIntegra on by Parts.

Example . . Integra on by Parts a er subs tu onEvaluate

∫

cos(ln x) dx.

S The integrand contains a composi on of func ons, leadingus to think Subs tu on would be beneficial. Le ng u = ln x, we have du =/x dx. This seems problema c, as we do not have a /x in the integrand. But

consider:du =

xdx ⇒ x · du = dx.

Since u = ln x, we can use inverse func ons and conclude that x = eu. Thereforewe have that

dx = x · du= eu du.

We can thus replace ln x with u and dx with eu du. Thus we rewrite our integralas ∫

cos(ln x) dx =∫

eu cos u du.

We evaluated this integral in Example . . . Using the result there, we have:∫

cos(ln x) dx =∫

eu cos u du

= eu(sin u+ cos u

)+ C

= eln x(sin(ln x) + cos(ln x)

)+ C

= x(sin(ln x) + cos(ln x)

)+ C.

Definite Integrals and Integra on By Parts

So far we have focused only on evalua ng indefinite integrals. Of course, wecan use Integra on by Parts to evaluate definite integrals as well, as Theorem

Notes:

Chapter Techniques of An differen a on

. . states. We do so in the next example.

Example . . Definite integra on using Integra on by Parts

Evaluate∫

x ln x dx.

S Our mnemonic suggests le ng u = ln x, hence dv = x dx.We then get du = ( /x) dx and v = x / as shown below.

u = ln x v = ?du = ? dv = x dx

⇒ u = ln x v = x /

du = /x dx dv = x dx

Figure . . : Se ng up Integra on by Parts.

The Integra on by Parts formula then gives∫

x ln x dx =x

ln x∣∣∣∣−∫

xxdx

=x

ln x∣∣∣∣−∫

xdx

=x

ln x∣∣∣∣− x

∣∣∣∣

=

(x

ln x− x) ∣∣∣∣

=

(

ln −)

−(

ln −)

= ln −

≈ . .

In general, Integra on by Parts is useful for integra ng certain products offunc ons, like

∫xex dx or

∫x sin x dx. It is also useful for integrals involving

logarithms and inverse trigonometric func ons.As stated before, integra on is generally more difficult than deriva on. We

are developing tools for handling a large array of integrals, and experience willtell us when one tool is preferable/necessary over another. For instance, con-sider the three similar–looking integrals

∫

xex dx,∫

xex dx and∫

xex dx.

Notes:

. Integra on by Parts

While the first is calculated easilywith Integra on by Parts, the second is bestapproached with Subs tu on. Taking things one step further, the third integralhas no answer in terms of elementary func ons, so none of the methods welearn in calculus will get us the exact answer.

Integra on by Parts is a very useful method, second only to Subs tu on. Inthe following sec ons of this chapter, we con nue to learn other integra ontechniques. The next sec on focuses on handling integrals containing trigono-metric func ons.

Notes:

Exercises .Terms and Concepts

. T/F: Integra on by Parts is useful in evalua ng integrandsthat contain products of func ons.

. T/F: Integra on by Parts can be thought of as the “oppositeof the Chain Rule.”

. For what is “LIATE” useful?

. T/F: If the integral that results from Integra on by Parts ap-pears to also need Integra on by Parts, then a mistake wasmade in the orginal choice of “u”.

ProblemsIn Exercises – , evaluate the given indefinite integral.

.∫

x sin x dx

.∫

xe−x dx

.∫

x sin x dx

.∫

x sin x dx

.∫

xex dx

.∫

x ex dx

.∫

xe− x dx

.∫

ex sin x dx

.∫

e x cos x dx

.∫

e x sin( x) dx

.∫

e x cos( x) dx

.∫

sin x cos x dx

.∫

sin− x dx

.∫

tan− ( x) dx

.∫

x tan− x dx

.∫

sin− x dx

.∫

x ln x dx

.∫

(x− ) ln x dx

.∫

x ln(x− ) dx

.∫

x ln(x ) dx

.∫

x ln x dx

.∫

(ln x) dx

.∫

(ln(x+ )) dx

.∫

x sec x dx

.∫

x csc x dx

.∫

x√x− dx

.∫

x√x − dx

.∫

sec x tan x dx

.∫

x sec x tan x dx

.∫

x csc x cot x dx

In Exercises – , evaluate the indefinite integral a er firstmaking a subs tu on.

.∫

sin(ln x) dx

.∫

e x cos(

ex)

dx

.∫

sin(√x) dx

.∫

ln(√x) dx

.∫

e√

x dx

.∫

eln x dx

In Exercises – , evaluate the definite integral. Note: thecorresponding indefinite integrals appear in Exercises – .

.∫ π

x sin x dx

.∫

−xe−x dx

.∫ π/

−π/

x sin x dx

.∫ π/

−π/

x sin x dx

.∫

√ln

xex dx

.∫

x ex dx

.∫

xe− x dx

.∫ π

ex sin x dx

.∫ π/

−π/

e x cos x dx

Chapter Techniques of An differen a on

. Trigonometric IntegralsFunc ons involving trigonometric func ons are useful as they are good at de-scribing periodic behavior. This sec on describes several techniques for findingan deriva ves of certain combina ons of trigonometric func ons.

Integrals of the form∫

sinm x cosn x dxIn learning the technique of Subs tu on, we saw the integral

∫sin x cos x dx

in Example . . . The integra onwas not difficult, and one could easily evaluatethe indefinite integral by le ng u = sin x or by le ng u = cos x. This integral iseasy since the power of both sine and cosine is .

Wegeneralize this integral and consider integrals of the form∫sinm x cosn x dx,

where m, n are nonnega ve integers. Our strategy for evalua ng these inte-grals is to use the iden ty cos x + sin x = to convert high powers of onetrigonometric func on into the other, leaving a single sine or cosine term in theintegrand. We summarize the general technique in the following Key Idea.

Key Idea . . Integrals Involving Powers of Sine and Cosine

Consider∫

sinm x cosn x dx, wherem, n are nonnega ve integers.

. Ifm is odd, thenm = k+ for some integer k. Rewrite

sinm x = sin k+ x = sin k x sin x = (sin x)k sin x = ( − cos x)k sin x.

Then∫

sinm x cosn x dx =∫

( − cos x)k sin x cosn x dx = −∫

( − u )kun du,

where u = cos x and du = − sin x dx.

. If n is odd, then using subs tu ons similar to that outlined above we have∫

sinm x cosn x dx =∫

um( − u )k du,

where u = sin x and du = cos x dx.

. If bothm and n are even, use the power–reducing iden es

cos x = + cos( x) and sin x = − cos( x)

to reduce the degree of the integrand. Expand the result and apply the principlesof this Key Idea again.

Notes:

. Trigonometric Integrals

We prac ce applying Key Idea . . in the next examples.

Example . . Integra ng powers of sine and cosineEvaluate

∫

sin x cos x dx.

S The power of the sine term is odd, so we rewrite sin x as

sin x = sin x sin x = (sin x) sin x = ( − cos x) sin x.

Our integral is now∫

( − cos x) cos x sin x dx. Let u = cos x, hence du =

− sin x dx. Making the subs tu on and expanding the integrand gives∫

( −cos ) cos x sin x dx = −∫

( −u ) u du = −∫(− u +u

)u du = −

∫(u − u +u

)du.

This final integral is not difficult to evaluate, giving

−∫(u − u + u

)du = − u + u − u + C

= − cos x+ cos x− cos x+ C.

Example . . Integra ng powers of sine and cosineEvaluate

∫

sin x cos x dx.

S Thepowers of both the sine and cosine terms are odd, there-fore we can apply the techniques of Key Idea . . to either power. We chooseto work with the power of the cosine term since the previous example used thesine term’s power.

We rewrite cos x as

cos x = cos x cos x= (cos x) cos x= ( − sin x) cos x= ( − sin x+ sin x− sin x+ sin x) cos x.

We rewrite the integral as∫

sin x cos x dx =∫

sin x(

− sin x+ sin x− sin x+ sin x)cos x dx.

Notes:

.....

g(x)

.f(x)

....− .

.

.

.

.

.

x

.

y

Figure . . : A plot of f(x) and g(x) fromExample . . and the Technology Note.

Chapter Techniques of An differen a on

Now subs tute and integrate, using u = sin x and du = cos x dx.∫

sin x(

− sin x+ sin x− sin x+ sin x)

cos x dx =∫

u ( − u + u − u + u ) du =

∫

(

u − u + u − u + u)

du

= u − u + u − u + u + C

= sin x− sin x+ sin x+ . . .

− sin x+ sin x+ C.

Technology Note: The work we are doing here can be a bit tedious, but theskills developed (problem solving, algebraic manipula on, etc.) are important.Nowadays problems of this sort are o en solved using a computer algebra sys-tem. The powerful programMathema ca® integrates

∫sin x cos x dx as

f(x) = − cos( x)− cos( x)+

cos( x)+cos( x)− cos( x)− cos( x)− cos( x)

,

which clearly has a different form than our answer in Example . . , which is

g(x) = sin x− sin x+ sin x− sin x+ sin x.

Figure . . shows a graph of f and g; they are clearly not equal, but they differonly by a constant. That is g(x) = f(x) + C for some constant C. So we havetwo different an deriva ves of the same func on, meaning both answers arecorrect.

Example . . Integra ng powers of sine and cosineEvaluate

∫

cos x sin x dx.

S The powers of sine and cosine are both even, so we employthe power–reducing formulas and algebra as follows.

∫

cos x sin x dx =∫ (

+ cos( x)) ( − cos( x)

)

dx

=

∫+ cos( x) + cos ( x) · − cos( x)

dx

=

∫(

+ cos( x)− cos ( x)− cos ( x))dx

The cos( x) term is easy to integrate, especiallywith Key Idea . . . The cos ( x)term is another trigonometric integral with an even power, requiring the power–reducing formula again. The cos ( x) term is a cosine func on with an oddpower, requiring a subs tu on as done before. We integrate each in turn below.

Notes:

. Trigonometric Integrals

∫

cos( x) dx = sin( x) + C.

∫

cos ( x) dx =∫

+ cos( x)dx =

(x+ sin( x)

)+ C.

Finally, we rewrite cos ( x) as

cos ( x) = cos ( x) cos( x) =(

− sin ( x))cos( x).

Le ng u = sin( x), we have du = cos( x) dx, hence∫

cos ( x) dx =∫(

− sin ( x))cos( x) dx

=

∫

( − u ) du

=(

u− u)

+ C

=(

sin( x)− sin ( x))

+ C

Pu ng all the pieces together, we have∫

cos x sin x dx =∫

(+ cos( x)− cos ( x)− cos ( x)

)dx

=[

x+ sin( x)−(x+ sin( x)

)−

(

sin( x)− sin ( x))]

+ C

=[

x− sin( x) + sin ( x)]

+ C

The process above was a bit long and tedious, but being able to work a prob-lem such as this from start to finish is important.

Integrals of the form∫

sin(mx) sin(nx) dx,∫

cos(mx) cos(nx) dx,

and∫

sin(mx) cos(nx) dx.

Func ons that contain products of sines and cosines of differing periods areimportant in many applica ons including the analysis of sound waves. Integralsof the form∫

sin(mx) sin(nx) dx,∫

cos(mx) cos(nx) dx and∫

sin(mx) cos(nx) dx

Notes:

Chapter Techniques of An differen a on

are best approached by first applying the Product to Sum Formulas found in theback cover of this text, namely

sin(mx) sin(nx) =[

cos((m− n)x

)− cos

((m+ n)x

)]

cos(mx) cos(nx) =[

cos((m− n)x

)+ cos

((m+ n)x

)]

sin(mx) cos(nx) =[

sin((m− n)x

)+ sin

((m+ n)x

)]

Example . . Integra ng products of sin(mx) and cos(nx)

Evaluate∫

sin( x) cos( x) dx.

S The applica on of the formula and subsequent integra onare straigh orward:

∫

sin( x) cos( x) dx =∫ [

sin( x) + sin( x)]

dx

= − cos( x)− cos( x) + C

Integrals of the form∫

tanm x secn x dx.

When evalua ng integrals of the form∫sinm x cosn x dx, the Pythagorean

Theorem allowed us to convert even powers of sine into even powers of cosine,and vise–versa. If, for instance, the power of sine was odd, we pulled out onesin x and converted the remaining even power of sin x into a func on using pow-ers of cos x, leading to an easy subs tu on.

The same basic strategy applies to integrals of the form∫tanm x secn x dx,

albeit a bit more nuanced. The following three facts will prove useful:

• ddx (tan x) = sec x,

• ddx (sec x) = sec x tan x , and

• + tan x = sec x (the Pythagorean Theorem).

If the integrand can be manipulated to separate a sec x term with the re-maining secant power even, or if a sec x tan x term can be separated with theremaining tan x power even, the Pythagorean Theorem can be employed, lead-ing to a simple subs tu on. This strategy is outlined in the following Key Idea.

Notes:

. Trigonometric Integrals

Key Idea . . Integrals Involving Powers of Tangent and Secant

Consider∫

tanm x secn x dx, wherem, n are nonnega ve integers.

. If n is even, then n = k for some integer k. Rewrite secn x as

secn x = sec k x = sec k− x sec x = ( + tan x)k− sec x.

Then∫

tanm x secn x dx =∫

tanm x( + tan x)k− sec x dx =∫

um( + u )k− du,

where u = tan x and du = sec x dx.

. Ifm is odd, thenm = k+ for some integer k. Rewrite tanm x secn x as

tanm x secn x = tan k+ x secn x = tan k x secn− x sec x tan x = (sec x− )k secn− x sec x tan x.

Then∫

tanm x secn x dx =∫

(sec x− )k secn− x sec x tan x dx =∫

(u − )kun− du,

where u = sec x and du = sec x tan x dx.

. If n is odd andm is even, thenm = k for some integer k. Convert tanm x to (sec x− )k. Expandthe new integrand and use Integra on By Parts, with dv = sec x dx.

. Ifm is even and n = , rewrite tanm x as

tanm x = tanm− x tan x = tanm− x(sec x− ) = tanm− sec x− tanm− x.

So ∫

tanm x dx =∫

tanm− sec x dx︸ ︷︷ ︸

apply rule #

−∫

tanm− x dx︸ ︷︷ ︸

apply rule # again

.

The techniques described in items and of Key Idea . . are rela velystraigh orward, but the techniques in items and can be rather tedious. Afew examples will help with these methods.

Notes:

Chapter Techniques of An differen a on

Example . . Integra ng powers of tangent and secantEvaluate

∫

tan x sec x dx.

S Since the power of secant is even, we use rule # from KeyIdea . . and pull out a sec x in the integrand. We convert the remaining pow-ers of secant into powers of tangent.

∫

tan x sec x dx =∫

tan x sec x sec x dx

=

∫

tan x(

+ tan x)

sec x dx

Now subs tute, with u = tan x, with du = sec x dx.

=

∫

u(

+ u)

du

We leave the integra on and subsequent subs tu on to the reader. The finalanswer is

= tan x+ tan x+ tan x+ C.

Example . . Integra ng powers of tangent and secantEvaluate

∫

sec x dx.

S We apply rule # from Key Idea . . as the power of secantis odd and the power of tangent is even ( is an even number). We use Integra-on by Parts; the rule suggests le ng dv = sec x dx, meaning that u = sec x.

u = sec x v = ?du = ? dv = sec x dx

⇒ u = sec x v = tan xdu = sec x tan x dx dv = sec x dx

Figure . . : Se ng up Integra on by Parts.

Employing Integra on by Parts, we have∫

sec x dx =∫

sec x︸︷︷︸

u

· sec x dx︸ ︷︷ ︸

dv

= sec x tan x−∫

sec x tan x dx.

Notes:

. Trigonometric Integrals

This new integral also requires applying rule # of Key Idea . . :

= sec x tan x−∫

sec x(sec x−

)dx

= sec x tan x−∫

sec x dx+∫

sec x dx

= sec x tan x−∫

sec x dx+ ln | sec x+ tan x|

In previous applica ons of Integra on by Parts, we have seen where the originalintegral has reappeared in our work. We resolve this by adding

∫sec x dx to

both sides, giving:∫

sec x dx = sec x tan x+ ln | sec x+ tan x|∫

sec x dx =(

sec x tan x+ ln | sec x+ tan x|)

+ C

We give one more example.

Example . . Integra ng powers of tangent and secantEvaluate

∫

tan x dx.

S We employ rule # of Key Idea . . .∫

tan x dx =∫

tan x tan x dx

=

∫

tan x(sec x−

)dx

=

∫

tan x sec x dx−∫

tan x dx

Integrate the first integral with subs tu on, u = tan x; integrate the second byemploying rule # again.

= tan x−∫

tan x tan x dx

= tan x−∫

tan x(sec x−

)dx

= tan x−∫

tan x sec x dx+∫

tan x dx

Notes:

Chapter Techniques of An differen a on

Again, use subs tu on for the first integral and rule # for the second.

= tan x− tan x+∫(sec x−

)dx

= tan x− tan x+ tan x− x+ C.

These la er examples were admi edly long, with repeated applica ons ofthe same rule. Try to not be overwhelmed by the length of the problem, butrather admire how robust this solu on method is. A trigonometric func on ofa high power can be systema cally reduced to trigonometric func ons of lowerpowers un l all an deriva ves can be computed.

The next sec on introduces an integra on technique known as Trigonomet-ric Subs tu on, a clever combina on of Subs tu on and the Pythagorean The-orem.

Notes:

Exercises .Terms and Concepts

. T/F:∫

sin x cos x dx cannot be evaluated using the tech-niques described in this sec on since both powers of sin xand cos x are even.

. T/F:∫

sin x cos x dx cannot be evaluated using the tech-niques described in this sec on since both powers of sin xand cos x are odd.

. T/F: This sec on addresses how to evaluate indefinite inte-grals such as

∫

sin x tan x dx.

. T/F: Some mes computer programs evaluate integrals in-volving trigonometric func ons differently than one wouldusing the techniques of this sec on. When this is the case,the techniques of this sec on have failed and one shouldonly trust the answer given by the computer.

ProblemsIn Exercises – , evaluate the indefinite integral.

.∫

sin x cos x dx

.∫

sin x cos x dx

.∫

sin x cos x dx

.∫

sin x cos x dx

.∫

sin x cos x dx

.∫

sin x cos x dx

.∫

sin x cos x dx

.∫

sin x cos x dx

.∫

sin( x) cos( x) dx

.∫

sin(x) cos( x) dx

.∫

sin( x) sin( x) dx

.∫

sin(πx) sin( πx) dx

.∫

cos(x) cos( x) dx

.∫

cos(π x

)

cos(πx) dx

.∫

tan x sec x dx

.∫

tan x sec x dx

.∫

tan x sec x dx

.∫

tan x sec x dx

.∫

tan x sec x dx

.∫

tan x sec x dx

.∫

tan x dx

.∫

sec x dx

.∫

tan x sec x dx

.∫

tan x sec x dx

In Exercises – , evaluate the definite integral. Note: thecorresponding indefinite integrals appear in the previous set.

.∫ π

sin x cos x dx

.∫ π

−π

sin x cos x dx

.∫ π/

−π/

sin x cos x dx

.∫ π/

sin( x) cos( x) dx

.∫ π/

−π/

cos(x) cos( x) dx

.∫ π/

tan x sec x dx

.∫ π/

−π/

tan x sec x dx

Chapter Techniques of An differen a on

. Trigonometric Subs tu onIn Sec on . we defined the definite integral as the “signed area under thecurve.” In that sec on we had not yet learned the Fundamental Theorem ofCalculus, so we only evaluated special definite integrals which described nice,geometric shapes. For instance, we were able to evaluate

∫

−

√

− x dx =π

( . )

as we recognized that f(x) =√

− x described the upper half of a circle withradius .

We have since learned a number of integra on techniques, including Sub-s tu on and Integra on by Parts, yet we are s ll unable to evaluate the aboveintegral without resor ng to a geometric interpreta on. This sec on introducesTrigonometric Subs tu on, amethod of integra on that fills this gap in our inte-gra on skill. This techniqueworks on the sameprinciple as Subs tu on as foundin Sec on . , though it can feel “backward.” In Sec on . , we set u = f(x), forsome func on f, and replaced f(x) with u. In this sec on, we will set x = f(θ),where f is a trigonometric func on, then replace x with f(θ).

We start by demonstra ng this method in evalua ng the integral in Equa on( . ). A er the example, wewill generalize themethod and givemore examples.

Example . . Using Trigonometric Subs tu on

Evaluate∫

−

√

− x dx.

S We begin by no ng that sin θ + cos θ = , and hencecos θ = − sin θ. If we let x = sin θ, then −x = − sin θ = cos θ.Se ng x = sin θ gives dx = cos θ dθ. We are almost ready to subs tute.

We also wish to change our bounds of integra on. The bound x = − corre-sponds to θ = −π/ (for when θ = −π/ , x = sin θ = − ). Likewise, thebound of x = is replaced by the bound θ = π/ . Thus

∫

−

√

− x dx =∫ π/

−π/

√

− sin θ( cos θ) dθ

=

∫ π/

−π/

√cos θ cos θ dθ

=

∫ π/

−π/

| cos θ| cos θ dθ.

On [−π/ , π/ ], cos θ is always posi ve, so we can drop the absolute value bars,then employ a power–reducing formula:

Notes:

. Trigonometric Subs tu on

=

∫ π/

−π/

cos θ dθ

=

∫ π/

−π/

(+ cos( θ)

)dθ

=(θ + sin( θ)

)

∣∣∣∣∣

π/

−π/

= π.

This matches our answer from before.

We now describe in detail Trigonometric Subs tu on. This method excelswhen dealing with integrands that contain

√a − x ,

√x − a and

√x + a .

The following Key Idea outlines the procedure for each case, followed by moreexamples. Each right triangle acts as a reference to help us understand the re-la onships between x and θ.

Key Idea . . Trigonometric Subs tu on

(a) For integrands containing√a − x :

Let x = a sin θ, dx = a cos θ dθ

Thus θ = sin− (x/a), for−π/ ≤ θ ≤ π/ .

On this interval, cos θ ≥ , so√a − x = a cos θ

..√

a2 − x2.

x

.

a

. θ

(b) For integrands containing√x + a :

Let x = a tan θ, dx = a sec θ dθ

Thus θ = tan− (x/a), for−π/ < θ < π/ .

On this interval, sec θ > , so√x + a = a sec θ

..a

.

x

.

√ x2 +a2

. θ

(c) For integrands containing√x − a :

Let x = a sec θ, dx = a sec θ tan θ dθ

Thus θ = sec− (x/a). If x/a ≥ , then ≤ θ < π/ ;if x/a ≤ − , then π/ < θ ≤ π.

We restrict our work to where x ≥ a, so x/a ≥ , and≤ θ < π/ . On this interval, tan θ ≥ , so

√x − a = a tan θ

..a

.

√

x2 − a2

.

x

. θ

Notes:

Chapter Techniques of An differen a on

Example . . Using Trigonometric Subs tu onEvaluate

∫

√+ x

dx.

S Using Key Idea . . (b), we recognize a =√

and set x =√tan θ. This makes dx =

√sec θ dθ. We will use the fact that

√+ x =√

+ tan θ =√

sec θ =√

sec θ. Subs tu ng, we have:∫

√+ x

dx =∫

√+ tan θ

√sec θ dθ

=

∫ √sec θ√sec θ

dθ

=

∫

sec θ dθ

= ln∣∣ sec θ + tan θ

∣∣+ C.

While the integra on steps are over, we are not yet done. The original problemwas stated in terms of x, whereas our answer is given in terms of θ. We mustconvert back to x.

The reference triangle given in Key Idea . . (b) helps. With x =√

tan θ,we have

tan θ =x√ and sec θ =

√x +√ .

This gives∫

√+ x

dx = ln∣∣ sec θ + tan θ

∣∣+ C

= ln

∣∣∣∣∣

√x +√ +

x√∣∣∣∣∣+ C.

We can leave this answer as is, or we can use a logarithmic iden ty to simplifyit. Note:

ln

∣∣∣∣∣

√x +√ +

x√∣∣∣∣∣+ C = ln

∣∣∣∣√(√

x + + x)∣∣∣∣+ C

= ln∣∣∣∣√∣∣∣∣+ ln

∣∣√

x + + x∣∣+ C

= ln∣∣√

x + + x∣∣+ C,

where the ln(/√ )

term is absorbed into the constant C. (In Sec on . wewill learn another way of approaching this problem.)

Notes:

. Trigonometric Subs tu on

Example . . Using Trigonometric Subs tu onEvaluate

∫√

x − dx.

S Westart by rewri ng the integrand so that it looks like√x − a

for some value of a:

√

x − =

√(

x −)

=

√

x −( )

.

So we have a = / , and following Key Idea . . (c), we set x = sec θ, andhence dx = sec θ tan θ dθ. We now rewrite the integral with these subs tu-ons:

∫√

x − dx =∫

√

x −( )

dx

=

∫ √

sec θ −(

sec θ tan θ)

dθ

=

∫ √

(sec θ − )(

sec θ tan θ)

dθ

=

∫ √

tan θ(

sec θ tan θ)

dθ

=

∫

tan θ sec θ dθ

=

∫ (

sec θ −)

sec θ dθ

=

∫(sec θ − sec θ

)dθ.

We integrated sec θ in Example . . , finding its an deriva ves to be∫

sec θ dθ =(

sec θ tan θ + ln | sec θ + tan θ|)

+ C.

Thus∫√

x − dx =∫(sec θ − sec θ

)dθ

=

( (

sec θ tan θ + ln | sec θ + tan θ|)

− ln | sec θ + tan θ|)

+ C

= (sec θ tan θ − ln | sec θ + tan θ|) + C.

Notes:

Chapter Techniques of An differen a on

We are not yet done. Our original integral is given in terms of x, whereas ourfinal answer, as given, is in terms of θ. We need to rewrite our answer in termsof x. With a = / , and x = sec θ, the reference triangle in Key Idea . . (c)shows that

tan θ =√

x − //

( / ) =√

x − / and sec θ = x.

Thus(

sec θ tan θ − ln∣

∣ sec θ + tan θ∣

∣

)

+ C =(

x ·√

x − / − ln∣

∣ x+√

x − /∣

∣

)

+ C

=(

x√

x − / − ln∣

∣ x+√

x − /∣

∣

)

+ C.

The final answer is given in the last line above, repeated here:∫√

x − dx =(

x√

x − / − ln∣∣ x+

√

x − /∣∣

)

+ C.

Example . . Using Trigonometric Subs tu on

Evaluate∫ √

− xx

dx.

S We use Key Idea . . (a) with a = , x = sin θ, dx =cos θ and hence

√− x = cos θ. This gives

∫ √− xx

dx =∫

cos θsin θ

( cos θ) dθ

=

∫

cot θ dθ

=

∫

(csc θ − ) dθ

= − cot θ − θ + C.

We need to rewrite our answer in terms of x. Using the reference triangle foundin Key Idea . . (a), we have cot θ =

√− x /x and θ = sin− (x/ ). Thus

∫ √− xx

dx = −√

− xx

− sin−( x)

+ C.

Trigonometric Subs tu on can be applied inmany situa ons, even those notof the form

√a − x ,

√x − a or

√x + a . In the following example, we ap-

ply it to an integral we already know how to handle.

Notes:

. Trigonometric Subs tu on

Example . . Using Trigonometric Subs tu onEvaluate

∫

x +dx.

S Weknow the answer already as tan− x+C. Weapply Trigono-metric Subs tu on here to show that we get the same answer without inher-ently relying on knowledge of the deriva ve of the arctangent func on.

Using Key Idea . . (b), let x = tan θ, dx = sec θ dθ and note that x + =tan θ + = sec θ. Thus

∫

x +dx =

∫

sec θsec θ dθ

=

∫

dθ

= θ + C.

Since x = tan θ, θ = tan− x, and we conclude that∫

x +dx = tan− x+C.

The next example is similar to the previous one in that it does not involve asquare–root. It shows how several techniques and iden es can be combinedto obtain a solu on.

Example . . Using Trigonometric Subs tu onEvaluate

∫

(x + x+ )dx.

S We start by comple ng the square, then make the subs tu-on u = x+ , followed by the trigonometric subs tu on of u = tan θ:∫

(x + x+ )dx =

∫

((x+ ) +

) dx =∫

(u + )du.

Now make the subs tu on u = tan θ, du = sec θ dθ:

=

∫

(tan θ + )sec θ dθ

=

∫

(sec θ)sec θ dθ

=

∫

cos θ dθ.

Notes:

Chapter Techniques of An differen a on

Applying a power reducing formula, we have

=

∫ (

+ cos( θ)

)

dθ

= θ + sin( θ) + C. ( . )

We need to return to the variable x. As u = tan θ, θ = tan− u. Using theiden ty sin( θ) = sin θ cos θ and using the reference triangle found in KeyIdea . . (b), we have

sin( θ) =u√

u +· √

u +=

uu +

.

Finally, we return to xwith the subs tu on u = x+ . We start with the expres-sion in Equa on ( . ):

θ + sin( θ) + C = tan− u+u

u ++ C

= tan− (x+ ) +x+

(x + x+ )+ C.

Sta ng our final result in one line,∫

(x + x+ )dx = tan− (x+ ) +

x+(x + x+ )

+ C.

Our last example returns us to definite integrals, as seen in our first example.Given a definite integral that can be evaluated using Trigonometric Subs tu on,we could first evaluate the corresponding indefinite integral (by changing froman integral in terms of x to one in terms of θ, then conver ng back to x) and thenevaluate using the original bounds. It is much more straigh orward, though, tochange the bounds as we subs tute.

Example . . Definite integra on and Trigonometric Subs tu on

Evaluate∫

x√x +

dx.

S Using Key Idea . . (b), we set x = tan θ, dx = sec θ dθ,and note that

√x + = sec θ. As we subs tute, we can also change the

bounds of integra on.The lower bound of the original integral is x = . As x = tan θ, we solve for

θ and find θ = tan− (x/ ). Thus the new lower bound is θ = tan− ( ) = . The

Notes:

. Trigonometric Subs tu on

original upper bound is x = , thus the new upper bound is θ = tan− ( / ) =π/ .

Thus we have∫

x√x +

dx =∫ π/ tan θ

sec θsec θ dθ

=

∫ π/

tan θ sec θ dθ.

We encountered this indefinite integral in Example . . where we found∫

tan θ sec θ dθ =(sec θ tan θ − ln | sec θ + tan θ|

).

So

∫ π/

tan θ sec θ dθ =(sec θ tan θ − ln | sec θ + tan θ|

)

∣∣∣∣∣

π/

=(√

− ln(√

+ ))

≈ . .

The following equali es are very usefulwhenevalua ng integrals using Trigono-metric Subs tu on.

Key Idea . . Useful Equali es with Trigonometric Subs tu on

. sin( θ) = sin θ cos θ

. cos( θ) = cos θ − sin θ = cos θ − = − sin θ

.∫

sec θ dθ =(

sec θ tan θ + ln∣∣ sec θ + tan θ

∣∣

)

+ C

.∫

cos θ dθ =

∫(

+ cos( θ))dθ =

(θ + sin θ cos θ

)+ C.

The next sec on introduces Par al Frac onDecomposi on, which is an alge-braic technique that turns “complicated” frac ons into sums of “simpler” frac-ons, making integra on easier.

Notes:

Exercises .Terms and Concepts. Trigonometric Subs tu on works on the same principles asIntegra on by Subs tu on, though it can feel “ ”.

. If one uses Trigonometric Subs tu on on an integrand con-taining

√− x , then one should set x = .

. Consider the Pythagorean Iden ty sin θ + cos θ = .

(a) What iden ty is obtained when both sides are di-vided by cos θ?

(b) Use the new iden ty to simplify tan θ + .

. Why does Key Idea . . (a) state that√a − x = a cos θ,

and not |a cos θ|?

ProblemsIn Exercises – , apply Trigonometric Subs tu on to eval-uate the indefinite integrals.

.∫ √

x + dx

.∫ √

x + dx

.∫ √

− x dx

.∫ √

− x dx

.∫ √

x − dx

.∫ √

x − dx

.∫ √

x + dx

.∫ √

− x dx

.∫ √

x − dx

.∫

√x +

dx

.∫

√− x

dx

.∫

√x −

dx

In Exercises – , evaluate the indefinite integrals. Somemay be evaluated without Trigonometric Subs tu on.

.∫

√x −x

dx

.∫

(x + )dx

.∫

x√x −

dx

.∫

x√

− x dx

.∫

x(x + ) /

dx

.∫

x√x −

dx

.∫

(x + x+ )dx

.∫

x ( − x )− / dx

.∫

√− xx

dx

.∫

x√x +

dx

In Exercises – , evaluate the definite integrals by mak-ing the proper trigonometric subs tu on and changing thebounds of integra on. (Note: each of the correspondingindefinite integrals has appeared previously in this Exerciseset.)

.∫

−

√− x dx

.∫ √

x − dx

.∫ √

x + dx

.∫

− (x + )dx

.∫

−

√− x dx

.∫

−x√

− x dx

. Par al Frac on Decomposi on

. Par al Frac on Decomposi on

In this sec onwe inves gate the an deriva ves of ra onal func ons. Recall thatra onal func ons are func ons of the form f(x) = p(x)

q(x) , where p(x) and q(x) arepolynomials and q(x) ̸= . Such func ons arise in many contexts, one of whichis the solving of certain fundamental differen al equa ons.

We begin with an example that demonstrates the mo va on behind thissec on. Consider the integral

∫

x − dx. We do not have a simple formula

for this (if the denominator were x + , we would recognize the an deriva veas being the arctangent func on). It can be solved using Trigonometric Subs -tu on, but note how the integral is easy to evaluate once we realize:

x − =/

x− − /

x+.

Thus

∫

x − dx =∫

/

x− dx−∫

/

x+dx

= ln |x− | − ln |x+ |+ C.

This sec on teaches how to decompose

x − into/

x− − /

x+.

We start with a ra onal func on f(x) = p(x)q(x) , where p and q do not have any

common factors and the degree of p is less than the degree of q. It can be shownthat any polynomial, and hence q, can be factored into a product of linear andirreducible quadra c terms. The following Key Idea states how to decompose ara onal func on into a sum of ra onal func ons whose denominators are all oflower degree than q.

Notes:

Chapter Techniques of An differen a on

Key Idea . . Par al Frac on Decomposi on

Letp(x)q(x)

be a ra onal func on, where the degree of p is less than the

degree of q.

. Linear Terms: Let (x−a) divide q(x), where (x−a)n is the highestpower of (x−a) that divides q(x). Then the decomposi on of p(x)

q(x)will contain the sum

A(x− a)

+A

(x− a)+ · · ·+ An

(x− a)n.

. Quadra c Terms: Let x +bx+ c divide q(x), where (x +bx+ c)nis the highest power of x + bx + c that divides q(x). Then thedecomposi on of p(x)

q(x) will contain the sum

B x+ Cx + bx+ c

+B x+ C

(x + bx+ c)+ · · ·+ Bnx+ Cn

(x + bx+ c)n.

To find the coefficients Ai, Bi and Ci:

. Mul ply all frac ons by q(x), clearing the denominators. Collectlike terms.

. Equate the resul ng coefficients of the powers of x and solve theresul ng system of linear equa ons.

The following examples will demonstrate how to put this Key Idea into prac-ce. Example . . stresses the decomposi on aspect of the Key Idea.

Example . . Decomposing into par al frac onsDecompose f(x) =

(x+ )(x− ) (x + x+ )(x + x+ )without solving

for the resul ng coefficients.

S The denominator is already factored, as both x + x+ andx + x + cannot be factored further. We need to decompose f(x) properly.Since (x+ ) is a linear term that divides the denominator, there will be a

Ax+

Notes:

. Par al Frac on Decomposi on

term in the decomposi on.As (x− ) divides the denominator, we will have the following terms in the

decomposi on:B

x− ,C

(x− )and

D(x− )

.

The x + x+ term in the denominator results in aEx+ F

x + x+term.

Finally, the (x + x+ ) term results in the terms

Gx+ Hx + x+

andIx+ J

(x + x+ ).

All together, we have

(x+ )(x− ) (x + x+ )(x + x+ )=

Ax+

+B

x− +C

(x− )+

D(x− )

+

Ex+ Fx + x+

+Gx+ H

x + x++

Ix+ J(x + x+ )

Solving for the coefficients A, B . . . J would be a bit tedious but not “hard.”

Example . . Decomposing into par al frac onsPerform the par al frac on decomposi on of

x − .

S The denominator factors into two linear terms: x − =(x− )(x+ ). Thus

x − =A

x− +B

x+.

To solve for A and B, first mul ply through by x − = (x− )(x+ ):

=A(x− )(x+ )

x− +B(x− )(x+ )

x+= A(x+ ) + B(x− )

= Ax+ A+ Bx− B

Now collect like terms.

= (A+ B)x+ (A− B).

The next step is key. Note the equality we have:

= (A+ B)x+ (A− B).

Notes:

Note: Equa on . offers a direct route tofinding the values of A, B and C. Since theequa on holds for all values of x, it holdsin par cular when x = . However, whenx = , the right hand side simplifies toA( + ) = A. Since the le hand sideis s ll , we have = A. HenceA = / .Likewise, the equality holds when x =− ; this leads to the equa on = − C.Thus C = − / .Knowing A and C, we can find the value ofB by choosing yet another value of x, suchas x = , and solving for B.

Chapter Techniques of An differen a on

For clarity’s sake, rewrite the le hand side as

x+ = (A+ B)x+ (A− B).

On the le , the coefficient of the x term is ; on the right, it is (A + B). Sinceboth sides are equal, we must have that = A+ B.

Likewise, on the le , we have a constant term of ; on the right, the constantterm is (A− B). Therefore we have = A− B.

We have two linear equa ons with two unknowns. This one is easy to solveby hand, leading to

A+ B =A− B =

⇒ A = /B = − /

.

Thus

x − =/

x− − /

x+.

Example . . Integra ng using par al frac onsUse par al frac on decomposi on to integrate

∫

(x− )(x+ )dx.

S Wedecompose the integrand as follows, as described by KeyIdea . . :

(x− )(x+ )=

Ax− +

Bx+

+C

(x+ ).

To solve for A, B and C, we mul ply both sides by (x− )(x+ ) and collect liketerms:

= A(x+ ) + B(x− )(x+ ) + C(x− ) ( . )= Ax + Ax+ A+ Bx + Bx− B+ Cx− C= (A+ B)x + ( A+ B+ C)x+ ( A− B− C)

We have

x + x+ = (A+ B)x + ( A+ B+ C)x+ ( A− B− C)

leading to the equa ons

A+ B = , A+ B+ C = and A− B− C = .

These three equa ons of three unknowns lead to a unique solu on:

A = / , B = − / and C = − / .

Notes:

Note: The values ofA andB can be quicklyfound using the technique described inthe margin of Example . . .

. Par al Frac on Decomposi on

Thus∫

(x− )(x+ )dx =

∫/

x− dx+∫ − /

x+dx+

∫ − /

(x+ )dx.

Each can be integrated with a simple subs tu onwith u = x− or u = x+(or by directly applying Key Idea . . as the denominators are linear func ons).The end result is

∫

(x− )(x+ )dx = ln |x− | − ln |x+ |+

(x+ )+ C.

Example . . Integra ng using par al frac ons

Use par al frac on decomposi on to integrate∫

x(x− )(x+ )

dx.

S Key Idea . . presumes that the degree of the numeratoris less than the degree of the denominator. Since this is not the case here, webegin by using polynomial division to reduce the degree of the numerator. Weomit the steps, but encourage the reader to verify that

x(x− )(x+ )

= x+ +x+

(x− )(x+ ).

Using Key Idea . . , we can rewrite the new ra onal func on as:

x+(x− )(x+ )

=A

x− +B

x+

for appropriate values of A and B. Clearing denominators, we have

x+ = A(x+ ) + B(x− )

= (A+ B)x+ ( A− B).

This implies that:

= A+ B= A− B.

Solving this system of linear equa ons gives

/ = A/ = B.

Notes:

Chapter Techniques of An differen a on

We can now integrate.∫

x(x− )(x+ )

dx =∫ (

x+ +/

x− +/

x+

)

dx

=x

+ x+ ln |x− |+ ln |x+ |+ C.

Example . . Integra ng using par al frac ons

Use par al frac on decomposi on to evaluate∫

x + x+(x+ )(x + x+ )

dx.

S The degree of the numerator is less than the degree of thedenominator so we begin by applying Key Idea . . . We have:

x + x+(x+ )(x + x+ )

=A

x++

Bx+ Cx + x+

.

Now clear the denominators.

x + x+ = A(x + x+ ) + (Bx+ C)(x+ )

= (A+ B)x + ( A+ B+ C)x+ ( A+ C).

This implies that:

= A+ B= A+ B+ C= A+ C.

Solving this system of linear equa ons gives the nice result of A = , B = andC = − . Thus

∫x + x+

(x+ )(x + x+ )dx =

∫ (

x++

x−x + x+

)

dx.

The first termof this new integrand is easy to evaluate; it leads to a ln |x+ |term. The second term is not hard, but takes several steps and uses subs tu ontechniques.

The integrandx−

x + x+has a quadra c in the denominator and a linear

term in the numerator. This leads us to try subs tu on. Let u = x + x+ , sodu = ( x+ ) dx. The numerator is x− , not x+ , but we can get a x+

Notes:

. Par al Frac on Decomposi on

term in the numerator by adding in the form of “ − .”

x−x + x+

=x− + −x + x+

=x+

x + x+−

x + x+.

Wecannow integrate the first termwith subs tu on, leading to a ln |x + x+ |term. The final term can be integrated using arctangent. First, complete thesquare in the denominator:

x + x+=

(x+ ) +.

An an deriva ve of the la er term can be found using Theorem . . and sub-s tu on: ∫

x + x+dx = √ tan−

(x+√

)

+ C.

Let’s start at the beginning and put all of the steps together.∫

x + x+(x+ )(x + x+ )

dx =∫(

x++

x−x + x+

)

dx

=

∫

x+dx+

∫

x+x + x+

dx−∫

x + x+dx

= ln |x+ |+ ln |x + x+ | − √ tan−(

x+√)

+ C.

As with many other problems in calculus, it is important to remember that oneis not expected to “see” the final answer immediately a er seeing the problem.Rather, given the ini al problem, we break it down into smaller problems thatare easier to solve. The final answer is a combina on of the answers of thesmaller problems.

Par al Frac on Decomposi on is an important tool when dealing with ra o-nal func ons. Note that at its heart, it is a technique of algebra, not calculus,as we are rewri ng a frac on in a new form. Regardless, it is very useful in therealm of calculus as it lets us evaluate a certain set of “complicated” integrals.

The next sec on introduces new func ons, called the Hyperbolic Func ons.They will allow us to make subs tu ons similar to those found when studyingTrigonometric Subs tu on, allowing us to approach evenmore integra onprob-lems.

Notes:

Exercises .Terms and Concepts. Fill in the blank: Par al Frac on Decomposi on is amethodof rewri ng func ons.

. T/F: It is some mes necessary to use polynomial divisionbefore using Par al Frac on Decomposi on.

. Decomposex − x

without solving for the coefficients, asdone in Example . . .

. Decompose − xx − without solving for the coefficients, as

done in Example . . .

. Decompose x−x − without solving for the coefficients, as

done in Example . . .

. Decompose x+x + x

without solving for the coefficients, asdone in Example . . .

ProblemsIn Exercises – , evaluate the indefinite integral.

.∫

x+x + x− dx

.∫

x−x + x

dx

.∫ −

x − dx

.∫

x+x + x+

dx

.∫

x+(x+ )

dx

.∫ − x−

(x+ )dx

.∫

x + x+x(x+ )

dx

.∫ − x − x+

(x− )(x+ )( − x)dx

.∫

x − x( x+ )( x− )( x− )

dx

.∫

x + x+x + x− dx

.∫

xx − x− dx

.∫

x − x+x − x+

dx

.∫

x + x + xdx

.∫

x + x+x + x+

dx

.∫

x + x+(x+ )( x + x− )

dx

.∫

x + x−(x− )(x + x+ )

dx

.∫

x + x+(x+ )(x + )

dx

.∫

x − x−(x− )(x + x+ )

dx

.∫

x − x+(x− )(x − x+ )

dx

.∫

x + x+(x+ )(x + x+ )

dx

In Exercises – , evaluate the definite integral.

.∫

x+(x+ )(x+ )

dx

.∫

x+( x+ )(x+ )

dx

.∫

−

x + x−(x− )(x + x+ )

dx

.∫

x(x+ )(x + x+ )

dx

.....

(cos θ,sin θ)

.

θ

2

.

x2 + y2 = 1

.

−1

.

1

. −1.

1

.

x

.

y

.....

(cosh θ,sinh θ)

.

θ2

.

x2 − y2 = 1

.

−2

.

2

.

−2

.

2

.

x

.

y

Figure . . : Using trigonometric func-ons to define points on a circle and hy-

perbolic func ons to define points on ahyperbola. The area of the shaded re-gions are included in them.

Pronuncia on Note:“cosh” rhymes with “gosh,”“sinh” rhymes with “pinch,” and“tanh” rhymes with “ranch.”

. Hyperbolic Func ons

. Hyperbolic Func onsThe hyperbolic func ons are a set of func ons that have many applica ons tomathema cs, physics, and engineering. Among many other applica ons, theyare used to describe the forma on of satellite rings around planets, to describethe shape of a rope hanging from two points, and have applica on to the theoryof special rela vity. This sec on defines the hyperbolic func ons and describesmany of their proper es, especially their usefulness to calculus.

These func ons are some mes referred to as the “hyperbolic trigonometricfunc ons” as there are many, many connec ons between them and the stan-dard trigonometric func ons. Figure . . demonstrates one such connec on.Just as cosine and sine are used to define points on the circle defined by x +y =, the func ons hyperbolic cosine and hyperbolic sine are used to define points

on the hyperbola x − y = .We begin with their defini on.

Defini on . . Hyperbolic Func ons

. cosh x =ex + e−x

. sinh x =ex − e−x

. tanh x =sinh xcosh x

. sech x =cosh x

. csch x =sinh x

. coth x =cosh xsinh x

These hyperbolic func ons are graphed in Figure . . . In the graphs ofcosh x and sinh x, graphs of ex/ and e−x/ are included with dashed lines. Asx gets “large,” cosh x and sinh x each act like ex/ ; when x is a large nega venumber, cosh x acts like e−x/ whereas sinh x acts like−e−x/ .

No ce the domains of tanh x and sech x are (−∞,∞), whereas both coth xand csch x have ver cal asymptotes at x = . Also note the ranges of thesefunc ons, especially tanh x: as x → ∞, both sinh x and cosh x approach ex/ ,hence tanh x approaches .

The following example explores some of the proper es of these func onsthat bear remarkable resemblance to the proper es of their trigonometric coun-terparts.

Notes:

Chapter Techniques of An differen a on

.....

f(x) = cosh x

.

−

.. −.

−5

.

5

..

x

.

y

.....

f(x) = sinh x

.

−

.. −.

−5

.

5

..

x

.

y

...

..

f(x) = tanh x

.

f(x) = coth x

.

−2

.

2

.

−2

.

2

.

x

.

y

...

..

f(x) = sech x

.

f(x) = csch x

.

−

..

−

.

−

.

−

...

.

x

.

y

Figure . . : Graphs of the hyperbolic func ons.

Example . . Exploring proper es of hyperbolic func onsUse Defini on . . to rewrite the following expressions.

. cosh x− sinh x

. tanh x+ sech x

. cosh x sinh x

. ddx

(cosh x

)

. ddx

(sinh x

)

. ddx

(tanh x

)

S

. cosh x− sinh x =(ex + e−x)

−(ex − e−x)

=e x + exe−x + e− x

− e x − exe−x + e− x

= = .

So cosh x− sinh x = .

Notes:

. Hyperbolic Func ons

. tanh x+ sech x =sinh xcosh x

+cosh x

=sinh x+cosh x

Now use iden ty from # .

=cosh xcosh x

= .

So tanh x+ sech x = .

. cosh x sinh x =(ex + e−x)(ex − e−x)

= · ex − e− x

=e x − e− x

= sinh( x).

Thus cosh x sinh x = sinh( x).

.ddx(cosh x

)=

ddx

(ex + e−x)

=ex − e−x

= sinh x.

So ddx

(cosh x

)= sinh x.

.ddx(sinh x

)=

ddx

(ex − e−x)

=ex + e−x

= cosh x.

So ddx

(sinh x

)= cosh x.

.ddx(tanh x

)=

ddx

(sinh xcosh x

)

=cosh x cosh x− sinh x sinh x

cosh x

=cosh x

= sech x.

So ddx

(tanh x

)= sech x.

Notes:

Chapter Techniques of An differen a on

The following Key Idea summarizes many of the important iden es relat-ing to hyperbolic func ons. Each can be verified by referring back to Defini on. . .

Key Idea . . Useful Hyperbolic Func on Proper es

Basic Iden es

. cosh x− sinh x =

. tanh x+ sech x =

. coth x− csch x =

. cosh x = cosh x+ sinh x

. sinh x = sinh x cosh x

. cosh x =cosh x+

. sinh x =cosh x−

Deriva ves

. ddx

(cosh x

)= sinh x

. ddx

(sinh x

)= cosh x

. ddx

(tanh x

)= sech x

. ddx

(sech x

)= − sech x tanh x

. ddx

(csch x

)= − csch x coth x

. ddx

(coth x

)= − csch x

Integrals

.∫

cosh x dx = sinh x+ C

.∫

sinh x dx = cosh x+ C

.∫

tanh x dx = ln(cosh x) + C

.∫

coth x dx = ln | sinh x |+ C

We prac ce using Key Idea . . .

Example . . Deriva ves and integrals of hyperbolic func onsEvaluate the following deriva ves and integrals.

.ddx(cosh x

)

.∫

sech ( t− ) dt

.∫ ln

cosh x dx

S

. Using the Chain Rule directly, we have ddx

(cosh x

)= sinh x.

Just to demonstrate that it works, let’s also use the Basic Iden ty found inKey Idea . . : cosh x = cosh x+ sinh x.

ddx(cosh x

)=

ddx(cosh x+ sinh x

)= cosh x sinh x+ sinh x cosh x

= cosh x sinh x.

Notes:

. Hyperbolic Func ons

Using another Basic Iden ty, we can see that cosh x sinh x = sinh x.We get the same answer either way.

. We employ subs tu on, with u = t − and du = dt. Applying KeyIdeas . . and . . we have:

∫

sech ( t− ) dt = tanh( t− ) + C.

.∫ ln

cosh x dx = sinh x∣∣∣

ln= sinh(ln )− sinh = sinh(ln ).

We can simplify this last expression as sinh x is based on exponen als:

sinh(ln ) =eln − e− ln

=− /

= .

Inverse Hyperbolic Func ons

Just as the inverse trigonometric func ons are useful in certain integra ons,the inverse hyperbolic func ons are useful with others. Figure . . shows therestric ons on the domains to make each func on one-to-one and the resul ngdomains and ranges of their inverse func ons. Their graphs are shown in Figure. . .

Because the hyperbolic func ons are defined in terms of exponen al func-ons, their inverses can be expressed in terms of logarithms as shown in Key Idea. . . It is o en more convenient to refer to sinh− x than to ln

(x+

√x +

),

especially when one is working on theory and does not need to compute actualvalues. On the other hand, when computa ons are needed, technology is o enhelpful but many hand-held calculators lack a convenient sinh− x bu on. (Of-ten it can be accessed under a menu system, but not conveniently.) In such asitua on, the logarithmic representa on is useful. The reader is not encouragedtomemorize these, but rather know they exist and know how to use themwhenneeded.

Notes:

Chapter Techniques of An differen a on

Func on Domain Rangecosh x [ ,∞) [ ,∞)sinh x (−∞,∞) (−∞,∞)tanh x (−∞,∞) (− , )sech x [ ,∞) ( , ]csch x (−∞, ) ∪ ( ,∞) (−∞, ) ∪ ( ,∞)coth x (−∞, ) ∪ ( ,∞) (−∞,− ) ∪ ( ,∞)

Func on Domain Rangecosh− x [ ,∞) [ ,∞)sinh− x (−∞,∞) (−∞,∞)tanh− x (− , ) (−∞,∞)sech− x ( , ] [ ,∞)csch− x (−∞, ) ∪ ( ,∞) (−∞, ) ∪ ( ,∞)coth− x (−∞,− ) ∪ ( ,∞) (−∞, ) ∪ ( ,∞)

Figure . . : Domains and ranges of the hyperbolic and inverse hyperbolic func ons.

.....

y = cosh−1 x

.

y = cosh x

. 5. 10.

5

.

10

.x

.

y

.....

y = sinh x

.

y = sinh−1 x

.

−10

.

−5

.

5

.

10

. −10.

−5

.

5

.

10

.

x

.

y

...

..

y = coth−1 x

.

y = tanh−1 x

.

−2

.

2

.

−2

.

2

.

x

.

y

.....

y = sech− x

.

y = csch− x

.

−

.. −.

−

.

−

....

x

.

y

Figure . . : Graphs of the hyperbolic func ons and their inverses.

Key Idea . . Logarithmic defini ons of Inverse Hyperbolic Func ons

. cosh− x = ln(x+

√

x −); x ≥

. tanh− x = ln(

+ x− x

)

; |x| <

. sech− x = ln

(

+√

− xx

)

; < x ≤

. sinh− x = ln(x+

√

x +)

. coth− x = ln(x+x−

)

; |x| >

. csch− x = ln

(

x+

√+ x|x|

)

; x ̸=

Notes:

. Hyperbolic Func ons

The following Key Ideas give the deriva ves and integrals rela ng to the in-verse hyperbolic func ons. In Key Idea . . , both the inverse hyperbolic andlogarithmic func on representa ons of the an deriva ve are given, based onKey Idea . . . Again, these la er func ons are o en more useful than the for-mer. Note how inverse hyperbolic func ons can be used to solve integrals weused Trigonometric Subs tu on to solve in Sec on . .

Key Idea . . Deriva ves Involving Inverse Hyperbolic Func ons

.ddx(cosh− x

)= √

x −; x >

.ddx(sinh− x

)= √

x +

.ddx(tanh− x

)= − x

; |x| <

.ddx(sech− x

)=

−x√

− x; < x <

.ddx(csch− x

)=

−|x|√

+ x; x ̸=

.ddx(coth− x

)= − x

; |x| >

Key Idea . . Integrals Involving Inverse Hyperbolic Func ons

.∫

√x − a

dx = cosh−( xa

)

+ C; < a < x = ln∣∣∣x+

√

x − a∣∣∣+ C

.∫

√x + a

dx = sinh−( xa

)

+ C; a > = ln∣∣∣x+

√

x + a∣∣∣+ C

.∫

a − xdx =

a tanh− ( x

a

)+ C x < a

a coth− ( x

a

)+ C a < x

=aln∣∣∣∣

a+ xa− x

∣∣∣∣+ C

.∫

x√a − x

dx = −asech−

( xa

)

+ C; < x < a =aln(

xa+

√a − x

)

+ C

.∫

x√x + a

dx = −acsch−

∣∣∣xa

∣∣∣+ C; x ̸= , a > =

aln∣∣∣∣

xa+

√a + x

∣∣∣∣+ C

We prac ce using the deriva ve and integral formulas in the following ex-ample.

Notes:

Chapter Techniques of An differen a on

Example . . Deriva ves and integrals involving inverse hyperbolicfunc ons

Evaluate the following.

.ddx

[

cosh−(

x− )]

.∫

x − dx

.∫

√x +

dx

S

. Applying Key Idea . . with the Chain Rule gives:

ddx

[

cosh−(

x− )]

= √( x− )

−· .

. Mul plying the numerator anddenominator by (− ) gives:∫

x − dx =∫ −

− xdx. The second integral can be solved with a direct applica on

of item # from Key Idea . . , with a = . Thus∫

x − dx = −∫

− xdx

=

− tanh− (x) + C x <

− coth− (x) + C < x

= − ln∣∣∣∣

x+x−

∣∣∣∣+ C

= ln∣∣∣∣

x−x+

∣∣∣∣+ C. ( . )

We should note that this exact problem was solved at the beginning ofSec on . . In that example the answer was given as ln |x− |− ln |x+|+ C. Note that this is equivalent to the answer given in Equa on . , as

ln(a/b) = ln a− ln b.

. This requires a subs tu on, then item # of Key Idea . . can be applied.Let u = x, hence du = dx. We have

∫

√x +

dx =∫

√u +

du.

Notes:

. Hyperbolic Func ons

Note a = , hence a =√

. Now apply the integral rule.

= sinh−(

x√)

+ C

= ln∣∣∣ x+

√

x +∣∣∣+ C.

This sec on covers a lot of ground. New func ons were introduced, alongwith some of their fundamental iden es, their deriva ves and an deriva ves,their inverses, and the deriva ves and an deriva ves of these inverses. FourKey Ideas were presented, each including quite a bit of informa on.

Do not view this sec on as containing a source of informa on to be memo-rized, but rather as a reference for future problem solving. Key Idea . . con-tains perhaps the most useful informa on. Know the integra on forms it helpsevaluate and understand how to use the inverse hyperbolic answer and the log-arithmic answer.

The next sec on takes a brief break from demonstra ng new integra ontechniques. It instead demonstrates a technique of evalua ng limits that re-turn indeterminate forms. This technique will be useful in Sec on . , wherelimits will arise in the evalua on of certain definite integrals.

Notes:

Exercises .Terms and Concepts

. In Key Idea . . , the equa on∫

tanh x dx = ln(cosh x)+C

is given. Why is “ln | cosh x|” not used – i.e., why are abso-lute values not necessary?

. The hyperbolic func ons are used to define points on theright hand por on of the hyperbola x − y = , as shownin Figure . . . How can we use the hyperbolic func ons todefine points on the le hand por on of the hyperbola?

ProblemsIn Exercises – , verify the given iden ty using Defini on. . , as done in Example . . .

. coth x− csch x =

. cosh x = cosh x+ sinh x

. cosh x = cosh x+

. sinh x = cosh x−

. ddx

[sech x] = − sech x tanh x

. ddx

[coth x] = − csch x

.∫

tanh x dx = ln(cosh x) + C

.∫

coth x dx = ln | sinh x|+ C

In Exercises – , find the deriva ve of the given func on.

. f(x) = sinh x

. f(x) = cosh x

. f(x) = tanh(x )

. f(x) = ln(sinh x)

. f(x) = sinh x cosh x

. f(x) = x sinh x− cosh x

. f(x) = sech− (x )

. f(x) = sinh− ( x)

. f(x) = cosh− ( x )

. f(x) = tanh− (x+ )

. f(x) = tanh− (cos x)

. f(x) = cosh− (sec x)

In Exercises – , find the equa on of the line tangent tothe func on at the given x-value.

. f(x) = sinh x at x =

. f(x) = cosh x at x = ln

. f(x) = tanh x at x = − ln

. f(x) = sech x at x = ln

. f(x) = sinh− x at x =

. f(x) = cosh− x at x =√

In Exercises – , evaluate the given indefinite integral.

.∫

tanh( x) dx

.∫

cosh( x− ) dx

.∫

sinh x cosh x dx

.∫

x cosh x dx

.∫

x sinh x dx

.∫

√x +

dx

.∫

√x −

dx

.∫

− xdx

.∫

x√x −

dx

.∫ √

x√+ x

dx

.∫

x − dx

.∫

x + xdx

.∫

ex

e x +dx

.∫

sinh− x dx

.∫

tanh− x dx

.∫

sech x dx (Hint: mul ply by cosh xcosh x ; set u = sinh x.)

In Exercises – , evaluate the given definite integral.

.∫

−sinh x dx

.∫ ln

− lncosh x dx

.∫

tanh− x dx

.∫

√x +

dx

Chapter Techniques of An differen a on

. L’Hôpital’s RuleWhile this chapter is devoted to learning techniques of integra on, this sec onis not about integra on. Rather, it is concerned with a technique of evalua ngcertain limits that will be useful in the following sec on, where integra on isonce more discussed.

Our treatment of limits exposed us to the no on of “ / ”, an indeterminateform. If lim

x→cf(x) = and lim

x→cg(x) = , we do not conclude that lim

x→cf(x)/g(x) is

/ ; rather, we use / as nota on to describe the fact that both the numeratorand denominator approach . The expression / has no numeric value; otherwork must be done to evaluate the limit.

Other indeterminate forms exist; they are: ∞/∞, ·∞,∞−∞, , ∞ and∞ . Just as “ / ” does not mean “divide by ,” the expression “∞/∞” doesnot mean “divide infinity by infinity.” Instead, it means “a quan ty is growingwithout bound and is being divided by another quan ty that is growing withoutbound.” We cannot determine from such a statement what value, if any, resultsin the limit. Likewise, “ ·∞” does not mean “mul ply zero by infinity.” Instead,it means “one quan ty is shrinking to zero, and is being mul plied by a quan tythat is growing without bound.” We cannot determine from such a descrip onwhat the result of such a limit will be.

This sec on introduces l’Hôpital’s Rule, amethod of resolving limits that pro-duce the indeterminate forms / and ∞/∞. We’ll also show how algebraicmanipula on can be used to convert other indeterminate expressions into oneof these two forms so that our new rule can be applied.

Theorem . . L’Hôpital’s Rule, Part

Let limx→c

f(x) = and limx→c

g(x) = , where f and g are differen able func-ons on an open interval I containing c, and g ′(x) ̸= on I except possi-

bly at c. Then

limx→c

f(x)g(x)

= limx→c

f ′(x)g ′(x)

.

We demonstrate the use of l’Hôpital’s Rule in the following examples; wewill o en use “LHR” as an abbrevia on of “l’Hôpital’s Rule.”

Notes:

. L’Hôpital’s Rule

Example . . Using l’Hôpital’s RuleEvaluate the following limits, using l’Hôpital’s Rule as needed.

. limx→

sin xx

. limx→

√x+ −

− x

. limx→

x− cos x

. limx→

x + x−x − x+

S

. We proved this limit is in Example . . using the Squeeze Theorem.Here we use l’Hôpital’s Rule to show its power.

limx→

sin xx

by LHR= lim

x→cos x

= .

. limx→

√x+ −

− x

by LHR= lim

x→

(x+ )− /

− = − .

. limx→

x− cos x

by LHR= lim

x→x

sin x.

This la er limit also evaluates to the / indeterminate form. To evaluateit, we apply l’Hôpital’s Rule again.

limx→

xsin x

by LHR=

cos x= .

Thus limx→

x− cos x

= .

. We already know how to evaluate this limit; first factor the numerator anddenominator. We then have:

limx→

x + x−x − x+

= limx→

(x− )(x+ )

(x− )(x− )= lim

x→x+x− = .

We now show how to solve this using l’Hôpital’s Rule.

limx→

x + x−x − x+

by LHR= lim

x→x+x− = .

Note that at each stepwhere l’Hôpital’s Rule was applied, it was needed: theini al limit returned the indeterminate form of “ / .” If the ini al limit returns,for example, / , then l’Hôpital’s Rule does not apply.

Notes:

Chapter Techniques of An differen a on

The following theorem extends our ini al version of l’Hôpital’s Rule in twoways. It allows the technique to be applied to the indeterminate form ∞/∞and to limits where x approaches±∞.

Theorem . . L’Hôpital’s Rule, Part

. Let limx→a

f(x) = ±∞ and limx→a

g(x) = ±∞, where f and g are differ-en able on an open interval I containing a. Then

limx→a

f(x)g(x)

= limx→a

f ′(x)g ′(x)

.

. Let f and g be differen able func ons on the open interval (a,∞)for some value a, where g ′(x) ̸= on (a,∞) and lim

x→∞f(x)/g(x)

returns either “ / ” or “∞/∞”. Then

limx→∞

f(x)g(x)

= limx→∞

f ′(x)g ′(x)

.

A similar statement can be made for limits where x approaches−∞.

Example . . Using l’Hôpital’s Rule with limits involving∞Evaluate the following limits.

. limx→∞

x − x+x + x− . lim

x→∞ex

x.

S

. We can evaluate this limit already using Theorem . . ; the answer is / .We apply l’Hôpital’s Rule to demonstrate its applicability.

limx→∞

x − x+x + x−

by LHR= lim

x→∞x−x+

by LHR= lim

x→∞= .

. limx→∞

ex

x

by LHR= lim

x→∞ex

x

by LHR= lim

x→∞ex

x

by LHR= lim

x→∞ex

= ∞.

Recall that this means that the limit does not exist; as x approaches ∞,the expression ex/x grows without bound. We can infer from this thatex grows “faster” than x ; as x gets large, ex is far larger than x . (This

Notes:

. L’Hôpital’s Rule

has important implica ons in compu ng when considering efficiency ofalgorithms.)

Indeterminate Forms · ∞ and∞−∞

L’Hôpital’s Rule can only be applied to ra os of func ons. When faced withan indeterminate form such as ·∞ or∞−∞, we can some mes apply algebrato rewrite the limit so that l’Hôpital’s Rule can be applied. We demonstrate thegeneral idea in the next example.

Example . . Applying l’Hôpital’s Rule to other indeterminate formsEvaluate the following limits.

. limx→ +

x · e /x

. limx→ −

x · e /x

. limx→∞

ln(x+ )− ln x

. limx→∞

x − ex

S

. As x → +, x → and e /x → ∞. Thus we have the indeterminate form

· ∞. We rewrite the expression x · e /x ase /x

/x; now, as x → +, we get

the indeterminate form∞/∞ to which l’Hôpital’s Rule can be applied.

limx→ +

x · e /x = limx→ +

e /x

/x

by LHR= lim

x→ +

(− /x )e /x

− /x= lim

x→ +e /x = ∞.

Interpreta on: e /x grows “faster” than x shrinks to zero, meaning theirproduct grows without bound.

. As x → −, x → and e /x → e−∞ → . The the limit evaluates to ·which is not an indeterminate form. We conclude then that

limx→ −

x · e /x = .

. This limit ini ally evaluates to the indeterminate form∞−∞. By applyinga logarithmic rule, we can rewrite the limit as

limx→∞

ln(x+ )− ln x = limx→∞

ln(x+x

)

.

As x → ∞, the argument of the ln term approaches ∞/∞, to which wecan apply l’Hôpital’s Rule.

limx→∞

x+x

by LHR= = .

Notes:

Chapter Techniques of An differen a on

Since x → ∞ impliesx+x

→ , it follows that

x → ∞ implies ln(x+x

)

→ ln = .

Thuslimx→∞

ln(x+ )− ln x = limx→∞

ln(x+x

)

= .

Interpreta on: since this limit evaluates to , it means that for large x,there is essen ally no difference between ln(x + ) and ln x; their differ-ence is essen ally .

. The limit limx→∞

x −ex ini ally returns the indeterminate form∞−∞. We

can rewrite the expression by factoring out x ; x − ex = x(

− ex

x

)

.

We need to evaluate how ex/x behaves as x → ∞:

limx→∞

ex

x

by LHR= lim

x→∞ex

x

by LHR= lim

x→∞ex

= ∞.

Thus limx→∞ x ( − ex/x ) evaluates to∞ · (−∞), which is not an inde-terminate form; rather, ∞ · (−∞) evaluates to −∞. We conclude thatlimx→∞

x − ex = −∞.

Interpreta on: as x gets large, the difference between x and ex growsvery large.

Indeterminate Forms , ∞ and∞

When faced with an indeterminate form that involves a power, it o en helpsto employ the natural logarithmic func on. The following Key Idea expresses theconcept, which is followed by an example that demonstrates its use.

Key Idea . . Evalua ng Limits Involving Indeterminate Forms, ∞ and∞

If limx→c

ln(f(x))= L, then lim

x→cf(x) = lim

x→celn(f(x)) = e L.

Notes:

. L’Hôpital’s Rule

Example . . Using l’Hôpital’s Rule with indeterminate forms involvingexponents

Evaluate the following limits.

. limx→∞

(

+x

)x

. limx→ +

xx.

S

. This is equivalent to a special limit given in Theorem . . ; these limitshave important applica ons within mathema cs and finance. Note thatthe exponent approaches∞ while the base approaches , leading to theindeterminate form ∞. Let f(x) = ( + /x)x; the problem asks to eval-uate lim

x→∞f(x). Let’s first evaluate lim

x→∞ln(f(x)).

limx→∞

ln(f(x))= lim

x→∞ln(

+x

)x

= limx→∞

x ln(

+x

)

= limx→∞

ln(

+ x

)

/x

This produces the indeterminate form / , so we apply l’Hôpital’s Rule.

= limx→∞

+ /x · (− /x )

(− /x )

= limx→∞ + /x

= .

Thus limx→∞

ln(f(x))= .We return to the original limit and apply Key Idea

. . .

limx→∞

(

+x

)x

= limx→∞

f(x) = limx→∞

eln(f(x)) = e = e.

. This limit leads to the indeterminate form . Let f(x) = xx and consider

Notes:

.....

f(x) = xx

.......x

.

y

Figure . . : A graph of f(x) = xx sup-por ng the fact that as x → +, f(x) → .

Chapter Techniques of An differen a on

first limx→ +

ln(f(x)).

limx→ +

ln(f(x))= lim

x→ +ln (xx)

= limx→ +

x ln x

= limx→ +

ln x/x

.

This produces the indeterminate form−∞/∞ soweapply l’Hôpital’s Rule.

= limx→ +

/x− /x

= limx→ +

−x

= .

Thus limx→ +

ln(f(x))= . We return to the original limit and apply Key Idea

. . .lim

x→ +xx = lim

x→ +f(x) = lim

x→ +eln(f(x)) = e = .

This result is supported by the graph of f(x) = xx given in Figure . . .

Our brief revisit of limits will be rewarded in the next sec on where we con-sider improper integra on. So far, we have only considered definite integrals

where the bounds are finite numbers, such as∫

f(x) dx. Improper integra on

considers integrals where one, or both, of the bounds are “infinity.” Such inte-grals have many uses and applica ons, in addi on to genera ng ideas that areenlightening.

Notes:

Exercises .Terms and Concepts. List the different indeterminate forms described in this sec-

on.

. T/F: l’Hôpital’s Rule provides a faster method of compu ngderiva ves.

. T/F: l’Hôpital’s Rule states that ddx

[

f(x)g(x)

]

=f ′(x)g′(x)

.

. Explain what the indeterminate form “ ∞” means.

. Fill in the blanks:

The Quo ent Rule is applied to f(x)g(x)

when taking;

l’Hôpital’s Rule is applied to f(x)g(x)

when taking certain.

. Create (but do not evaluate!) a limit that returns “∞ ”.

. Create a func on f(x) such that limx→

f(x) returns “ ”.

. Create a func on f(x) such that limx→∞

f(x) returns “ · ∞”.

ProblemsIn Exercises – , evaluate the given limit.

. limx→

x + x−x−

. limx→

x + x−x − x+

. limx→π

sin xx− π

. limx→π/

sin x− cos xcos( x)

. limx→

sin( x)x

. limx→

sin( x)x+

. limx→

sin( x)sin( x)

. limx→

sin(ax)sin(bx)

. limx→ +

ex −x

. limx→ +

ex − x−x

. limx→ +

x− sin xx − x

. limx→∞

xex

. limx→∞

√x

ex

. limx→∞ x

ex

. limx→∞

ex√x

. limx→∞

exx

. limx→∞

exx

. limx→

x − x + x+x − x + x−

. limx→−

x + x + xx + x + x+

. limx→∞

ln xx

. limx→∞

ln(x )

x

. limx→∞

(

ln x)

x

. limx→ +

x · ln x

. limx→ +

√x · ln x

. limx→ +

xe /x

. limx→∞

x − x

. limx→∞

√x− ln x

. limx→−∞

xex

. limx→ + x

e− /x

. limx→ +

( + x) /x

. limx→ +

( x)x

. limx→ +

( /x)x

. limx→ +

(sin x)x Hint: use the Squeeze Theorem.

. limx→ +

( − x) −x

. limx→∞

(x) /x

. limx→∞

( /x)x

. limx→ +

(ln x) −x

. limx→∞

( + x) /x

. limx→∞

( + x ) /x

. limx→π/

tan x cos x

. limx→π/

tan x sin( x)

. limx→ + ln x

−x−

. limx→ + x − − x

x−

. limx→∞

x tan( /x)

. limx→∞

(ln x)x

. limx→

x + x−ln x

..... 5. 10.

0.5

.

1

.x

.

y

Figure . . : Graphing f(x) =+ x

.

. Improper Integra on

. Improper Integra onWe begin this sec on by considering the following definite integrals:

•∫

+ xdx ≈ . ,

•∫

+ xdx ≈ . ,

•∫ ,

+ xdx ≈ . .

No ce how the integrand is /( + x ) in each integral (which is sketched inFigure . . ). As the upper bound gets larger, one would expect the “area underthe curve” would also grow. While the definite integrals do increase in value asthe upper bound grows, they are not increasing by much. In fact, consider:

∫ b

+ xdx = tan− x

∣∣∣

b= tan− b− tan− = tan− b.

As b → ∞, tan− b → π/ . Therefore it seems that as the upper bound b grows,

the value of the definite integral∫ b

+ xdx approaches π/ ≈ . . This

should strike the reader as being a bit amazing: even though the curve extends“to infinity,” it has a finite amount of area underneath it.

Whenwe defined the definite integral∫ b

af(x) dx, wemade two s pula ons:

. The interval over which we integrated, [a, b], was a finite interval, and

. The func on f(x) was con nuous on [a, b] (ensuring that the range of fwas finite).

In this sec on we consider integrals where one or both of the above condi-ons do not hold. Such integrals are called improper integrals.

Notes:

.....

f(x) =x

.. 5..

.5

..x

.

y

Figure . . : A graph of f(x) = x in Ex-ample . . .

Chapter Techniques of An differen a on

Improper Integrals with Infinite Bounds

Defini on . . Improper Integrals with Infinite Bounds; Converge,Diverge

. Let f be a con nuous func on on [a,∞). Define∫ ∞

af(x) dx to be lim

b→∞

∫ b

af(x) dx.

. Let f be a con nuous func on on (−∞, b]. Define∫ b

−∞f(x) dx to be lim

a→−∞

∫ b

af(x) dx.

. Let f be a con nuous func on on (−∞,∞). Let c be any real num-ber; define∫ ∞

−∞f(x) dx to be lim

a→−∞

∫ c

af(x) dx + lim

b→∞

∫ b

cf(x) dx.

An improper integral is said to converge if its corresponding limit exists;otherwise, it diverges. The improper integral in part converges if andonly if both of its limits exist.

Example . . Evalua ng improper integralsEvaluate the following improper integrals.

.∫ ∞

xdx

.∫ ∞

xdx

.∫

−∞ex dx

.∫ ∞

−∞ + xdx

S

.∫ ∞

xdx = lim

b→∞

∫ b

xdx = lim

b→∞

−x

∣∣∣

b

= limb→∞

−b

+

= .

A graph of the area defined by this integral is given in Figure . . .

Notes:

.....

f(x) =1x

. 1. 5. 10.

0.5

.

1

.x

.

y

Figure . . : A graph of f(x) = x in Exam-ple . . .

.....

f(x) = ex

.−1.

−5.−10.

1

.x

.

y

Figure . . : A graph of f(x) = ex in Ex-ample . . .

.....

f(x) =+ x

.−.

−5. 5...x

.

y

Figure . . : A graph of f(x) =+x in

Example . . .

. Improper Integra on

.∫ ∞

xdx = lim

b→∞

∫ b

xdx

= limb→∞

ln |x|∣∣∣

b

= limb→∞

ln(b)

= ∞.

The limit does not exist, hence the improper integral∫ ∞

xdx diverges.

Compare the graphs in Figures . . and . . ; no ce how the graph off(x) = /x is no ceably larger. This difference is enough to cause theimproper integral to diverge.

.∫

−∞ex dx = lim

a→−∞

∫

aex dx

= lima→−∞

ex∣∣∣a

= lima→−∞

e − ea

= .

A graph of the area defined by this integral is given in Figure . . .

. We will need to break this into two improper integrals and choose a valueof c as in part of Defini on . . . Any value of c is fine; we choose c = .

∫ ∞

−∞ + xdx = lim

a→−∞

∫

a + xdx+ lim

b→∞

∫ b

+ xdx

= lima→−∞

tan− x∣∣∣a+ lim

b→∞tan− x

∣∣∣

b

= lima→−∞

(tan− − tan− a

)+ lim

b→∞

(tan− b− tan−

)

=

(

− −π)

+(π −

)

.

Each limit exists, hence the original integral converges and has value:

= π.

A graph of the area defined by this integral is given in Figure . . .

Notes:

.....

f(x) =ln xx

..5

..

.

.

.4

.

x

.

y

Figure . . : A graph of f(x) = ln xx in Ex-

ample . . .

Chapter Techniques of An differen a on

The previous sec on introduced l’Hôpital’s Rule, a method of evalua ng lim-its that return indeterminate forms. It is not uncommon for the limits resul ngfrom improper integrals to need this rule as demonstrated next.

Example . . Improper integra on and l’Hôpital’s RuleEvaluate the improper integral

∫ ∞ ln xx

dx.

S This integral will require the use of Integra on by Parts. Letu = ln x and dv = /x dx. Then

∫ ∞ ln xx

dx = limb→∞

∫ b ln xx

dx

= limb→∞

(

− ln xx

∣∣∣

b+

∫ b

xdx

)

= limb→∞

(

− ln xx

−x

)∣∣∣∣

b

= limb→∞

(

− ln bb

−b− (− ln − )

)

.

The /b and ln terms go to , leaving limb→∞

− ln bb

+ . We need to evaluate

limb→∞

ln bb

with l’Hôpital’s Rule. We have:

limb→∞

ln bb

by LHR= lim

b→∞

/b

= .

Thus the improper integral evaluates as:

∫ ∞ ln xx

dx = .

Improper Integrals with Infinite Range

We have just considered definite integrals where the interval of integra onwas infinite. We now consider another type of improper integra on, where therange of the integrand is infinite.

Notes:

Note: In Defini on . . , c can be one ofthe endpoints (a or b). In that case, thereis only one limit to consider as part of thedefini on.

.....

f(x) =1√

x

.0.5

.1

.

5

.

10

. x.

y

Figure . . : A graph of f(x) = √x in Ex-

ample . . .

.....

f(x) =x

.−

.− .5

..5

..

5

.. x.

y

Figure . . : A graph of f(x) = x in Ex-ample . . .

. Improper Integra on

Defini on . . Improper Integra on with Infinite Range

Let f(x) be a con nuous func on on [a, b] except at c, a ≤ c ≤ b, wherex = c is a ver cal asymptote of f. Define

∫ b

af(x) dx = lim

t→c−

∫ t

af(x) dx+ lim

t→c+

∫ b

tf(x) dx.

Example . . Improper integra on of func ons with infinite rangeEvaluate the following improper integrals:

.

∫

√xdx .

∫

− xdx.

S

. A graph of f(x) = /√x is given in Figure . . . No ce that f has a ver cal

asymptote at x = ; in some sense, we are trying to compute the area ofa region that has no “top.” Could this have a finite value?

∫

√xdx = lim

a→ +

∫

a√xdx

= lima→ +

√x∣∣∣a

= lima→ +

(√−√a)

= .

It turns out that the region does have a finite area even though it has noupper bound (strange things can occur in mathema cs when consideringthe infinite).

. The func on f(x) = /x has a ver cal asymptote at x = , as shownin Figure . . , so this integral is an improper integral. Let’s eschew usinglimits for amoment and proceedwithout recognizing the improper natureof the integral. This leads to:

∫

− xdx = −

x

∣∣∣−

= − − ( )

= − . (!)

Notes:

.....

f(x) =1x q

.f(x) =

1x p

.

p < 1 < q

.1

. x.

y

Figure . . : Plo ng func ons of theform /x p in Example . . .

Chapter Techniques of An differen a on

Clearly the area in ques on is above the x-axis, yet the area is supposedlynega ve! Why does our answer not match our intui on? To answer this,evaluate the integral using Defini on . . .

∫

− xdx = lim

t→ −

∫ t

− xdx+ lim

t→ +

∫

t xdx

= limt→ −

−x

∣∣∣

t

−+ lim

t→ +−x

∣∣∣t

= limt→ −

−t− + lim

t→ +− +

t

⇒(

∞−)

+(

− +∞)

.

Neither limit converges hence the original improper integral diverges. Thenonsensical answer we obtained by ignoring the improper nature of theintegral is just that: nonsensical.

Understanding Convergence and Divergence

O en mes we are interested in knowing simply whether or not an improperintegral converges, and not necessarily the value of a convergent integral. Weprovide here several tools that help determine the convergence or divergenceof improper integrals without integra ng.

Our first tool is to understand the behavior of func ons of the formxp

.

Example . . Improper integra on of /xp

Determine the values of p for which∫ ∞

xpdx converges.

S We begin by integra ng and then evalua ng the limit.∫ ∞

xpdx = lim

b→∞

∫ b

xpdx

= limb→∞

∫ bx−p dx (assume p ̸= )

= limb→∞ −p+

x−p+∣∣∣

b

= limb→∞ − p

(b −p − −p).

When does this limit converge – i.e., when is this limit not ∞? This limit con-verges precisely when the power of b is less than : when − p < ⇒ < p.

Notes:

Note: We used the upper and lowerbound of “ ” in Key Idea . . for conve-nience. It can be replaced by any awherea > .

. Improper Integra on

Our analysis shows that if p > , then∫ ∞

xpdx converges. When p <

the improper integral diverges; we showed in Example . . that when p =the integral also diverges.

Figure . . graphs y = /x with a dashed line, along with graphs of y =/xp, p < , and y = /xq, q > . Somehow the dashed line forms a dividing

line between convergence and divergence.

The result of Example . . provides an important tool in determining theconvergence of other integrals. A similar result is proved in the exercises about

improper integrals of the form∫

xpdx. These results are summarized in the

following Key Idea.

Key Idea . . Convergence of Improper Integrals∫ ∞

xpdx and

∫

xpdx.

. The improper integral∫ ∞

xpdx converges when p > and diverges when p ≤ .

. The improper integral∫

xpdx converges when p < and diverges when p ≥ .

A basic technique in determining convergence of improper integrals is tocompare an integrand whose convergence is unknown to an integrand whoseconvergence is known. We o en use integrands of the form /xp to compareto as their convergence on certain intervals is known. This is described in thefollowing theorem.

Theorem . . Direct Comparison Test for Improper Integrals

Let f and g be con nuous on [a,∞) where ≤ f(x) ≤ g(x) for all x in[a,∞).

. If∫ ∞

ag(x) dx converges, then

∫ ∞

af(x) dx converges.

. If∫ ∞

af(x) dx diverges, then

∫ ∞

ag(x) dx diverges.

Notes:

.....

f(x) = e−x

.

f(x) =x

.....

.

..x

.

y

Figure . . : Graphs of f(x) = e−x andf(x) = /x in Example . . .

.....

f(x) = √

x − x

.

f(x) =x

..4

.6

.

.

.

.4

.

x

.

y

Figure . . : Graphs of f(x) =/√x − x and f(x) = /x in Example

. . .

Chapter Techniques of An differen a on

Example . . Determining convergence of improper integralsDetermine the convergence of the following improper integrals.

.∫ ∞

e−x dx .∫ ∞

√x − x

dx

S

. The func on f(x) = e−x does not have an an deriva ve expressible interms of elementary func ons, so we cannot integrate directly. It is com-parable to g(x) = /x , and as demonstrated in Figure . . , e−x <

/x on [ ,∞). We know from Key Idea . . that∫ ∞

xdx converges,

hence∫ ∞

e−x dx also converges.

. Note that for large values of x, √x − x

≈ √x

=x. We know from Key

Idea . . and the subsequent note that∫ ∞

xdx diverges, so we seek

to compare the original integrand to /x.

It is easy to see that when x > , we have x =√x >

√x − x. Taking

reciprocals reverses the inequality, giving

x< √

x − x.

Using Theorem . . , we conclude that since∫ ∞

xdxdiverges,

∫ ∞√x − x

dx

diverges as well. Figure . . illustrates this.

Being able to compare “unknown” integrals to “known” integrals is very use-ful in determining convergence. However, some of our examples were a li le“too nice.” For instance, it was convenient that

x< √

x − x, but what if the

“−x” were replaced with a “+ x+ ”? That is, what can we say about the con-

vergence of∫ ∞

√x + x+

dx? We havex> √

x + x+, so we cannot

use Theorem . . .In cases like this (and many more) it is useful to employ the following theo-

rem.

Notes:

.....

f(x) = √

x + x + 5

.

f(x) =x

.5

..5

..

.

.

x

.

y

Figure . . : Graphing f(x) = √x + x+

and f(x) = x in Example . . .

. Improper Integra on

Theorem . . Limit Comparison Test for Improper Integrals

Let f and g be con nuous func ons on [a,∞)where f(x) > and g(x) >for all x. If

limx→∞

f(x)g(x)

= L, < L < ∞,

then ∫ ∞

af(x) dx and

∫ ∞

ag(x) dx

either both converge or both diverge.

Example . . Determining convergence of improper integralsDetermine the convergence of

∫ ∞√x + x+

dx.

S As x gets large, the denominator of the integrand will beginto behave much like y = x. So we compare √

x + x+to

xwith the Limit

Comparison Test:

limx→∞

/√x + x+

/x= lim

x→∞x√

x + x+.

The immediate evalua onof this limit returns∞/∞, an indeterminate form.Using l’Hôpital’s Rule seems appropriate, but in this situa on, it does not leadto useful results. (We encourage the reader to employ l’Hôpital’s Rule at leastonce to verify this.)

The trouble is the square root func on. To get rid of it, we employ the fol-lowing fact: If lim

x→cf(x) = L, then lim

x→cf(x) = L . (This is true when either c or L

is∞.) So we consider now the limit

limx→∞

xx + x+

.

This converges to , meaning the original limit also converged to . As x getsvery large, the func on √

x + x+looks verymuch like

x. Sincewe know that

∫ ∞

xdxdiverges, by the Limit Comparison Testwe know that

∫ ∞

√x + x+

dx

also diverges. Figure . . graphs f(x) = /√x + x+ and f(x) = /x, il-

lustra ng that as x gets large, the func ons become indis nguishable.

Notes:

Chapter Techniques of An differen a on

Both the Direct and Limit Comparison Tests were given in terms of integralsover an infinite interval. There are versions that apply to improper integrals withan infinite range, but as they are a bit wordy and a li le more difficult to employ,they are omi ed from this text.

This chapter has explored many integra on techniques. We learned Subs -tu on, which “undoes” the Chain Rule of differen a on, as well as Integra onby Parts, which “undoes” the Product Rule. We learned specialized techniquesfor handling trigonometric func ons and introduced the hyperbolic func ons,which are closely related to the trigonometric func ons. All techniques effec-vely have this goal in common: rewrite the integrand in a new way so that the

integra on step is easier to see and implement.As stated before, integra on is, in general, hard. It is easy to write a func on

whose an deriva ve is impossible to write in terms of elementary func ons,and evenwhen a func on does have an an deriva ve expressible by elementaryfunc ons, it may be really hard to discover what it is. The powerful computeralgebra systemMathema ca® has approximately , pages of code dedicatedto integra on.

Do not let this difficulty discourage you. There is great value in learning in-tegra on techniques, as they allow one to manipulate an integral in ways thatcan illuminate a concept for greater understanding. There is also great valuein understanding the need for good numerical techniques: the Trapezoidal andSimpson’s Rules are just the beginning of powerful techniques for approximat-ing the value of integra on.

The next chapter stresses the uses of integra on. We generally do not findan deriva ves for an deriva ve’s sake, but rather because they provide the so-lu on to some typeof problem. The following chapter introduces us to a numberof different problems whose solu on is provided by integra on.

Notes:

Exercises .Terms and Concepts. The definite integral was defined with what two s pula-

ons?

. If limb→∞

∫ b

f(x) dx exists, then the integral∫ ∞

f(x) dx is

said to .

. If∫ ∞

f(x) dx = , and ≤ g(x) ≤ f(x) for all x, then we

know that∫ ∞

g(x) dx .

. For what values of p will∫ ∞

xpdx converge?

. For what values of p will∫ ∞

xpdx converge?

. For what values of p will∫

xpdx converge?

ProblemsIn Exercises – , evaluate the given improper integral.

.∫ ∞

e − x dx

.∫ ∞

xdx

.∫ ∞

x− dx

.∫ ∞

−∞ x +dx

.∫

−∞

x dx

.∫

−∞

( )x

dx

.∫ ∞

−∞

xx +

dx

.∫ ∞

x − dx

.∫ ∞

(x− )dx

.∫

(x− )dx

.∫ ∞

x− dx

.∫

x− dx

.∫

− xdx

.∫

x− dx

.∫ π

sec x dx

.∫

−

√

|x|dx

.∫ ∞

xe−x dx

.∫ ∞

xe−x dx

.∫ ∞

−∞xe−x dx

.∫ ∞

−∞ ex + e−x dx

.∫

x ln x dx

.∫

x ln x dx

.∫ ∞ ln x

xdx

.∫

ln x dx

.∫ ∞ ln x

xdx

.∫ ∞ ln x√

xdx

.∫ ∞

e−x sin x dx

.∫ ∞

e−x cos x dx

In Exercises – , use the Direct Comparison Test or theLimit Comparison Test to determine whether the given def-inite integral converges or diverges. Clearly state what testis being used and what func on the integrand is being com-pared to.

.∫ ∞

√x + x−

dx

.∫ ∞

√x − x

dx

.∫ ∞ √

x+√x − x + x+

dx

.∫ ∞

e−x ln x dx

.∫ ∞

e−x + x+ dx

.∫ ∞ √

xex

dx

.∫ ∞

x + sin xdx

.∫ ∞ x

x + cos xdx

.∫ ∞

x+ exdx

.∫ ∞

ex − xdx

: AI

We begin this chapter with a reminder of a few key concepts from Chapter .Let f be a con nuous func on on [a, b]which is par oned into n equally spacedsubintervals as

a = x < x < · · · < xn < xn+ = b.

Let ∆x = (b − a)/n denote the length of the subintervals, and let ci be anyx-value in the i th subinterval. Defini on . . states that the sum

n∑

i=

f(ci)∆x

is a Riemann Sum. Riemann Sums are o en used to approximate some quan-ty (area, volume, work, pressure, etc.). The approxima on becomes exact by

taking the limit

limn→∞

n∑

i=

f(ci)∆x.

Theorem . . connects limits of Riemann Sums to definite integrals:

limn→∞

n∑

i=

f(ci)∆x =∫ b

af(x) dx.

Finally, the Fundamental Theorem of Calculus states how definite integrals canbe evaluated using an deriva ves.

This chapter employs the following technique to a variety of applica ons.Suppose the value Q of a quan ty is to be calculated. We first approximate thevalue ofQ using a Riemann Sum, then find the exact value via a definite integral.We spell out this technique in the following Key Idea.

Key Idea . . Applica on of Definite Integrals Strategy

Let a quan ty be given whose value Q is to be computed.

. Divide the quan ty into n smaller “subquan es” of value Qi.

. Iden fy a variable x and func on f(x) such that each subquan tycan be approximated with the product f(ci)∆x, where ∆x repre-sents a small change in x. Thus Qi ≈ f(ci)∆x. A sample approxi-ma on f(ci)∆x of Qi is called a differen al element.

. Recognize that Q =

n∑

i=

Qi ≈n∑

i=

f(ci)∆x, which is a Riemann

Sum.

. Taking the appropriate limit gives Q =

∫ b

af(x) dx

This Key Idea will make more sense a er we have had a chance to use itseveral mes. We begin with Area Between Curves, which we addressed brieflyin Sec on . .

.....

f(x)

.

g(x)

. a. b.x

.

y

(a)

.....

f(x)

.

g(x)

. a. b.x

.

y

(b)

.....

f(x)

.

g(x)

. a. b.x

.

y

(c)

Figure . . : Subdividing a region intover cal slices and approxima ng the ar-eas with rectangles.

Chapter Applica ons of Integra on

. Area Between CurvesWe are o en interested in knowing the area of a region. Forget momentarilythat we addressed this already in Sec on . and approach it instead using thetechnique described in Key Idea . . .

LetQ be the area of a region bounded by con nuous func ons f and g. If webreak the region into many subregions, we have an obvious equa on:

Total Area = sum of the areas of the subregions.The issue to address next is how to systema cally break a region into subre-

gions. A graph will help. Consider Figure . . (a) where a region between twocurves is shaded. While there are many ways to break this into subregions, onepar cularly efficient way is to “slice” it ver cally, as shown in Figure . . (b),into n equally spaced slices.

We now approximate the area of a slice. Again, we have many op ons, butusing a rectangle seems simplest. Picking any x-value ci in the i th slice, we setthe height of the rectangle to be f(ci)− g(ci), the difference of the correspond-ing y-values. The width of the rectangle is a small difference in x-values, whichwe represent with ∆x. Figure . . (c) shows sample points ci chosen in eachsubinterval and appropriate rectangles drawn. (Each of these rectangles rep-resents a differen al element.) Each slice has an area approximately equal to(f(ci)− g(ci)

)∆x; hence, the total area is approximately the Riemann Sum

Q =

n∑

i=

(f(ci)− g(ci)

)∆x.

Taking the limit as n → ∞ gives the exact area as∫ ba

(f(x)− g(x)

)dx.

Theorem . . Area Between Curves(restatement of Theorem . . )

Let f(x) and g(x) be con nuous func ons defined on [a, b] where f(x) ≥g(x) for all x in [a, b]. The area of the region bounded by the curvesy = f(x), y = g(x) and the lines x = a and x = b is

∫ b

a

(f(x)− g(x)

)dx.

Example . . Finding area enclosed by curvesFind the area of the region bounded by f(x) = sin x+ , g(x) = cos( x) − ,x = and x = π, as shown in Figure . . .

Notes:

.....

f(x)

.g(x)

.

5

..−

..

4π

.

x

.

y

Figure . . : Graphing an enclosed regionin Example . . .

...

......

−

.

−

.

.

x

.

y

Figure . . : Graphing a region enclosedby two func ons in Example . . .

.....

y =√

x +

.

y = −(x − ) +

...... x.

y

Figure . . : Graphing a region for Exam-ple . . .

. Area Between Curves

S The graph verifies that the upper boundary of the region isgiven by f and the lower bound is given by g. Therefore the area of the region isthe value of the integral

∫ π (f(x)− g(x)

)dx =

∫ π (

sin x+ −(

cos( x)−))

dx

= − cos x− sin( x) + x∣∣∣

π

= π ≈ . units .

Example . . Finding total area enclosed by curvesFind the total area of the region enclosed by the func ons f(x) = − x+ andg(x) = x − x + x− as shown in Figure . . .

S A quick calcula on shows that f = g at x = , and . One

can proceed thoughtlessly by compu ng∫(f(x) − g(x)

)dx, but this ignores

the fact that on [ , ], g(x) > f(x). (In fact, the thoughtless integra on returns− / , hardly the expected value of an area.) Thus we compute the total area bybreaking the interval [ , ] into two subintervals, [ , ] and [ , ] and using theproper integrand in each.

Total Area =

∫(g(x)− f(x)

)dx+

∫(f(x)− g(x)

)dx

=

∫(x − x + x−

)dx+

∫(− x + x − x+

)dx

= / + /

= / = . units .

The previous example makes note that we are expec ng area to be posi ve.When first learning about the definite integral, we interpreted it as “signed areaunder the curve,” allowing for “nega ve area.” That doesn’t apply here; area isto be posi ve.

The previous example also demonstrates that we o en have to break a givenregion into subregions before applying Theorem . . . The following exampleshows another situa on where this is applicable, along with an alternate viewof applying the Theorem.

Example . . Finding area: integra ng with respect to yFind the area of the region enclosed by the func ons y =

√x + , y = −(x −

) + and y = , as shown in Figure . . .

Notes:

.....

x = (y − )

.

x =√

− y +

...... x.

y

Figure . . : The region used in Example. . with boundaries relabeled as func-ons of y.

Chapter Applica ons of Integra on

S We give two approaches to this problem. In the first ap-proach, we no ce that the region’s “top” is defined by two different curves.On [ , ], the top func on is y =

√x + ; on [ , ], the top func on is y =

−(x− ) + . Thus we compute the area as the sum of two integrals:

Total Area =

∫ ((√x+

)−)

dx+∫ ((

− (x− ) +)−)

dx

= / + /

= / .

The second approach is clever and very useful in certain situa ons. We areused to viewing curves as func ons of x; we input an x-value and a y-value is re-turned. Some curves can also be described as func ons of y: input a y-value andan x-value is returned. We can rewrite the equa ons describing the boundaryby solving for x:

y =√x+ ⇒ x = (y− )

y = −(x− ) + ⇒ x =√

− y+ .

Figure . . shows the region with the boundaries relabeled. A differen alelement, a horizontal rectangle, is also pictured. The width of the rectangle isa small change in y: ∆y. The height of the rectangle is a difference in x-values.The “top” x-value is the largest value, i.e., the rightmost. The “bo om” x-valueis the smaller, i.e., the le most. Therefore the height of the rectangle is

(√

− y+)− (y− ) .

The area is found by integra ng the above func on with respect to y withthe appropriate bounds. We determine these by considering the y-values theregion occupies. It is bounded below by y = , and bounded above by y = .That is, both the “top” and “bo om” func ons exist on the y interval [ , ]. Thus

Total Area =

∫(√

− y+ − (y− ))dy

=(

− ( − y) / + y− (y− ))∣∣∣

= / .

This calculus–based technique of finding area can be useful evenwith shapesthat we normally think of as “easy.” Example . . computes the area of a trian-gle. While the formula “ × base× height” is well known, in arbitrary trianglesit can be nontrivial to compute the height. Calculus makes the problem simple.

Notes:

.....

y = x +

.

y = − x + 7

.

y = − x + 5

....... x.

y

Figure . . : Graphing a triangular regionin Example . . .

. (a)

.....

.

.

.

..

.

.

.

.

..................... x.

y

(b)

Figure . . : (a) A sketch of a lake, and (b)the lake with length measurements.

. Area Between Curves

Example . . Finding the area of a triangleCompute the area of the regions bounded by the linesy = x+ , y = − x+ and y = − x+ , as shown in Figure . . .

S Recognize that there are two “top” func ons to this region,causing us to use two definite integrals.

Total Area =

∫((x+ )− (− x+ )

)dx+

∫((− x+ )− (− x+ )

)dx

= / + /

= / .

We can also approach this by conver ng each func on into a func on of y. Thisalso requires integrals, so there isn’t really any advantage to doing so. We doit here for demonstra on purposes.

The “top” func on is always x = −y while there are two “bo om” func-ons. Being mindful of the proper integra on bounds, we have

Total Area =

∫( − y − ( − y)

)dy+

∫( − y − (y− )

)dy

= / + /

= / .

Of course, the final answer is the same. (It is interes ng to note that the area ofall subregions used is / . This is coincidental.)

Whilewehave focused on producing exact answers, we are also able tomakeapproxima ons using the principle of Theorem . . . The integrand in the theo-rem is a distance (“top minus bo om”); integra ng this distance func on givesan area. By taking discrete measurements of distance, we can approximate anarea using numerical integra on techniques developed in Sec on . . The fol-lowing example demonstrates this.

Example . . Numerically approxima ng areaTo approximate the area of a lake, shown in Figure . . (a), the “length” of thelake is measured at -foot increments as shown in Figure . . (b), where thelengths are given in hundreds of feet. Approximate the area of the lake.

S The measurements of length can be viewed as measuring“top minus bo om” of two func ons. The exact answer is found by integra ng∫

(f(x) − g(x)

)dx, but of course we don’t know the func ons f and g. Our

discrete measurements instead allow us to approximate.

Notes:

Chapter Applica ons of Integra on

We have the following data points:

( , ), ( , . ), ( , . ), ( , . ), ( , . ), ( , . ), ( , ).

We also have that∆x = b−an = , so Simpson’s Rule gives

Area ≈(

· + · . + · . + · . + · . + · . + ·)

= . units .

Since the measurements are in hundreds of feet, units = ( ) =, , giving a total area of , . (Since we are approxima ng, we’d

likely say the area was about , , which is a li le more than acres.)

In the next sec on we apply our applica ons–of–integra on techniques tofinding the volumes of certain solids.

Notes:

Exercises .Terms and Concepts

. T/F: The area between curves is always posi ve.

. T/F: Calculus can be used to find the area of basic geometricshapes.

. In your own words, describe how to find the total area en-closed by y = f(x) and y = g(x).

. Describe a situa on where it is advantageous to find anarea enclosed by curves through integra on with respectto y instead of x.

Problems

In Exercises – , find the area of the shaded region in thegiven graph.

.

.....

y = cos x +

.

y = x +

...

6

.π

.π

. x.

y

.

.....y = x + x −

.

y = − x + x +

.

−

..−

....

x

.

y

.

.....

y = 1

.

y = 2

.

1

.

2

.π

.π/2

.

x

.

y

.

...

..

y = sin x

.

y = sin x + 1

.

1

.

2

.

π

.

π/2.

x

.

y

.

...

..

y = sin(4x)

.

y = sec2 x

.

1

.

2

.π/4

.π/8.

x.

y

.

.....

y = sin x

.

y = cos x

.− .

− .

.

.

..

π/

.

π/

.

π/

.

π

.

π/

.

x

.

y

.

.....

y =x

.

y =x

..

...... x.

y

.

y =√

x + 1 y =√

2 − x + 1

y = 1

1 2

1

2

x

y

In Exercises – , find the total area enclosed by the func-ons f and g.

. f(x) = x + x− , g(x) = x + x−

. f(x) = x − x+ , g(x) = − x+

. f(x) = sin x, g(x) = x/π

. f(x) = x − x + x− , g(x) = −x + x−

. f(x) = x, g(x) =√x

. f(x) = −x + x + x+ , g(x) = x + x+

. The func ons f(x) = cos(x) and g(x) = sin x intersectinfinitely many mes, forming an infinite number of re-peated, enclosed regions. Find the areas of these regions.

. The func ons f(x) = cos( x) and g(x) = sin x intersectinfinitely many mes, forming an infinite number of re-peated, enclosed regions. Find the areas of these regions.

In Exercises – , find the area of the enclosed region intwo ways:

. by trea ng the boundaries as func ons of x, and

. by trea ng the boundaries as func ons of y.

.

..........

y = x +

.

y = (x − ) +

.

y =

. x.

y

.

.....

y =√

x

.

y = − x +

.

y = − x

... −.

− .5

.

.5

..

x

.

y

.

.....

y = x2

.

y = x + 2

.−1

.1

.2

.

2

.

4

. x.

y

.

x =12 y

2

x = −

12 y + 1

1 2

−2

−1

1

x

y

.

.....

y = x /

.

y =

√

x − /

. .5..

.5

..x

.

y

.

y =√

x + 1 y =√

2 − x + 1

y = 1

1 2

1

2

x

y

In Exercises – , find the area triangle formed by the giventhree points.

. ( , ), ( , ), and ( , )

. (− , ), ( , ), and ( ,− )

. ( , ), ( , ), and ( , )

. ( , ), ( , ), and ( , )

. Use the Trapezoidal Rule to approximate the area of thepictured lake whose lengths, in hundreds of feet, are mea-sured in -foot increments.

..

.9

.

.

. 7.. .

. Use Simpson’s Rule to approximate the area of the picturedlake whose lengths, in hundreds of feet, are measured in

-foot increments.

..

.2

.

.

.

.

.

.

.

.9

Figure . . : The volume of a generalright cylinder

Figure . . : Orien ng a pyramid alongthe x-axis in Example . . .

Chapter Applica ons of Integra on

. VolumebyCross-Sec onal Area; Disk andWasherMethods

The volume of a general right cylinder, as shown in Figure . . , isArea of the base× height.

We can use this fact as the building block in finding volumes of a variety ofshapes.

Given an arbitrary solid, we can approximate its volume by cu ng it into nthin slices. When the slices are thin, each slice can be approximated well by ageneral right cylinder. Thus the volume of each slice is approximately its cross-sec onal area× thickness. (These slices are the differen al elements.)

By orien ng a solid along the x-axis, we can let A(xi) represent the cross-sec onal area of the i th slice, and let∆xi represent the thickness of this slice (thethickness is a small change in x). The total volume of the solid is approximately:

Volume ≈n∑

i=

[

Area × thickness]

=

n∑

i=

A(xi)∆xi.

Recognize that this is a Riemann Sum. By taking a limit (as the thickness ofthe slices goes to ) we can find the volume exactly.

Theorem . . Volume By Cross-Sec onal Area

The volume V of a solid, oriented along the x-axis with cross-sec onalarea A(x) from x = a to x = b, is

V =

∫ b

aA(x) dx.

Example . . Finding the volume of a solidFind the volume of a pyramidwith a square base of side length in and a heightof in.

S There are many ways to “orient” the pyramid along the x-axis; Figure . . gives one such way, with the pointed top of the pyramid at theorigin and the x-axis going through the center of the base.

Each cross sec on of the pyramid is a square; this is a sample differen alelement. To determine its area A(x), we need to determine the side lengths of

Notes:

Figure . . : Cu ng a slice in the pyramidin Example . . at x = .

. Volume by Cross-Sec onal Area; Disk and Washer Methods

the square.When x = , the square has side length ; when x = , the square has side

length . Since the edges of the pyramid are lines, it is easy to figure that eachcross-sec onal square has side length x, giving A(x) = ( x) = x .

If one were to cut a slice out of the pyramid at x = , as shown in Figure. . , one would have a shape with square bo om and top with sloped sides. If

the slice were thin, both the bo om and top squares would have sides lengthsof about , and thus the cross–sec onal area of the bo om and top would beabout in . Le ng∆xi represent the thickness of the slice, the volume of thisslice would then be about ∆xiin .

Cu ng the pyramid into n slices divides the total volume into n equally–spaced smaller pieces, each with volume ( xi) ∆x, where xi is the approximateloca on of the slice along the x-axis and ∆x represents the thickness of eachslice. One can approximate total volume of the pyramid by summing up thevolumes of these slices:

Approximate volume =n∑

i=

( xi) ∆x.

Taking the limit as n → ∞ gives the actual volume of the pyramid; recoginizingthis sum as a Riemann Sum allows us to find the exact answer using a definiteintegral, matching the definite integral given by Theorem . . .

We have

V = limn→∞

n∑

i=

( xi) ∆x

=

∫

x dx

= x∣∣∣

= in ≈ . in .

We can check our work by consul ng the general equa on for the volume of apyramid (see the back cover under “Volume of A General Cone”):

× area of base× height.Certainly, using this formula from geometry is faster than our new method, butthe calculus–based method can be applied to much more than just cones.

An important special case of Theorem . . is when the solid is a solid ofrevolu on, that is, when the solid is formed by rota ng a shape around an axis.

Start with a func on y = f(x) from x = a to x = b. Revolving this curveabout a horizontal axis creates a three-dimensional solid whose cross sec ons

Notes:

(a)

(b)

Figure . . : Sketching a solid in Example. . .

Chapter Applica ons of Integra on

are disks (thin circles). Let R(x) represent the radius of the cross-sec onal disk atx; the area of this disk is πR(x) . Applying Theorem . . gives the DiskMethod.

Key Idea . . The Disk Method

Let a solid be formed by revolving the curve y = f(x) from x = a to x = baround a horizontal axis, and let R(x) be the radius of the cross-sec onaldisk at x. The volume of the solid is

V = π

∫ b

aR(x) dx.

Example . . Finding volume using the Disk MethodFind the volume of the solid formed by revolving the curve y = /x, from x =to x = , around the x-axis.

S A sketch can help us understand this problem. In Figure. . (a) the curve y = /x is sketched along with the differen al element – a

disk – at xwith radius R(x) = /x. In Figure . . (b) the whole solid is pictured,along with the differen al element.

The volume of the differen al element shown in part (a) of the figure is ap-proximately πR(xi) ∆x, where R(xi) is the radius of the disk shown and ∆x isthe thickness of that slice. The radius R(xi) is the distance from the x-axis to thecurve, hence R(xi) = /xi.

Slicing the solid into n equally–spaced slices, we can approximate the totalvolume by adding up the approximate volume of each slice:

Approximate volume =

n∑

i=

π

(

xi

)

∆x.

Taking the limit of the above sum as n → ∞ gives the actual volume; recog-nizing this sum as a Riemann sum allows us to evaluate the limit with a definiteintegral, which matches the formula given in Key Idea . . :

V = limn→∞

n∑

i=

π

(

xi

)

∆x

= π

∫ (

x

)

dx

= π

∫

xdx

Notes:

(a)

(b)

Figure . . : Sketching a solid in Example. . .

(a)

(b)

Figure . . : Establishing the WasherMethod; see also Figure . . .

. Volume by Cross-Sec onal Area; Disk and Washer Methods

= π

[

−x

] ∣∣∣

= π

[

− − (− )

]

=πunits .

While Key Idea . . is given in terms of func ons of x, the principle involvedcan be applied to func ons of y when the axis of rota on is ver cal, not hori-zontal. We demonstrate this in the next example.

Example . . Finding volume using the Disk MethodFind the volume of the solid formed by revolving the curve y = /x, from x =to x = , about the y-axis.

S Since the axis of rota on is ver cal, we need to convert thefunc on into a func on of y and convert the x-bounds to y-bounds. Since y =/x defines the curve, we rewrite it as x = /y. The bound x = corresponds to

the y-bound y = , and the bound x = corresponds to the y-bound y = / .Thus we are rota ng the curve x = /y, from y = / to y = about the

y-axis to form a solid. The curve and sample differen al element are sketched inFigure . . (a), with a full sketch of the solid in Figure . . (b). We integrateto find the volume:

V = π

∫

/ ydy

= −π

y

∣∣∣/

= π units .

We can also compute the volume of solids of revolu on that have a hole inthe center. The general principle is simple: compute the volume of the solidirrespec ve of the hole, then subtract the volume of the hole. If the outsideradius of the solid is R(x) and the inside radius (defining the hole) is r(x), thenthe volume is

V = π

∫ b

aR(x) dx− π

∫ b

ar(x) dx = π

∫ b

a

(R(x) − r(x)

)dx.

One can generate a solid of revolu on with a hole in the middle by revolvinga region about an axis. Consider Figure . . (a), where a region is sketched along

Notes:

Figure . . : Establishing the WasherMethod; see also Figure . . .

(a)

(b)

(c)

Figure . . : Sketching the differen al el-ement and solid in Example . . .

Chapter Applica ons of Integra on

with a dashed, horizontal axis of rota on. By rota ng the region about the axis, asolid is formed as sketched in Figure . . (b). The outside of the solid has radiusR(x), whereas the inside has radius r(x). Each cross sec on of this solid will bea washer (a disk with a hole in the center) as sketched in Figure . . . This leadsus to the Washer Method.

Key Idea . . The Washer Method

Let a region bounded by y = f(x), y = g(x), x = a and x = b be ro-tated about a horizontal axis that does not intersect the region, forminga solid. Each cross sec on at x will be a washer with outside radius R(x)and inside radius r(x). The volume of the solid is

V = π

∫ b

a

(

R(x) − r(x))

dx.

Even though we introduced it first, the Disk Method is just a special case ofthe Washer Method with an inside radius of r(x) = .

Example . . Finding volume with the Washer MethodFind the volume of the solid formed by rota ng the region bounded by y =x − x+ and y = x− about the x-axis.

S A sketch of the region will help, as given in Figure . . (a).Rota ng about the x-axis will produce cross sec ons in the shape of washers, asshown in Figure . . (b); the complete solid is shown in part (c). The outsideradius of this washer is R(x) = x+ ; the inside radius is r(x) = x − x+ . Asthe region is bounded from x = to x = , we integrate as follows to computethe volume.

V = π

∫ (

( x− ) − (x − x+ ))

dx

= π

∫(− x + x − x + x−

)dx

= π[

− x + x − x + x − x]∣∣∣

= π ≈ . units .

When rota ng about a ver cal axis, the outside and inside radius func onsmust be func ons of y.

Notes:

(a)

(b)

(c)

Figure . . : Sketching the solid in Exam-ple . . .

. Volume by Cross-Sec onal Area; Disk and Washer Methods

Example . . Finding volume with the Washer MethodFind the volume of the solid formed by rota ng the triangular region with ver-ces at ( , ), ( , ) and ( , ) about the y-axis.

S The triangular region is sketched in Figure . . (a); the dif-feren al element is sketched in (b) and the full solid is drawn in (c). They help usestablish the outside and inside radii. Since the axis of rota on is ver cal, eachradius is a func on of y.

The outside radius R(y) is formed by the line connec ng ( , ) and ( , ); itis a constant func on, as regardless of the y-value the distance from the line tothe axis of rota on is . Thus R(y) = .

The inside radius is formedby the line connec ng ( , ) and ( , ). The equa-on of this line is y = x− , but we need to refer to it as a func on of y. Solving

for x gives r(y) = (y+ ).We integrate over the y-bounds of y = to y = . Thus the volume is

V = π

∫ (

−(

(y+ )) )

dy

= π

∫ (

− y − y+)

dy

= π[

− y − y + y]∣∣∣

= π ≈ . units .

This sec on introduced a new applica on of the definite integral. Our de-fault view of the definite integral is that it gives “the area under the curve.” How-ever, we can establish definite integrals that represent other quan es; in thissec on, we computed volume.

The ul mate goal of this sec on is not to compute volumes of solids. Thatcan be useful, but what ismore useful is the understanding of this basic principleof integral calculus, outlined in Key Idea . . : to find the exact value of somequan ty,

• we start with an approxima on (in this sec on, slice the solid and approx-imate the volume of each slice),

• then make the approxima on be er by refining our original approxima-on (i.e., use more slices),

• then use limits to establish a definite integral which gives the exact value.We prac ce this principle in the next sec on where we find volumes by slic-

ing solids in a different way.

Notes:

Exercises .Terms and Concepts

. T/F: A solid of revolu on is formed by revolving a shapearound an axis.

. In your ownwords, explain how the Disk andWasherMeth-ods are related.

. Explain the how the units of volume are found in the inte-gral of Theorem . . : if A(x) has units of in , how does∫

A(x) dx have units of in ?

. A fundamental principle of this sec on is “ can befound by integra ng an area func on.”

ProblemsIn Exercises – , a region of the Cartesian plane is shaded.Use the Disk/Washer Method to find the volume of the solidof revolu on formed by revolving the region about the x-axis.

.

.....

y = − x

.−

.−

...... x.

y

.

.....

y = 5x

..5

...5

..

5

.. x.

y

.

.....

y = cos x

. 0.5. 1. 1.5.

0.5

.

1

.x

.

y

.

.....

y =

√

x

.

y = x

. 0.5. 1.

0.5

.

1

.x

.

y

In Exercises – , a region of the Cartesian plane is shaded.Use the Disk/Washer Method to find the volume of the solidof revolu on formed by revolving the region about the y-axis.

.

.....

y = − x

.−

.−

...... x.

y

.

.....

y = 5x

..5

...5

..

5

.. x.

y

.

.....

y = cos x

. 0.5. 1. 1.5.

0.5

.

1

.x

.

y

(Hint: Integra on By Parts will be necessary, twice. First letu = arccos x, then let u = arccos x.)

.

.....

y =

√

x

.

y = x

. 0.5. 1.

0.5

.

1

.x

.

y

In Exercises – , a region of the Cartesian plane is de-scribed. Use the Disk/Washer Method to find the volume ofthe solid of revolu on formed by rota ng the region abouteach of the given axes.

. Region bounded by: y =√x, y = and x = .

Rotate about:

(a) the x-axis(b) y =

(c) the y-axis(d) x =

. Region bounded by: y = − x and y = .Rotate about:

(a) the x-axis(b) y =

(c) y = −(d) x =

. The triangle with ver ces ( , ), ( , ) and ( , ).Rotate about:

(a) the x-axis(b) y =

(c) the y-axis(d) x =

. Region bounded by y = x − x+ and y = x− .Rotate about:

(a) the x-axis(b) y =

(c) y =

. Region bounded by y = /√x + , x = − , x = and

the x-axis.Rotate about:

(a) the x-axis(b) y =

(c) y = −

. Region bounded by y = x, y = x and x = .Rotate about:

(a) the x-axis(b) y =

(c) the y-axis(d) x =

In Exercises – , a solid is described. Orient the solid alongthe x-axis such that a cross-sec onal area func on A(x) canbe obtained, then apply Theorem . . to find the volume ofthe solid.

. A right circular cone with height of and base radius of .

. A skew right circular cone with height of and base radiusof . (Hint: all cross-sec ons are circles.)

. A right triangular cone with height of and whose base isa right, isosceles triangle with side length .

. A solid with length with a rectangular base and triangu-lar top, wherein one end is a square with side length andthe other end is a triangle with base and height of .

(a)

(b)

(c)

Figure . . : Introducing the ShellMethod.

Chapter Applica ons of Integra on

. The Shell MethodO en a given problem can be solved in more than one way. A par cular methodmay be chosen out of convenience, personal preference, or perhaps necessity.Ul mately, it is good to have op ons.

The previous sec on introduced the Disk and Washer Methods, which com-puted the volume of solids of revolu on by integra ng the cross–sec onal areaof the solid. This sec on develops another method of compu ng volume, theShell Method. Instead of slicing the solid perpendicular to the axis of rota oncrea ng cross-sec ons, we now slice it parallel to the axis of rota on, crea ng“shells.”

Consider Figure . . , where the region shown in (a) is rotated around they-axis forming the solid shown in (b). A small slice of the region is drawn in (a),parallel to the axis of rota on. When the region is rotated, this thin slice formsa cylindrical shell, as pictured in part (c) of the figure. The previous sec onapproximated a solid with lots of thin disks (or washers); we now approximatea solid with many thin cylindrical shells.

To compute the volume of one shell, first consider the paper label on a soupcan with radius r and height h. What is the area of this label? A simple way ofdetermining this is to cut the label and lay it out flat, forming a rectangle withheight h and length πr. Thus the area is A = πrh; see Figure . . (a).

Do a similar process with a cylindrical shell, with height h, thickness∆x, andapproximate radius r. Cu ng the shell and laying it flat forms a rectangular solidwith length πr, height h and depth ∆x. Thus the volume is V ≈ πrh∆x; seeFigure . . (b). (We say “approximately” since our radius was an approxima-on.)By breaking the solid into n cylindrical shells, we can approximate the volume

of the solid as

V ≈n∑

i=

πrihi∆xi,

where ri, hi and∆xi are the radius, height and thickness of the i th shell, respec-vely.This is a Riemann Sum. Taking a limit as the thickness of the shells ap-

proaches leads to a definite integral.

Notes:

.....

h(x)

.︸ ︷︷ ︸

r(x).

y =1

1 + x2

. 1.

1

. x.x

.

y

Figure . . : Graphing a region in Exam-ple . . .

. The Shell Method

...

.

h

.

cuth

ere

. r.2πr

.

h

.

A = 2πrh

...

..

h

.

cuth

ere

. r.∆x

.. 2πr

.

.

h.

. ∆x.

V ≈ 2πrh∆x

(a) (b)

Figure . . : Determining the volume of a thin cylindrical shell.

Key Idea . . The Shell Method

Let a solid be formed by revolving a region R, bounded by x = a andx = b, around a ver cal axis. Let r(x) represent the distance from the axisof rota on to x (i.e., the radius of a sample shell) and let h(x) representthe height of the solid at x (i.e., the height of the shell). The volume ofthe solid is

V = π

∫ b

ar(x)h(x) dx.

Special Cases:

. When the region R is bounded above by y = f(x) and below by y = g(x),then h(x) = f(x)− g(x).

. When the axis of rota on is the y-axis (i.e., x = ) then r(x) = x.

Let’s prac ce using the Shell Method.

Example . . Finding volume using the Shell MethodFind the volume of the solid formed by rota ng the region bounded by y = ,y = /( + x ), x = and x = about the y-axis.

S This is the region used to introduce the Shell Method in Fig-ure . . , but is sketched again in Figure . . for closer reference. A line isdrawn in the region parallel to the axis of rota on represen ng a shell that will

Notes:

.....

y=

x+

.

h(x).

.

︸ ︷︷ ︸

r(x)

....... x. x.

y

(a)

(b)

(c)

Figure . . : Graphing a region in Exam-ple . . .

Chapter Applica ons of Integra on

be carved out as the region is rotated about the y-axis. (This is the differen alelement.)

The distance this line is from the axis of rota on determines r(x); as thedistance from x to the y-axis is x, we have r(x) = x. The height of this linedetermines h(x); the top of the line is at y = /( + x ), whereas the bo omof the line is at y = . Thus h(x) = /( + x )− = /( + x ). The region isbounded from x = to x = , so the volume is

V = π

∫x+ x

dx.

This requires subs tu on. Let u = + x , so du = x dx. We also change thebounds: u( ) = and u( ) = . Thus we have:

= π

∫

udu

= π ln u∣∣∣

= π ln ≈ . units .

Note: in order to find this volume using the Disk Method, two integrals wouldbe needed to account for the regions above and below y = / .

With the Shell Method, nothing special needs to be accounted for to com-pute the volume of a solid that has a hole in the middle, as demonstrated next.

Example . . Finding volume using the Shell MethodFind the volumeof the solid formed by rota ng the triangular region determinedby the points ( , ), ( , ) and ( , ) about the line x = .

S The region is sketched in Figure . . (a) along with the dif-feren al element, a line within the region parallel to the axis of rota on. In part(b) of the figure, we see the shell traced out by the differen al element, and inpart (c) the whole solid is shown.

The height of the differen al element is the distance from y = to y = x+, the line that connects the points ( , ) and ( , ). Thus h(x) = x+ − = x.

The radius of the shell formed by the differen al element is the distance fromx to x = ; that is, it is r(x) = − x. The x-bounds of the region are x = to

Notes:

.....

x =y−

.

︸ ︷︷ ︸

h(y)

.

r(y)

.....

y

. x.

y

(a)

(b)

(c)

Figure . . : Graphing a region in Exam-ple . . .

. The Shell Method

x = , giving

V = π

∫

( − x)( x) dx

= π

∫(x− x ) dx

= π

(

x − x) ∣∣∣

= π ≈ . units .

When revolving a region around a horizontal axis, we must consider the ra-dius and height func ons in terms of y, not x.

Example . . Finding volume using the Shell MethodFind the volume of the solid formed by rota ng the region given in Example . .about the x-axis.

S The region is sketched in Figure . . (a) with a sample dif-feren al element. In part (b) of the figure the shell formed by the differen alelement is drawn, and the solid is sketched in (c). (Note that the triangular re-gion looks “short and wide” here, whereas in the previous example the sameregion looked “tall and narrow.” This is because the bounds on the graphs aredifferent.)

The height of the differen al element is an x-distance, between x = y−and x = . Thus h(y) = −( y− ) = − y+ . The radius is the distance fromy to the x-axis, so r(y) = y. The y bounds of the region are y = and y = ,leading to the integral

V = π

∫ [

y(

− y+)]

dy

= π

∫ [

− y + y]

dy

= π

[

− y + y] ∣∣∣

= π

[

−]

= π ≈ . units .

Notes:

.....

h(x)

.

r(x)︷ ︸︸ ︷

.

1

. x.π2. π.

x

.

y

(a)

(b)

(c)

Figure . . : Graphing a region in Exam-ple . . .

Chapter Applica ons of Integra on

At the beginning of this sec on it was stated that “it is good to have op ons.”The next example finds the volume of a solid rather easily with the ShellMethod,but using the Washer Method would be quite a chore.

Example . . Finding volume using the Shell MethodFind the volumeof the solid formedby revolving the region bounded by y = sin xand the x-axis from x = to x = π about the y-axis.

S The region and a differen al element, the shell formed bythis differen al element, and the resul ng solid are given in Figure . . . Theradius of a sample shell is r(x) = x; the height of a sample shell is h(x) = sin x,each from x = to x = π. Thus the volume of the solid is

V = π

∫ π

x sin x dx.

This requires Integra on By Parts. Set u = x and dv = sin x dx; we leave it tothe reader to fill in the rest. We have:

= π[

− x cos x∣∣∣

π

+

∫ π

cos x dx]

= π[

π + sin x∣∣∣

π ]

= π[

π +]

= π ≈ . units .

Note that in order to use the Washer Method, we would need to solve y =sin x for x, requiring the use of the arcsine func on. We leave it to the readerto verify that the outside radius func on is R(y) = π − arcsin y and the insideradius func on is r(y) = arcsin y. Thus the volume can be computed as

π

∫ [

(π − arcsin y) − (arcsin y)]

dy.

This integral isn’t terrible given that the arcsin y terms cancel, but it is moreonerous than the integral created by the Shell Method.

We end this sec on with a table summarizing the usage of the Washer andShell Methods.

Notes:

. The Shell Method

Key Idea . . Summary of the Washer and Shell Methods

Let a region R be given with x-bounds x = a and x = b and y-boundsy = c and y = d.

Washer Method Shell Method

HorizontalAxis

π

∫ b

a

(R(x) − r(x)

)dx π

∫ d

cr(y)h(y) dy

Ver calAxis

π

∫ d

c

(R(y) − r(y)

)dy π

∫ b

ar(x)h(x) dx

As in the previous sec on, the real goal of this sec on is not to be able tocompute volumes of certain solids. Rather, it is to be able to solve a problemby first approxima ng, then using limits to refine the approxima on to give theexact value. In this sec on, we approximate the volume of a solid by cu ng itinto thin cylindrical shells. By summing up the volumes of each shell, we get anapproxima on of the volume. By taking a limit as the number of equally spacedshells goes to infinity, our summa on can be evaluated as a definite integral,giving the exact value.

We use this same principle again in the next sec on, where we find thelength of curves in the plane.

Notes:

Exercises .Terms and Concepts

. T/F: A solid of revolu on is formed by revolving a shapearound an axis.

. T/F: The Shell Method can only be used when the WasherMethod fails.

. T/F: The Shell Method works by integra ng cross–sec onalareas of a solid.

. T/F: When finding the volume of a solid of revolu on thatwas revolved around a ver cal axis, the Shell Method inte-grates with respect to x.

Problems

In Exercises – , a region of the Cartesian plane is shaded.Use the Shell Method to find the volume of the solid of revo-lu on formed by revolving the region about the y-axis.

.

.....

y = − x

.−

.−

...... x.

y

.

.....

y = 5x

..5

...5

..

5

.. x.

y

.

.....

y = cos x

. 0.5. 1. 1.5.

0.5

.

1

.x

.

y

.

.....

y =

√

x

.

y = x

. 0.5. 1.

0.5

.

1

.x

.

y

In Exercises – , a region of the Cartesian plane is shaded.Use the Shell Method to find the volume of the solid of revo-lu on formed by revolving the region about the x-axis.

.

.....

y = − x

.−

.−

...... x.

y

.

.....

y = 5x

..5

...5

..

5

.. x.

y

.

.....

y = cos x

. 0.5. 1. 1.5.

0.5

.

1

.x

.

y

.

.....

y =

√

x

.

y = x

. 0.5. 1.

0.5

.

1

.x

.

y

In Exercises – , a region of the Cartesian plane is de-scribed. Use the Shell Method to find the volume of the solidof revolu on formed by rota ng the region about each of thegiven axes.

. Region bounded by: y =√x, y = and x = .

Rotate about:

(a) the y-axis(b) x =

(c) the x-axis(d) y =

. Region bounded by: y = − x and y = .Rotate about:

(a) x =

(b) x = −(c) the x-axis(d) y =

. The triangle with ver ces ( , ), ( , ) and ( , ).Rotate about:

(a) the y-axis(b) x =

(c) the x-axis(d) y =

. Region bounded by y = x − x+ and y = x− .Rotate about:

(a) the y-axis(b) x =

(c) x = −

. Region bounded by y = /√x + , x = and the x and

y-axes.Rotate about:

(a) the y-axis (b) x =

. Region bounded by y = x, y = x and x = .Rotate about:

(a) the y-axis(b) x =

(c) the x-axis(d) y =

.....

.

..π

.π

. π. π.x

.

y

(a)

.....

.

..π

.π

. π. π.

√

.x

.

y

(b)

Figure . . : Graphing y = sin x on [ , π]and approxima ng the curve with linesegments.

.....

∆yi

.

∆xi

. xi. xi+1.

yi

.

yi+1

.x

.

y

Figure . . : Zooming in on the i th subin-terval [xi, xi+ ] of a par on of [a, b].

Chapter Applica ons of Integra on

. Arc Length and Surface AreaIn previous sec onswe have used integra on to answer the following ques ons:

. Given a region, what is its area?

. Given a solid, what is its volume?

In this sec on, we address a related ques on: Given a curve, what is itslength? This is o en referred to as arc length.

Consider the graph of y = sin x on [ , π] given in Figure . . (a). How long isthis curve? That is, if we were to use a piece of string to exactly match the shapeof this curve, how long would the string be?

As we have done in the past, we start by approxima ng; later, we will refineour answer using limits to get an exact solu on.

The length of straight–line segments is easy to compute using the DistanceFormula. We can approximate the length of the given curve by approxima ngthe curve with straight lines and measuring their lengths.

In Figure . . (b), the curve y = sin x has been approximated with linesegments (the interval [ , π] has been divided into equally–lengthed subinter-vals). It is clear that these four line segments approximate y = sin x very wellon the first and last subinterval, though not so well in the middle. Regardless,the sum of the lengths of the line segments is . , so we approximate the arclength of y = sin x on [ , π] to be . .

In general, we can approximate the arc length of y = f(x) on [a, b] in thefollowing manner. Let a = x < x < . . . < xn < xn+ = b be a par onof [a, b] into n subintervals. Let ∆xi represent the length of the i th subinterval[xi, xi+ ].

Figure . . zooms in on the i th subinterval where y = f(x) is approximatedby a straight line segment. The dashed lines show that we can view this line seg-ment as the hypotenuse of a right triangle whose sides have length∆xi and∆yi.Using the Pythagorean Theorem, the length of this line segment is

√

∆xi +∆yi .Summing over all subintervals gives an arc length approxima on

L ≈n∑

i=

√

∆xi +∆yi .

As shown here, this is not a Riemann Sum. While we could conclude thattaking a limit as the subinterval length goes to zero gives the exact arc length,we would not be able to compute the answer with a definite integral. We needfirst to do a li le algebra.

Notes:

Note: This is our first use of differen a-bility on a closed interval since Sec on. .

The theorem also requires that f ′ be con-nuous on [a, b]; while examples are ar-

cane, it is possible for f to be differen-able yet f ′ is not con nuous.

. Arc Length and Surface Area

In the above expression factor out a∆xi term:

n∑

i=

√

∆xi +∆yi =n∑

i=

√

∆xi

(

+∆yi∆xi

)

.

Now pull the∆xi term out of the square root:

=n∑

i=

√

+∆yi∆xi

∆xi.

This is nearly a Riemann Sum. Consider the ∆yi /∆xi term. The expression∆yi/∆xi measures the “change in y/change in x,” that is, the “rise over run” off on the i th subinterval. The Mean Value Theorem of Differen a on (Theorem. . ) states that there is a ci in the i th subinterval where f ′(ci) = ∆yi/∆xi. Thus

we can rewrite our above expression as:

=

n∑

i=

√

+ f ′(ci) ∆xi.

This is a Riemann Sum. As long as f ′ is con nuous, we can invoke Theorem . .and conclude

=

∫ b

a

√

+ f ′(x) dx.

Theorem . . Arc Length

Let f be differen able on [a, b], where f ′ is also con nuous on [a, b]. Thenthe arc length of f from x = a to x = b is

L =∫ b

a

√

+ f ′(x) dx.

As the integrand contains a square root, it is o en difficult to use the formulain Theorem . . to find the length exactly. When exact answers are difficult tocome by, we resort to using numerical methods of approxima ng definite inte-grals. The following examples will demonstrate this.

Notes:

.....2

.4

.

2

.

4

.

6

.

8

. x.

y

Figure . . : A graph of f(x) = x / fromExample . . .

Chapter Applica ons of Integra on

Example . . Finding arc lengthFind the arc length of f(x) = x / from x = to x = .

S We find f ′(x) = x / ; note that on [ , ], f is differen ableand f ′ is also con nuous. Using the formula, we find the arc length L as

L =∫√

+

(

x /

)

dx

=

∫ √

+ x dx

=

∫ (

+ x) /

dx

= · ·(

+ x) / ∣

∣∣

=(

/ −)

≈ . units.

A graph of f is given in Figure . . .

Example . . Finding arc lengthFind the arc length of f(x) = x − ln x from x = to x = .

S This func on was chosen specifically because the resul ngintegral can be evaluated exactly. We begin by finding f ′(x) = x/ − /x. Thearc length is

L =∫√

+

(x −

x

)

dx

=

∫ √

+x − +

xdx

=

∫ √

x+ +

xdx

=

∫√(x+

x

)

dx

Notes:

........

.5

..

x

.

y

Figure . . : A graph of f(x) = x − ln xfrom Example . . .

x√

+ cos x√

π/√

/π/

π/√

/

π√

Figure . . : A table of values of y =√+ cos x to evaluate a definite inte-

gral in Example . . .

. Arc Length and Surface Area

=

∫ (x+

x

)

dx

=

(x

+ ln x)∣∣∣∣∣

= + ln ≈ . units.

A graph of f is given in Figure . . ; the por on of the curve measured in thisproblem is in bold.

The previous examples found the arc length exactly through careful choiceof the func ons. In general, exact answers are much more difficult to come byand numerical approxima ons are necessary.

Example . . Approxima ng arc length numericallyFind the length of the sine curve from x = to x = π.

S This is somewhat of a mathema cal curiosity; in Example. . we found the area under one “hump” of the sine curve is square units;

now we are measuring its arc length.The setup is straigh orward: f(x) = sin x and f ′(x) = cos x. Thus

L =∫ π√

+ cos x dx.

This integral cannot be evaluated in terms of elementary func ons sowewill ap-proximate it with Simpson’s Method with n = . Figure . . gives

√+ cos x

evaluated at evenly spaced points in [ , π]. Simpson’s Rule then states that∫ π√

+ cos x dx ≈ π −·

(√+√

/ + ( ) +√

/ +√ )

= . .

Using a computer with n = the approxima on is L ≈ . ; our approxi-ma on with n = is quite good.

Notes:

...

..

a

.

xi

.

xi+1

.

b

.

x

.

y

(a)

(b)

Figure . . : Establishing the formula forsurface area.

Chapter Applica ons of Integra on

Surface Area of Solids of Revolu on

We have already seen how a curve y = f(x) on [a, b] can be revolved aroundan axis to form a solid. Instead of compu ng its volume, we now consider itssurface area.

We begin as we have in the previous sec ons: we par on the interval [a, b]with n subintervals, where the i th subinterval is [xi, xi+ ]. On each subinterval,we can approximate the curve y = f(x) with a straight line that connects f(xi)and f(xi+ ) as shown in Figure . . (a). Revolving this line segment about the x-axis creates part of a cone (called a frustumof a cone) as shown in Figure . . (b).The surface area of a frustum of a cone is

π · length · average of the two radii R and r.

The length is given by L; we use the material just covered by arc length tostate that

L ≈√

+ f ′(ci) ∆xi

for some ci in the i th subinterval. The radii are just the func on evaluated at theendpoints of the interval. That is,

R = f(xi+ ) and r = f(xi).

Thus the surface area of this sample frustum of the cone is approximately

πf(xi) + f(xi+ )√

+ f ′(ci) ∆xi.

Since f is a con nuous func on, the IntermediateValue Theoremstates there

is some di in [xi, xi+ ] such that f(di) =f(xi) + f(xi+ )

; we can use this to rewritethe above equa on as

πf(di)√

+ f ′(ci) ∆xi.

Summing over all the subintervals we get the total surface area to be approxi-mately

Surface Area ≈n∑

i=

πf(di)√

+ f ′(ci) ∆xi,

which is a Riemann Sum. Taking the limit as the subinterval lengths go to zerogives us the exact surface area, given in the following theorem.

Notes:

Figure . . : Revolving y = sin x on [ , π]about the x-axis.

. Arc Length and Surface Area

Theorem . . Surface Area of a Solid of Revolu on

Let f be differen able on [a, b], where f ′ is also con nuous on [a, b].

. The surface area of the solid formed by revolving the graph of y =f(x), where f(x) ≥ , about the x-axis is

Surface Area = π

∫ b

af(x)√

+ f ′(x) dx.

. The surface area of the solid formed by revolving the graph of y =f(x) about the y-axis, where a, b ≥ , is

Surface Area = π

∫ b

ax√

+ f ′(x) dx.

(When revolving y = f(x) about the y-axis, the radii of the resul ng frustumare xi and xi+ ; their average value is simply the midpoint of the interval. In thelimit, this midpoint is just x. This gives the second part of Theorem . . .)

Example . . Finding surface area of a solid of revolu onFind the surface area of the solid formed by revolving y = sin x on [ , π] aroundthe x-axis, as shown in Figure . . .

S The setup is rela vely straigh orward. Using Theorem . . ,we have the surface area SA is:

SA = π

∫ π

sin x√

+ cos x dx

= − π(

sinh− (cos x) + cos x√

+ cos x)∣∣∣

π

= π(√

+ sinh−)

≈ . units .

The integra on step above is nontrivial, u lizing an integra on method calledTrigonometric Subs tu on.

It is interes ng to see that the surface area of a solid, whose shape is definedby a trigonometric func on, involves both a square root and an inverse hyper-bolic trigonometric func on.

Example . . Finding surface area of a solid of revolu onFind the surface area of the solid formed by revolving the curve y = x on [ , ]about the x-axis and the y-axis.

Notes:

(a)

(b)

Figure . . : The solids used in Example. . .

Figure . . : A graph of Gabriel’s Horn.

Chapter Applica ons of Integra on

S About the x-axis: the integral is straigh orward to setup:

SA = π

∫

x√

+ ( x) dx.

Like the integral in Example . . , this requires Trigonometric Subs tu on.

=π (

( x + x)√

+ x − sinh− ( x))∣∣∣

=π ( √

− sinh−)

≈ . units .

The solid formed by revolving y = x around the x-axis is graphed in Figure . .(a).

About the y-axis: since we are revolving around the y-axis, the “radius” ofthe solid is not f(x) but rather x. Thus the integral to compute the surface areais:

SA = π

∫

x√

+ ( x) dx.

This integral can be solved using subs tu on. Set u = + x ; the new boundsare u = to u = . We then have

=π∫ √

u du

=π

u /

∣∣∣∣

=π ( √

−)

≈ . units .

The solid formed by revolving y = x about the y-axis is graphed in Figure . .(b).

Our final example is a famous mathema cal “paradox.”

Example . . The surface area and volume of Gabriel’s HornConsider the solid formed by revolving y = /x about the x-axis on [ ,∞). Findthe volume and surface area of this solid. (This shape, as graphed in Figure . . ,is known as “Gabriel’s Horn” since it looks like a very long horn that only a su-pernatural person, such as an angel, could play.)

Notes:

. Arc Length and Surface Area

S To compute the volume it is natural to use the Disk Method.We have:

V = π

∫ ∞

xdx

= limb→∞

π

∫ b

xdx

= limb→∞

π

(−x

)∣∣∣∣

b

= limb→∞

π

(

−b

)

= π units .

Gabriel’s Horn has a finite volume of π cubic units. Since we have already seenthat regions with infinite length can have a finite area, this is not too difficult toaccept.

We now consider its surface area. The integral is straigh orward to setup:

SA = π

∫ ∞

x√

+ /x dx.

Integra ng this expression is not trivial. We can, however, compare it to otherimproper integrals. Since <

√

+ /x on [ ,∞), we can state that

π

∫ ∞

xdx < π

∫ ∞

x√

+ /x dx.

By Key Idea . . , the improper integral on the le diverges. Since the integralon the right is larger, we conclude it also diverges, meaning Gabriel’s Horn hasinfinite surface area.

Hence the “paradox”: we can fill Gabriel’s Hornwith a finite amount of paint,but since it has infinite surface area, we can never paint it.

Somehow this paradox is striking when we think about it in terms of vol-ume and area. However, we have seen a similar paradox before, as referencedabove. We know that the area under the curve y = /x on [ ,∞) is finite, yetthe shape has an infinite perimeter. Strange things can occur when we deal withthe infinite.

A standard equa on from physics is “Work = force × distance”, when theforce applied is constant. In the next sec on we learn how to compute workwhen the force applied is variable.

Notes:

Exercises .Terms and Concepts

. T/F: The integral formula for compu ng Arc Length wasfound by first approxima ng arc length with straight linesegments.

. T/F: The integral formula for compu ng Arc Length includesa square–root, meaning the integra on is probably easy.

ProblemsIn Exercises – , find the arc length of the func on on thegiven interval.

. f(x) = x on [ , ].

. f(x) =√

x on [− , ].

. f(x) = x / − x / on [ , ].

. f(x) = x +xon [ , ].

. f(x) = x / −√x on [ , ].

. f(x) = cosh x on [− ln , ln ].

. f(x) =(

ex + e−x) on [ , ln ].

. f(x) = x +x

on [. , ].

. f(x) = ln(

sin x)

on [π/ , π/ ].

. f(x) = ln(

cos x)

on [ , π/ ].

In Exercises – , set up the integral to compute the arclength of the func on on the given interval. Do not evaluatethe integral.

. f(x) = x on [ , ].

. f(x) = x on [ , ].

. f(x) =√x on [ , ].

. f(x) = ln x on [ , e].

. f(x) =√

− x on [− , ]. (Note: this describes the tophalf of a circle with radius .)

. f(x) =√

− x / on [− , ]. (Note: this describes the tophalf of an ellipse with a major axis of length and a minoraxis of length .)

. f(x) =xon [ , ].

. f(x) = sec x on [−π/ , π/ ].

In Exercises – , use Simpson’s Rule, with n = , to ap-proximate the arc length of the func on on the given interval.Note: these are the same problems as in Exercises – .

. f(x) = x on [ , ].

. f(x) = x on [ , ].

. f(x) =√x on [ , ]. (Note: f ′(x) is not defined at x = .)

. f(x) = ln x on [ , e].

. f(x) =√

− x on [− , ]. (Note: f ′(x) is not defined atthe endpoints.)

. f(x) =√

− x / on [− , ]. (Note: f ′(x) is not definedat the endpoints.)

. f(x) =xon [ , ].

. f(x) = sec x on [−π/ , π/ ].

In Exercises – , find the surface area of the describedsolid of revolu on.

. The solid formed by revolving y = x on [ , ] about thex-axis.

. The solid formed by revolving y = x on [ , ] about they-axis.

. The solid formed by revolving y = x on [ , ] about thex-axis.

. The solid formed by revolving y =√x on [ , ] about the

x-axis.

. The sphere formed by revolving y =√

− x on [− , ]about the x-axis.

Note: Mass and weight are closely re-lated, yet different, concepts. The massm of an object is a quan ta ve measureof that object’s resistance to accelera on.The weight w of an object is a measure-ment of the force applied to the object bythe accelera on of gravity g.Since the two measurements are pro-

por onal, w = m · g, they are o enused interchangeably in everyday conver-sa on. When compu ng work, one mustbe careful to note which is being referredto. When mass is given, it must be mul -plied by the accelera on of gravity to ref-erence the related force.

. Work

. WorkWork is the scien fic term used to describe the ac on of a force which movesan object. When a constant force F is applied to move an object a distance d,the amount of work performed isW = F · d.

The SI unit of force is the Newton, (kg·m/s ), and the SI unit of distance isa meter (m). The fundamental unit of work is one Newton–meter, or a joule(J). That is, applying a force of one Newton for one meter performs one jouleof work. In Imperial units (as used in the United States), force is measured inpounds (lb) and distance is measured in feet ( ), hence work is measured in–lb.When force is constant, the measurement of work is straigh orward. For

instance, li ing a lb object performs · = –lb of work.What if the force applied is variable? For instance, imagine a climber pulling

a rope up a ver cal face. The rope becomes lighter as more is pulled in,requiring less force and hence the climber performs less work.

In general, let F(x) be a force func on on an interval [a, b]. We want to mea-sure the amount of work done applying the force F from x = a to x = b. We canapproximate the amount of work being done by par oning [a, b] into subinter-vals a = x < x < · · · < xn+ = b and assuming that F is constant on eachsubinterval. Let ci be a value in the i th subinterval [xi, xi+ ]. Then the work doneon this interval is approximatelyWi ≈ F(ci) · (xi+ − xi) = F(ci)∆xi, a constantforce× the distance over which it is applied. The total work is

W =

n∑

i=

Wi ≈n∑

i=

F(ci)∆xi.

This, of course, is a Riemann sum. Taking a limit as the subinterval lengths goto zero gives an exact value of work which can be evaluated through a definiteintegral.

Key Idea . . Work

Let F(x) be a con nuous func on on [a, b] describing the amount of forcebeing applied to an object in the direc on of travel from distance x = ato distance x = b. The total workW done on [a, b] is

W =

∫ b

aF(x) dx.

Notes:

Chapter Applica ons of Integra on

Example . . Compu ng work performed: applying variable forceA m climbing rope is hanging over the side of a tall cliff. How much workis performed in pulling the rope up to the top, where the rope has a mass of

g/m?

S Weneed to create a force func on F(x)on the interval [ , ].To do so, we must first decide what x is measuring: it is the length of the ropes ll hanging or is it the amount of rope pulled in? As long as we are consistent,either approach is fine. We adopt for this example the conven on that x is theamount of rope pulled in. This seems to match intui on be er; pulling up thefirst meters of rope involves x = to x = instead of x = to x = .

As x is the amount of rope pulled in, the amount of rope s ll hanging is −x.This length of rope has a mass of g/m, or . kg/m. The mass of the ropes ll hanging is . ( − x) kg; mul plying this mass by the accelera on ofgravity, . m/s , gives our variable force func on

F(x) = ( . )( . )( − x) = . ( − x).

Thus the total work performed in pulling up the rope is

W =

∫

. ( − x) dx = , . J.

By comparison, consider the work done in li ing the en re rope meters.The ropeweighs × . × . = . N, so thework applying this force for

meters is × . = , . J. This is exactly twice the work calculatedbefore (and we leave it to the reader to understand why.)

Example . . Compu ng work performed: applying variable forceConsider again pulling a m rope up a cliff face, where the rope has a mass of

g/m. At what point is exactly half the work performed?

S From Example . . we know the total work performed is, . J. We want to find a height h such that the work in pulling the rope

from a height of x = to a height of x = h is . , half the total work. Thuswe want to solve the equa on

∫ h. ( − x) dx = .

for h.

Notes:

Note: In Example . . , we find that halfof the work performed in pulling up am rope is done in the last . m. Why isit not coincidental that /

√= . ?

. Work

∫ h. ( − x) dx = .

(. x− . x

)∣∣∣

h= .

. h− . h = .

− . h + . h− . = .

Apply the Quadra c Formula:

h = . and .

As the rope is only m long, the only sensible answer is h = . . Thus abouthalf the work is done pulling up the first . m the other half of the work is donepulling up the remaining . m.

Example . . Compu ng work performed: applying variable forceA box of lb of sand is being pulled up at a uniform rate a distance ofover minute. The sand is leaking from the box at a rate of lb/s. The box itselfweighs lb and is pulled by a rope weighing . lb/ .

. How much work is done li ing just the rope?

. How much work is done li ing just the box and sand?

. What is the total amount of work performed?

S

. We start by forming the force func on Fr(x) for the rope (where the sub-script denotes we are considering the rope). As in the previous example,let x denote the amount of rope, in feet, pulled in. (This is the same assaying x denotes the height of the box.) The weight of the rope with xfeet pulled in is Fr(x) = . ( − x) = − . x. (Note that we do nothave to include the accelera on of gravity here, for theweight of the ropeper foot is given, not its mass per meter as before.) The work performedli ing the rope is

Wr =

∫

( − . x) dx = –lb.

Notes:

Chapter Applica ons of Integra on

. The sand is leaving the box at a rate of lb/s. As the ver cal trip is to takeoneminute, we know that lbwill have le when the box reaches its finalheight of . Again le ng x represent the height of the box, we havetwo points on the line that describes the weight of the sand: when x = ,the sand weight is lb, producing the point ( , ); when x = , thesand in the box weighs lb, producing the point ( , ). The slope ofthis line is −

− = − . , giving the equa on of the weight of the sandat height x as w(x) = − . x+ . The box itself weighs a constant lb,so the total force func on is Fb(x) = − . x+ . Integra ng from x =to x = gives the work performed in li ing box and sand:

Wb =

∫

(− . x+ ) dx = –lb.

. The total work is the sum of Wr and Wb: + = –lb. Wecan also arrive at this via integra on:

W =

∫

(Fr(x) + Fb(x)) dx

=

∫

( − . x− . x+ ) dx

=

∫

(− . x+ ) dx

= –lb.

Hooke’s Law and Springs

Hooke’s Law states that the force required to compress or stretch a spring xunits from its natural length is propor onal to x; that is, this force is F(x) = kxfor some constant k. For example, if a force of N stretches a given springcm, then a force of N will stretch the spring cm. Conver ng the dis-

tances to meters, we have that stretching this spring . m requires a forceof F( . ) = k( . ) = N, hence k = / . = N/m.

Example . . Compu ng work performed: stretching a springA force of lb stretches a spring from a natural length of inches to a lengthof inches. How much work was performed in stretching the spring to thislength?

S In many ways, we are not at all concerned with the actuallength of the spring, only with the amount of its change. Hence, we do not care

Notes:

Fluid lb/ kg/mConcreteFuel Oil . .Gasoline . .Iodine

Methanol . .MercuryMilk . – . –Water .

Figure . . : Weight and Mass densi es

. Work

that lb of force stretches the spring to a length of inches, but rather thata force of lb stretches the spring by in. This is illustrated in Figure . . ;we only measure the change in the spring’s length, not the overall length of thespring.

.

.

.F

..............

.

Figure . . : Illustra ng the important aspects of stretching a spring in compu ng workin Example . . .

Conver ng the units of length to feet, we have

F( / ) = / k = lb.

Thus k = lb/ and F(x) = x.We compute the total work performed by integra ng F(x) from x = to

x = / :

W =

∫ /

x dx

= x∣∣∣

/

= / ≈ . –lb.

Pumping Fluids

Another useful example of the applica on of integra on to compute workcomes in the pumping of fluids, o en illustrated in the context of emptying astorage tank by pumping the fluid out the top. This situa on is different thanour previous examples for the forces involved are constant. A er all, the forcerequired to move one cubic foot of water (about . lb) is the same regardlessof its loca on in the tank. What is variable is the distance that cubic foot ofwater has to travel; water closer to the top travels less distance than water atthe bo om, producing less work.

We demonstrate how to compute the total work done in pumping a fluid outof the top of a tank in the next two examples.

Notes:

..

y

.0 .

30

.

35

.

35−

y i

.

10

.

yi

.

yi+1

.

}

∆yi

..

Figure . . : Illustra ng a water tank inorder to compute the work required toempty it in Example . . .

Chapter Applica ons of Integra on

Example . . Compu ng work performed: pumping fluidsA cylindrical storage tank with a radius of and a height of is filled withwater, which weighs approximately . lb/ . Compute the amount of workperformed by pumping the water up to a point feet above the top of the tank.

S Wewill refer o en to Figure . . which illustrates the salientaspects of this problem.

We start aswe o en do: we par on an interval into subintervals. We orientour tank ver cally since this makes intui ve sense with the base of the tank aty = . Hence the top of the water is at y = , meaning we are interested insubdividing the y-interval [ , ] into n subintervals as

= y < y < · · · < yn+ = .

Consider the workWi of pumping only the water residing in the i th subinterval,illustrated in Figure . . . The force required to move this water is equal to itsweight which we calculate as volume × density. The volume of water in thissubinterval is Vi = π∆yi; its density is . lb/ . Thus the required force is

π∆yi lb.We approximate the distance the force is applied by using any y-value con-

tained in the i th subinterval; for simplicity, we arbitrarily use yi for now (it willnot ma er later on). The water will be pumped to a point feet above the topof the tank, that is, to the height of y = . Thus the distance the water atheight yi travels is − yi .

In all, the approximate work Wi peformed in moving the water in the i thsubinterval to a point feet above the tank is

Wi ≈ π∆yi( − yi).

To approximate the total work performed in pumping out all the water from thetank, we sum all the workWi performed in pumping the water from each of then subintervals of [ , ]:

W ≈n∑

i=

Wi =

n∑

i=

π∆yi( − yi).

This is a Riemann sum. Taking the limit as the subinterval length goes to gives

W =

∫

π( − y) dy

= π(

y− / y)∣∣∣

= , , –lb≈ . × –lb.

Notes:

..

y

.0 .

30

.

35

.

y

.

35−

y

.

10

.

.

V(y) = 100πdy

Figure . . : A simplified illustra on forcompu ng work.

..

y

.−

...

y

.

−y

.

.

.V(y) = π( y

5 + ) dy

Figure . . : A graph of the conical watertank in Example . . .

. Work

We can “streamline” the above process a bit as we may now recognize whatthe important features of the problem are. Figure . . shows the tank fromExample . . without the i th subinterval iden fied. Instead, we just draw onedifferen al element. This helps establish the height a small amount of watermust travel along with the force required to move it (where the force is volume× density).

We demonstrate the concepts again in the next examples.

Example . . Compu ng work performed: pumping fluidsA conicalwater tank has its top at ground level and its base feet belowground.The radius of the cone at ground level is . It is filled with water weighing .lb/ and is to be emp ed by pumping thewater to a spigot feet above groundlevel. Find the total amount of work performed in emptying the tank.

S The conical tank is sketched in Figure . . . We can orientthe tank in a variety of ways; we could let y = represent the base of the tankand y = represent the top of the tank, but we choose to keep the conven onof the wording given in the problem and let y = represent ground level andhence y = − represents the bo om of the tank. The actual “height” of thewater does not ma er; rather, we are concerned with the distance the watertravels.

The figure also sketches a differen al element, a cross–sec onal circle. Theradius of this circle is variable, depending on y. When y = − , the circle hasradius ; when y = , the circle has radius . These two points, (− , ) and( , ), allow us to find the equa on of the line that gives the radius of the cross–sec onal circle, which is r(y) = / y + . Hence the volume of water at thisheight is V(y) = π( / y + ) dy, where dy represents a very small height ofthe differen al element. The force required to move the water at height y isF(y) = . × V(y).

The distance the water at height y travels is given by h(y) = − y. Thus thetotal work done in pumping the water from the tank is

W =

∫

−. π( / y+ ) ( − y) dy

= . π

∫

−

(

− y − y − y+)

dy

= . π · ≈ , –lb.

Notes:

...

..

.

.

.6 .

.

5

Figure . . : The cross–sec on of a swim-ming pool filled with water in Example. . .

..

y

.0 .y

.3

.

6

.

8

.(10, 0).

(15, 3).

x

.

0

.

10

.

15

Figure . . : Orien ng the pool andshowing differen al elements for Exam-ple . . .

Chapter Applica ons of Integra on

Example . . Compu ng work performed: pumping fluidsA rectangular swimming pool is wide and has a “shallow end” and a“deep end.” It is to have its water pumped out to a point above the currenttop of the water. The cross–sec onal dimensions of the water in the pool aregiven in Figure . . ; note that the dimensions are for the water, not the poolitself. Compute the amount of work performed in draining the pool.

S For the purposes of this problem we choose to set y =to represent the bo om of the pool, meaning the top of the water is at y = .Figure . . shows the pool oriented with this y-axis, along with differen alelements as the pool must be split into two different regions.

The top region lies in the y-interval of [ , ], where the length of the differen-al element is as shown. As the pool is wide, this differen al element

represents a thin slice of water with volume V(y) = · · dy. The water isto be pumped to a height of y = , so the height func on is h(y) = − y. Thework done in pumping this top region of water is

Wt = .

∫

( − y) dy = , –lb.

The bo om region lies in the y-interval of [ , ]; we need to compute thelength of the differen al element in this interval.

One end of the differen al element is at x = and the other is along the linesegment joining the points ( , ) and ( , ). The equa on of this line is y =/ (x− ); as we will be integra ng with respect to y, we rewrite this equa on

as x = / y + . So the length of the differen al element is a difference ofx-values: x = and x = / y+ , giving a length of x = / y+ .

Again, as the pool is wide, this differen al element represents a thinslice of water with volume V(y) = · ( / y + ) · dy; the height func on isthe same as before at h(y) = − y. The work performed in emptying this partof the pool is

Wb = .

∫

( / y+ )( − y) dy = , –lb.

The total work in empy ng the pool is

W = Wb +Wt = , + , = , –lb.

No ce how the emptying of the bo om of the pool performs almost as muchwork as emptying the top. The top por on travels a shorter distance but hasmore water. In the end, this extra water produces more work.

The next sec on introduces one final applica on of the definite integral, thecalcula on of fluid force on a plate.

Notes:

Exercises .Terms and Concepts

. What are the typical units of work?

. If a man has a mass of kg on Earth, will his mass on themoon be bigger, smaller, or the same?

. If a woman weighs lb on Earth, will her weight on themoon be bigger, smaller, or the same?

. Fill in the blanks:Some integrals in this sec on are set up by mul plying avariable by a constant distance; others are setup by mul plying a constant force by a variable .

Problems

. A rope, weighing . lb/ , hangs over the edge of atall building.

(a) Howmuchwork is done pulling the en re rope to thetop of the building?

(b) How much rope is pulled in when half of the totalwork is done?

. A m rope, with a mass density of . kg/m, hangs overthe edge of a tall building.

(a) Howmuchwork is done pulling the en re rope to thetop of the building?

(b) How much work is done pulling in the first m?

. A rope of length ℓ hangs over the edge of tall cliff. (As-sume the cliff is taller than the length of the rope.) Therope has a weight density of d lb/ .

(a) Howmuchwork is done pulling the en re rope to thetop of the cliff?

(b) What percentage of the total work is done pulling inthe first half of the rope?

(c) How much rope is pulled in when half of the totalwork is done?

. A m rope with mass density of . kg/m hangs over theedge of a m building. How much work is done pullingthe rope to the top?

. A crane li s a , lb load ver cally with a ” cableweighing . lb/ .

(a) How much work is done li ing the cable alone?

(b) How much work is done li ing the load alone?

(c) Could one conclude that the work done li ing the ca-ble is negligible compared to thework done li ing theload?

. A lb bag of sand is li ed uniformly in oneminute.Sand leaks from the bag at a rate of / lb/s. What is thetotal work done in li ing the bag?

. A boxweighing lb li s lb of sand ver cally . A crackin the box allows the sand to leak out such that lb of sandis in the box at the end of the trip. Assume the sand leakedout at a uniform rate. What is the total work done in li ingthe box and sand?

. A force of lb compresses a spring in. Howmuchworkis performed in compressing the spring?

. A force of N stretches a spring cm. How much work isperformed in stretching the spring?

. A force of lb compresses a spring froma natural length ofin to in. Howmuchwork is performed in compressing

the spring?

. A force of lb stretches a spring from a natural length ofin to in. How much work is performed in stretching the

spring?

. A force of N stretches a spring from a natural length ofcm to cm. How much work is performed in stretchingthe spring from a length of cm to cm?

. A force of f N stretches a spring dm from its natural length.How much work is performed in stretching the spring?

. A lb weight is a ached to a spring. The weight rests onthe spring, compressing the spring from a natural length of

to in.How much work is done in li ing the box . (i.e, thespring will be stretched beyond its natural length)?

. A lb weight is a ached to a spring. The weight rests onthe spring, compressing the spring from a natural length of

to in.How much work is done in li ing the box in (i.e, bringingthe spring back to its natural length)?

. A m tall cylindrical tank with radius of m is filled withm of gasoline, with a mass density of . kg/m . Com-pute the total work performed in pumping all the gasolineto the top of the tank.

. A cylindrical tank with a radius of is filled with wa-ter, which has a weight density of . lb/ . The water isto be pumped to a point above the top of the tank.

(a) How much work is performed in pumping all the wa-ter from the tank?

(b) How much work is performed in pumping of wa-ter from the tank?

(c) At what point is / of the total work done?

. A gasoline tanker is filled with gasoline with a weight den-sity of . lb/ . The dispensing valve at the base isjammed shut, forcing the operator to empty the tank viapumping the gas to a point above the top of the tank.Assume the tank is a perfect cylinder, long with a di-ameter of . . How much work is performed in pumpingall the gasoline from the tank?

. A fuel oil storage tank is deep with trapezoidal sides,at the top and at the bo om, and is wide (see

diagram below). Given that fuel oil weighs . lb/ , findthe work performed in pumping all the oil from the tank toa point above the top of the tank.

. A conical water tank is m deep with a top radius of m.(This is similar to Example . . .) The tank is filledwith purewater, with a mass density of kg/m .

(a) Find the work performed in pumping all the water tothe top of the tank.

(b) Find the work performed in pumping the top . mof water to the top of the tank.

(c) Find the work performed in pumping the top half ofthe water, by volume, to the top of the tank.

. A water tank has the shape of a truncated cone, with di-mensions given below, and is filledwithwaterwith aweightdensity of . lb/ . Find the work performed in pumpingall water to a point above the top of the tank.

. A water tank has the shape of an inverted pyramid, with di-mensions given below, and is filled with water with a massdensity of kg/m . Find the work performed in pump-ing all water to a point m above the top of the tank.

m

m

m

. A water tank has the shape of an truncated, inverted pyra-mid, with dimensions given below, and is filled with wa-ter with a mass density of kg/m . Find the work per-formed in pumping all water to a point m above the topof the tank.

m

m

m

m

m

.

..

..

(a)

...............

.

.5. 5

.

..

(b)

Figure . . : The cylindrical and rectan-gular tank in Example . . .

. Fluid Forces

. Fluid ForcesIn the unfortunate situa on of a car driving into a body of water, the conven-onal wisdom is that the water pressure on the doors will quickly be so great

that they will be effec vely unopenable. (Survival techniques suggest immedi-ately opening the door, rolling down or breaking the window, or wai ng un lthe water fills up the interior at which point the pressure is equalized and thedoor will open. See Mythbusters episode # to watch Adam Savage test theseop ons.)

How can this be true? How much force does it take to open the door ofa submerged car? In this sec on we will find the answer to this ques on byexamining the forces exerted by fluids.

We start with pressure, which is related to force by the following equa ons:

Pressure =ForceArea

⇔ Force = Pressure× Area.

In the context of fluids, we have the following defini on.

Defini on . . Fluid Pressure

Let w be the weight–density of a fluid. The pressure p exerted on anobject at depth d in the fluid is p = w · d.

We use this defini on to find the force exerted on a horizontal sheet by con-sidering the sheet’s area.

Example . . Compu ng fluid force

. A cylindrical storage tank has a radius of and holds of a fluidwith a weight–density of lb/ . (See Figure . . (a).) What is the forceexerted on the base of the cylinder by the fluid?

. A rectangular tank whose base is a square has a circular hatch at thebo om with a radius of . The tank holds of a fluid with a weight–density of lb/ . (See Figure . . (b).) What is the force exerted onthe hatch by the fluid?

S

. Using Defini on . . , we calculate that the pressure exerted on the cylin-der’s base isw · d = lb/ × = lb/ . The area of the base is

Notes:

..

}

∆yi

.ℓ(ci)

.

di

Figure . . : A thin, ver cally orientedplate submerged in a fluid with weight–density w.

Chapter Applica ons of Integra on

π · = π . So the force exerted by the fluid is

F = × π = lb.

Note that we effec vely just computed theweight of the fluid in the tank.

. The dimensions of the tank in this problem are irrelevant. All we are con-cerned with are the dimensions of the hatch and the depth of the fluid.Since the dimensions of the hatch are the same as the base of the tankin the previous part of this example, as is the depth, we see that the fluidforce is the same. That is, F = lb.A key concept to understand here is that we are effec vely measuring theweight of a column of water above the hatch. The size of the tankholding the fluid does not ma er.

The previous example demonstrates that compu ng the force exerted on ahorizontally oriented plate is rela vely easy to compute. What about a ver callyoriented plate? For instance, supposewe have a circular porthole located on theside of a submarine. How do we compute the fluid force exerted on it?

Pascal’s Principle states that the pressure exerted by a fluid at a depth isequal in all direc ons. Thus the pressure on any por on of a plate that isbelow the surface of water is the same no ma er how the plate is oriented.(Thus a hollow cube submerged at a great depth will not simply be “crushed”from above, but the sides will also crumple in. The fluid will exert force on allsides of the cube.)

So consider a ver cally oriented plate as shown in Figure . . submerged ina fluid with weight–densityw. What is the total fluid force exerted on this plate?We find this force by first approxima ng the force on small horizontal strips.

Let the top of the plate be at depth b and let the bo om be at depth a. (Fornow we assume that surface of the fluid is at depth , so if the bo om of theplate is under the surface, we have a = − . We will come back to this later.)We par on the interval [a, b] into n subintervals

a = y < y < · · · < yn+ = b,

with the i th subinterval having length ∆yi. The force Fi exerted on the plate inthe i th subinterval is Fi = Pressure× Area.

The pressure is depth ×w. We approximate the depth of this thin strip bychoosing any value di in [yi, yi+ ]; the depth is approximately−di. (Our conven-on has di being a nega ve number, so−di is posi ve.) For convenience, we let

di be an endpoint of the subinterval; we let di = yi.The area of the thin strip is approximately length×width. The width is∆yi.

The length is a func on of some y-value ci in the i th subinterval. We state the

Notes:

..

4

.

4

Figure . . : A thin plate in the shape ofan isosceles triangle in Example . . .

. Fluid Forces

length is ℓ(ci). Thus

Fi = Pressure× Area= −yi · w× ℓ(ci) ·∆yi.

To approximate the total force, we add up the approximate forces on each ofthe n thin strips:

F =n∑

i=

Fi ≈n∑

i=

−w · yi · ℓ(ci) ·∆yi.

This is, of course, another Riemann Sum. We can find the exact force by takinga limit as the subinterval lengths go to ; we evaluate this limit with a definiteintegral.

Key Idea . . Fluid Force on a Ver cally Oriented Plate

Let a ver cally oriented plate be submerged in a fluid with weight–density w where the top of the plate is at y = b and the bo om is aty = a. Let ℓ(y) be the length of the plate at y.

. If y = corresponds to the surface of the fluid, then the forceexerted on the plate by the fluid is

F =∫ b

aw · (−y) · ℓ(y) dy.

. In general, let d(y) represent the distance between the surface ofthe fluid and the plate at y. Then the force exerted on the plate bythe fluid is

F =∫ b

aw · d(y) · ℓ(y) dy.

Example . . Finding fluid forceConsider a thin plate in the shape of an isosceles triangle as shown in Figure. . submerged in water with a weight–density of . lb/ . If the bo om

of the plate is below the surface of the water, what is the total fluid forceexerted on this plate?

S We approach this problem in two different ways to illustratethe different ways Key Idea . . can be implemented. First we will let y =represent the surface of the water, then we will consider an alternate conven-on.

Notes:

..

( ,−6)

.

(− ,−6)

.( ,− )

.

y

.

y

.

x

.

−

.

−

...−

.

−8

.

−4

.

−

.

water line

.

d(y)

=−y

Figure . . : Sketching the triangularplate in Example . . with the conven-on that the water level is at y = .

..

( , 4)

.

(− , 4)

.

y

.

y

. x.−

.−

....

8

.

6

..

water line

.

d(y)

=−

y

Figure . . : Sketching the triangularplate in Example . . with the conven-on that the base of the triangle is at

( , ).

Chapter Applica ons of Integra on

. We let y = represent the surface of the water; therefore the bo om ofthe plate is at y = − . We center the triangle on the y-axis as shownin Figure . . . The depth of the plate at y is −y as indicated by the KeyIdea. We now consider the length of the plate at y.We need to find equa ons of the le and right edges of the plate. Theright hand side is a line that connects the points ( ,− ) and ( ,− ):that line has equa on x = / (y+ ). (Find the equa on in the familiary = mx+b format and solve for x.) Likewise, the le hand side is describedby the line x = − / (y + ). The total length is the distance betweenthese two lines: ℓ(y) = / (y+ )− (− / (y+ )) = y+ .

The total fluid force is then:

F =∫ −

−. (−y)(y+ ) dy

= . · ≈ . lb.

. Some mes it seems easier to orient the thin plate nearer the origin. Forinstance, consider the conven on that the bo om of the triangular plateis at ( , ), as shown in Figure . . . The equa ons of the le and righthand sides are easy to find. They are y = x and y = − x, respec vely,which we rewrite as x = / y and x = − / y. Thus the length func onis ℓ(y) = / y− (− / y) = y.As the surface of the water is above the base of the plate, we havethat the surface of the water is at y = . Thus the depth func on is thedistance between y = and y; d(y) = − y. We compute the totalfluid force as:

F =∫

. ( − y)(y) dy

≈ . lb.

The correct answer is, of course, independent of the placement of the plate inthe coordinate plane as long as we are consistent.

Example . . Finding fluid forceFind the total fluid force on a car door submerged up to the bo omof its windowin water, where the car door is a rectangle ” long and ” high (based on thedimensions of a Fiat Grande Punto.)

S The car door, as a rectangle, is drawn in Figure . . . Itslength is / and its height is . . We adopt the conven on that the top

Notes:

.. (3.3, 0).

(3.3,−2.25)

.

(0,−2.25)

.(0, 0) .

y

.

y

. x

Figure . . : Sketching a submerged cardoor in Example . . .

..y

.

y

. x.−

.−

...

−

.

−

...

5

.

water line

.

not to scale

.

d(y)

=5

−y

Figure . . : Measuring the fluid force onan underwater porthole in Example . . .

. Fluid Forces

of the door is at the surface of the water, both of which are at y = . Using theweight–density of water of . lb/ , we have the total force as

F =∫

− .

. (−y) / dy

=

∫

− .

− y dy

= − y∣∣∣− .

= . lb.

Most adults would find it very difficult to apply over lb of force to a cardoor while seated inside, making the door effec vely impossible to open. Thisis counter–intui ve as most assume that the door would be rela vely easy toopen. The truth is that it is not, hence the survival ps men oned at the begin-ning of this sec on.

Example . . Finding fluid forceAn underwater observa on tower is being built with circular viewing portholesenabling visitors to see underwater life. Each ver cally oriented porthole is tohave a diameter whose center is to be located underwater. Find thetotal fluid force exerted on each porthole. Also, compute the fluid force on ahorizontally oriented porthole that is under of water.

S We place the center of the porthole at the origin, meaningthe surface of thewater is at y = and the depth func onwill be d(y) = −y;see Figure . .

The equa on of a circle with a radius of . is x + y = . ; solving forx we have x = ±

√

. − y , where the posi ve square root corresponds tothe right side of the circle and the nega ve square root corresponds to the leside of the circle. Thus the length func on at depth y is ℓ(y) =

√

. − y .Integra ng on [− . , . ] we have:

F = .

∫ .

− .

( − y)√

. − y dy

= .

∫ .

− .

( √

. − y − y√

. − y)dy

=

∫ .

− .

(√

. − y)dy− .

∫ .

− .

(y√

. − y)dy.

Notes:

Chapter Applica ons of Integra on

The second integral above can be evaluated using subs tu on. Let u = . −ywith du = − y dy. The new bounds are: u(− . ) = and u( . ) = ; the newintegral will integrate from u = to u = , hence the integral is .

The first integral above finds the area of half a circle of radius . , thus thefirst integral evaluates to · π · . / = , . Thus the total fluid forceon a ver cally oriented porthole is , lb.

Finding the force on a horizontally oriented porthole ismore straigh orward:

F = Pressure× Area = . · × π · . = , lb.

That these two forces are equal is not coincidental; it turns out that the fluidforce applied to a ver cally oriented circle whose center is at depth d is thesame as force applied to a horizontally oriented circle at depth d.

We end this chapter with a reminder of the true skills meant to be developedhere. We are not truly concerned with an ability to find fluid forces or the vol-umes of solids of revolu on. Work done by a variable force is important, thoughmeasuring the work done in pulling a rope up a cliff is probably not.

What we are actually concerned with is the ability to solve certain problemsby first approxima ng the solu on, then refining the approxima on, then recog-nizing if/when this refining process results in a definite integral through a limit.Knowing the formulas found inside the special boxes within this chapter is ben-eficial as it helps solve problems found in the exercises, and other mathema calskills are strengthened by properly applying these formulas. However, more im-portantly, understand how each of these formulas was constructed. Each is theresult of a summa on of approxima ons; each summa on was a Riemann sum,allowing us to take a limit and find the exact answer through a definite integral.

The next chapter addresses an en rely different topic: sequences and series.In short, a sequence is a list of numbers, where a series is the summa on of a listof numbers. These seemingly–simple ideas lead to very powerful mathema cs.

Notes:

Exercises .Terms and Concepts

. State in your own words Pascal’s Principle.

. State in your own words how pressure is different fromforce.

Problems

In Exercises – , find the fluid force exerted on the givenplate, submerged in water with a weight density of .lb/ .

.

.

.

.

.

.

.

.

.

.

In Exercises – , the side of a container is pictured. Findthe fluid force exerted on this plate when the container is fullof:

. water, with a weight density of . lb/ , and

. concrete, with a weight density of lb/ .

.

.

y = x

.

y = − x

.

y = −√

− x

.

y =√

− x

.

y = −√

− x

. How deep must the center of a ver cally oriented circularplate with a radius of be submerged in water, with aweight density of . lb/ , for the fluid force on the plateto reach , lb?

. How deep must the center of a ver cally oriented squareplate with a side length of be submerged in water, witha weight density of . lb/ , for the fluid force on theplate to reach , lb?

Nota on: WeuseN to describe the set ofnatural numbers, that is, the integers , ,, …

Factorial: The expression ! refers to thenumber · · · = .

In general, n! = n·(n− )·(n− ) · · · · ,where n is a natural number.

We define ! = . While this does notimmediately make sense, it makes manymathema cal formulas work properly.

: S SThis chapter introduces sequences and series, important mathema cal con-struc ons that are useful when solving a large variety of mathema cal prob-lems. The content of this chapter is considerably different from the content ofthe chapters before it. While the material we learn here definitely falls underthe scope of “calculus,” we will make very li le use of deriva ves or integrals.Limits are extremely important, though, especially limits that involve infinity.

One of the problems addressed by this chapter is this: suppose we knowinforma on about a func on and its deriva ves at a point, such as f( ) = ,f ′( ) = , f ′′( ) = − , f ′′′( ) = , and so on. What can I say about f(x) itself?Is there any reasonable approxima on of the value of f( )? The topic of TaylorSeries addresses this problem, and allows us to make excellent approxima onsof func ons when limited knowledge of the func on is available.

. SequencesWe commonly refer to a set of events that occur one a er the other as a se-quence of events. In mathema cs, we use the word sequence to refer to anordered set of numbers, i.e., a set of numbers that “occur one a er the other.”

For instance, the numbers , , , , …, form a sequence. The order is impor-tant; the first number is , the second is , etc. It seems natural to seek a formulathat describes a given sequence, and o en this can be done. For instance, thesequence above could be described by the func on a(n) = n, for the values ofn = , , . . . To find the th term in the sequence, we would compute a( ).This leads us to the following, formal defini on of a sequence.

Defini on . . Sequence

A sequence is a func on a(n) whose domain is N. The range of asequence is the set of all dis nct values of a(n).

The terms of a sequence are the values a( ), a( ), …, which are usuallydenoted with subscripts as a , a , ….

A sequence a(n) is o en denoted as {an}.

.....

an =

n

n!

.......... n.

y

(a)

.....

an = + (− )n

.......... n.

y

(b)

.....

an =(− )n(n+ )/

n

......− .

/

.

/

.

n

.

y

(c)

Figure . . : Plo ng sequences in Exam-ple . . .

Chapter Sequences and Series

Example . . Lis ng terms of a sequenceList the first four terms of the following sequences.

. {an} =

{ n

n!

}

. {an} = { +(− )n} . {an} =

{(− )n(n+ )/

n

}

S

. a =!= ; a =

!= ; a =

!= ; a =

!=

We can plot the terms of a sequence with a sca er plot. The “x”-axis isused for the values of n, and the values of the terms are plo ed on they-axis. To visualize this sequence, see Figure . . (a).

. a = + (− ) = ; a = + (− ) = ;

a = +(− ) = ; a = +(− ) = . Note that the range of thissequence is finite, consis ng of only the values and . This sequence isplo ed in Figure . . (b).

. a =(− ) ( )/

= − ; a =(− ) ( )/

= −

a =(− ) ( )/

= a =(− ) ( )/

= ;

a =(− ) ( )/

= − .

We gave one extra term to begin to show the pa ern of signs is “−,−,+,+,−,−, . . .” due to the fact that the exponent of− is a special quadra c.This sequence is plo ed in Figure . . (c).

Example . . Determining a formula for a sequenceFind the nth term of the following sequences, i.e., find a func on that describeseach of the given sequences.

. , , , , , . . .

. ,− , ,− , ,− , . . .

. , , , , , , , . . .

. , , , , , . . .

Notes:

. Sequences

S Weshould first note that there is never exactly one func on thatdescribes a finite set of numbers as a sequence. There are many sequencesthat start with , then , as our first example does. We are looking for a simpleformula that describes the terms given, knowing there is possiblymore than oneanswer.

. Note how each term is more than the previous one. This implies a linearfunc on would be appropriate: a(n) = an = n+b for some appropriatevalue of b. As we want a = , we set b = − . Thus an = n− .

. First no ce how the sign changes from term to term. This is most com-monly accomplished bymul plying the terms by either (− )n or (− )n+ .Using (− )n mul plies the odd terms by (− ); using (− )n+ mul pliesthe even terms by (− ). As this sequence has nega ve even terms, wewill mul ply by (− )n+ .

A er this, we might feel a bit stuck as to how to proceed. At this point,we are just looking for a pa ern of some sort: what do the numbers , ,

, , etc., have in common? There are many correct answers, but theone that we’ll use here is that each is one more than a perfect square.That is, = + , = + , = + , etc. Thus our formula isan = (− )n+ (n + ).

. One who is familiar with the factorial func on will readily recognize thesenumbers. They are !, !, !, !, etc. Since our sequences start with n = ,we cannot write an = n!, for this misses the ! term. Instead, we shi by, and write an = (n− )!.

. This one may appear difficult, especially as the first two terms are thesame, but a li le “sleuthing” will help. No ce how the terms in the nu-merator are always mul ples of , and the terms in the denominator arealways powers of . Does something as simple as an = n

n work?

When n = , we see that we indeed get / as desired. When n = ,we get / = / . Further checking shows that this formula indeedmatches the other terms of the sequence.

A common mathema cal endeavor is to create a new mathema cal object(for instance, a sequence) and then apply previously knownmathema cs to thenew object. We do so here. The fundamental concept of calculus is the limit, sowe will inves gate what it means to find the limit of a sequence.

Notes:

Chapter Sequences and Series

Defini on . . Limit of a Sequence, Convergent, Divergent

Let {an} be a sequence and let L be a real number. Given any ε > , ifanm can be found such that |an − L| < ε for all n > m, then we say thelimit of {an}, as n approaches infinity, is L, denoted

limn→∞

an = L.

If limn→∞

an exists, we say the sequence converges; otherwise, the se-quence diverges.

This defini on states, informally, that if the limit of a sequence is L, then ifyou go far enough out along the sequence, all subsequent terms will be reallyclose to L. Of course, the terms “far enough” and “really close” are subjec veterms, but hopefully the intent is clear.

This defini on is reminiscent of the ε–δ proofs of Chapter . In that chapterwe developed other tools to evaluate limits apart from the formal defini on; wedo so here as well.

Theorem . . Limit of a Sequence

Let {an} be a sequence and let f(x) be a func onwhose domain containsthe posi ve real numbers where f(n) = an for all n in N.

If limx→∞

f(x) = L, then limn→∞

an = L.

Theorem . . allows us, in certain cases, to apply the tools developed inChapter to limits of sequences. Note two things not stated by the theorem:

. If limx→∞

f(x) does not exist, we cannot conclude that limn→∞

an does not exist.It may, or may not, exist. For instance, we can define a sequence {an} ={cos( πn)}. Let f(x) = cos( πx). Since the cosine func on oscillatesover the real numbers, the limit lim

x→∞f(x) does not exist.

However, for every posi ve integer n, cos( πn) = , so limn→∞

an = .

. If we cannot find a func on f(x) whose domain contains the posi ve realnumbers where f(n) = an for all n inN, we cannot conclude lim

n→∞an does

not exist. It may, or may not, exist.

Notes:

an =n − n +

n −

8

−

−

n

y

(a)

........

8

..− .

− .

.

.

..

n

.

y

.

an = cos n

(b)

.....

an =(− )n

n

.

5

..

5

..− .

− .5

.

.5

..

n

.

y

(c)

Figure . . : Sca er plots of the se-quences in Example . . .

. Sequences

Example . . Determining convergence/divergence of a sequenceDetermine the convergence or divergence of the following sequences.

. {an} =

{n − n+n −

}

. {an} = {cos n} . {an} =

{(− )n

n

}

S

. Using Theorem . . , we can state that limx→∞

x − x+x − = . (We could

have also directly applied l’Hôpital’s Rule.) Thus the sequence {an} con-verges, and its limit is . A sca er plot of every values of an is given inFigure . . (a). The values of an vary widely near n = , ranging fromabout− to , but as n grows, the values approach .

. The limit limx→∞

cos x does not exist as cos x oscillates (and takes on everyvalue in [− , ] infinitely many mes). Thus we cannot apply Theorem. . .

The fact that the cosine func on oscillates strongly hints that cos n, whenn is restricted toN, will also oscillate. Figure . . (b), where the sequenceis plo ed, implies that this is true. Because only discrete values of cosineare plo ed, it does not bear strong resemblance to the familiar cosinewave. The proof of the following statement is beyond the scope of thistext, but it is true: there are infinitely many integers n that are arbitrarily(i.e., very) close to an even mul ple of π, so that cos n ≈ . Similarly,there are infinitely many integers m that are arbitrarily close to an oddmul ple of π, so that cosm ≈ − . As the sequence takes on values nearand− infinitely many mes, we conclude that lim

n→∞an does not exist.

. We cannot actually apply Theorem . . here, as the func on f(x) =

(− )x/x is not well defined. (What does (− )√

mean? In actuality, thereis an answer, but it involves complex analysis, beyond the scope of thistext.)

Instead, we invoke the defini on of the limit of a sequence. By looking atthe plot in Figure . . (c), we would like to conclude that the sequenceconverges to L = . Let ε > be given. We can find a natural numberm

Notes:

Chapter Sequences and Series

such that /m < ε. Let n > m, and consider |an − L|:

|an − L| =∣∣∣∣

(− )n

n−∣∣∣∣

=n

<m

(since n > m)

< ε.

We have shown that by pickingm large enough, we can ensure that an isarbitrarily close to our limit, L = , hence by the defini on of the limit ofa sequence, we can say lim

n→∞an = .

In the previous example we used the defini on of the limit of a sequence todetermine the convergence of a sequence as we could not apply Theorem . . .In general, we like to avoid invoking the defini on of a limit, and the followingtheorem gives us tool that we could use in that example instead.

Theorem . . Absolute Value Theorem

Let {an} be a sequence. If limn→∞

|an| = , then limn→∞

an =

Example . . Determining the convergence/divergence of a sequenceDetermine the convergence or divergence of the following sequences.

. {an} =

{(− )n

n

}

. {an} =

{(− )n(n+ )

n

}

S

. This appeared in Example . . . We want to apply Theorem . . , so con-sider the limit of {|an|}:

limn→∞

|an| = limn→∞

∣∣∣∣

(− )n

n

∣∣∣∣

= limn→∞ n

= .

Since this limit is , we can apply Theorem . . and state that limn→∞

an =.

Notes:

.....

an =(− )n(n + )

n

.

5

..

5

..

−

.−

...

n

.

y

Figure . . : A plot of a sequence in Ex-ample . . , part .

. Sequences

. Because of the alterna ng nature of this sequence (i.e., every other term

ismul plied by− ), we cannot simply look at the limit limx→∞

(− )x(x+ )

x.

We can try to apply the techniques of Theorem . . :

limn→∞

|an| = limn→∞

∣∣∣∣

(− )n(n+ )

n

∣∣∣∣

= limn→∞

n+n

= .

Wehave concluded thatwhenwe ignore the alterna ng sign, the sequenceapproaches . This means we cannot apply Theorem . . ; it states thethe limit must be in order to conclude anything.Since we know that the signs of the terms alternate and we know thatthe limit of |an| is , we know that as n approaches infinity, the termswill alternate between values close to and − , meaning the sequencediverges. A plot of this sequence is given in Figure . . .

We con nue our study of the limits of sequences by considering some of theproper es of these limits.

Theorem . . Proper es of the Limits of Sequences

Let {an} and {bn} be sequences such that limn→∞

an = L, limn→∞

bn = K, andlet c be a real number.

. limn→∞

(an ± bn) = L± K

. limn→∞

(an · bn) = L · K

. limn→∞

(an/bn) = L/K, K ̸=

. limn→∞

c · an = c · L

Example . . Applying proper es of limits of sequencesLet the following sequences, and their limits, be given:

• {an} =

{n+n

}

, and limn→∞

an = ;

• {bn} =

{(

+n

)n}

, and limn→∞

bn = e; and

• {cn} ={n · sin( /n)

}, and lim

n→∞cn = .

Notes:

Chapter Sequences and Series

Evaluate the following limits.

. limn→∞

(an + bn) . limn→∞

(bn · cn) . limn→∞

( · an)

S We will use Theorem . . to answer each of these.

. Since limn→∞

an = and limn→∞

bn = e, we conclude that limn→∞

(an + bn) =

+ e = e. So even though we are adding something to each term of thesequence bn, we are adding something so small that the final limit is thesame as before.

. Since limn→∞

bn = e and limn→∞

cn = , we conclude that limn→∞

(bn · cn) =

e · = e.

. Since limn→∞

an = , we have limn→∞

an = · = . It does notma er that wemul ply each term by ; the sequence s ll approaches. (It just takes longer to get close to .)

There is more to learn about sequences than just their limits. We will alsostudy their range and the rela onships terms have with the terms that follow.We start with some defini ons describing proper es of the range.

Defini on . . Bounded and Unbounded Sequences

A sequence {an} is said to be bounded if there exist real numbers mandM such thatm < an < M for all n in N.

A sequence {an} is said to be unbounded if it is not bounded.

A sequence {an} is said to be bounded above if there exists an M suchthat an < M for all n in N; it is bounded below if there exists anm suchthatm < an for all n in N.

It follows from this defini on that an unbounded sequencemay be boundedabove or bounded below; a sequence that is both bounded above and below issimply a bounded sequence.

Example . . Determining boundedness of sequencesDetermine the boundedness of the following sequences.

. {an} =

{

n

}

. {an} = { n}

Notes:

.....

an =

n

............

/

.

/

.

/

.n

.

y

(a)

.....

an =

n

..4

.6

.8

... n.

y

(b)

Figure . . : A plot of {an} = { /n} and{an} = { n} from Example . . .

Note: Keep in mind what Theorem . .does not say. It does not say thatbounded sequences must converge, nordoes it say that if a sequence does notconverge, it is not bounded.

. Sequences

S

. The terms of this sequence are always posi ve but are decreasing, so wehave < an < for all n. Thus this sequence is bounded. Figure . . (a)illustrates this.

. The terms of this sequence obviously grow without bound. However, it isalso true that these terms are all posi ve, meaning < an. Thus we cansay the sequence is unbounded, but also bounded below. Figure . . (b)illustrates this.

The previous example produces some interes ng concepts. First, we canrecognize that the sequence { /n} converges to . This says, informally, that“most” of the terms of the sequence are “really close” to . This implies that thesequence is bounded, using the following logic. First, “most” terms are near ,so we could find some sort of bound on these terms (using Defini on . . , thebound is ε). That leaves a “few” terms that are not near (i.e., a finite numberof terms). A finite list of numbers is always bounded.

This logic implies that if a sequence converges, it must be bounded. This isindeed true, as stated by the following theorem.

Theorem . . Convergent Sequences are Bounded

Let {an} be a convergent sequence. Then {an} is bounded.

In Example . . we saw the sequence {bn} ={( + /n)n

}, where it was

stated that limn→∞

bn = e. (Note that this is simply resta ng part of Theorem. . .) Even though it may be difficult to intui vely grasp the behavior of this

sequence, we know immediately that it is bounded.

Another interes ng concept to come out of Example . . again involvesthe sequence { /n}. We stated, without proof, that the terms of the sequencewere decreasing. That is, that an+ < an for all n. (This is easy to show. Clearlyn < n + . Taking reciprocals flips the inequality: /n > /(n + ). This is thesame as an > an+ .) Sequences that either steadily increase or decrease areimportant, so we give this property a name.

Notes:

Note: It is some mes useful to calla monotonically increasing sequencestrictly increasing if an < an+ for alln; i.e, we remove the possibility thatsubsequent terms are equal.A similar statement holds for strictly de-creasing.

.....

an =n +

n

.5

.... n.

y

Figure . . : A plot of {an} = {(n +)/n} in Example . . .

Chapter Sequences and Series

Defini on . . Monotonic Sequences

. A sequence {an} is monotonically increasing if an ≤ an+ for alln, i.e.,

a ≤ a ≤ a ≤ · · · an ≤ an+ · · ·

. A sequence {an} is monotonically decreasing if an ≥ an+ for alln, i.e.,

a ≥ a ≥ a ≥ · · · an ≥ an+ · · ·

. A sequence ismonotonic if it is monotonically increasing ormono-tonically decreasing.

Example . . Determining monotonicityDetermine the monotonicity of the following sequences.

. {an} =

{n+n

}

. {an} =

{n +

n+

}

. {an} =

{n −

n − n+

}

. {an} =

{nn!

}

S In each of the following, wewill examine an+ −an. If an+ −an ≥ , we conclude that an ≤ an+ and hence the sequence is increasing. Ifan+ − an ≤ , we conclude that an ≥ an+ and the sequence is decreasing. Ofcourse, a sequence need not be monotonic and perhaps neither of the abovewill apply.

We also give a sca er plot of each sequence. These are useful as they sug-gest a pa ern of monotonicity, but analy c work should be done to confirm agraphical trend.

. an+ − an =n+n+

− n+n

=(n+ )(n)− (n+ )

(n+ )n

=−

n(n+ )

< for all n.

Since an+ −an < for all n, we conclude that the sequence is decreasing.

Notes:

.....

an =n +

n +

. 5..

5

..n

.

y

(a)

.....

an =n − 9

n − n + 6

. 5..

5

..

5

. n.

y

(b)

.....

an =nn!

.5

..

.5

..

.5

.. n.

y

(c)

Figure . . : Plots of sequences in Exam-ple . . .

. Sequences

. an+ − an =(n+ ) +

n+− n +

n+

=

((n+ ) +

)(n+ )− (n + )(n+ )

(n+ )(n+ )

=n + n+

(n+ )(n+ )

> for all n.Since an+ − an > for all n, we conclude the sequence is increasing; seeFigure . . (a).

. We can clearly see in Figure . . (b), where the sequence is plo ed, thatit is not monotonic. However, it does seem that a er the first termsit is decreasing. To understand why, perform the same analysis as donebefore:

an+ − an =(n+ ) −

(n+ ) − (n+ ) +− n −

n − n+

=n + n−n − n+

− n −n − n+

=(n + n− )(n − n+ )− (n − )(n − n+ )

(n − n+ )(n − n+ )

=− n + n−

(n − n+ )(n − n+ ).

We want to know when this is greater than, or less than, . The denomi-nator is always posi ve, therefore we are only concerned with the numer-ator. For small values of n, the numerator is posi ve. As n grows large,the numerator is dominated by − n , meaning the en re frac on willbe nega ve; i.e., for large enough n, an+ − an < . Using the quadra cformula we can determine that the numerator is nega ve for n ≥ .In short, the sequence is simply not monotonic, though it is useful to notethat for n ≥ , the sequence is monotonically decreasing.

. Again, the plot in Figure . . (c) shows that the sequence is not mono-tonic, but it suggests that it is monotonically decreasing a er the firstterm. We perform the usual analysis to confirm this.

an+ − an =(n+ )

(n+ )!− n

n!

=(n+ ) − n (n+ )

(n+ )!

=−n + n+

(n+ )!

Notes:

Chapter Sequences and Series

When n = , the above expression is > ; for n ≥ , the above expres-sion is < . Thus this sequence is not monotonic, but it is monotonicallydecreasing a er the first term.

Knowing that a sequence is monotonic can be useful. Consider, for example,a sequence that is monotonically decreasing and is bounded below. We knowthe sequence is always ge ng smaller, but that there is a bound to how small itcan become. This is enough to prove that the sequence will converge, as statedin the following theorem.

Theorem . . Bounded Monotonic Sequences are Convergent

. Let {an} be a monotonically increasing sequence that is boundedabove. Then {an} converges.

. Let {an} be a monotonically decreasing sequence that is boundedbelow. Then {an} converges.

Consider once again the sequence {an} = { /n}. It is easy to show it ismonotonically decreasing and that it is always posi ve (i.e., bounded below by). Therefore we can conclude by Theorem . . that the sequence converges.

We already knew this by other means, but in the following sec on this theoremwill become very useful.

We can replace Theorem . . with the statement “Let {an} be a bounded,monotonic sequence. Then {an} converges; i.e., lim

n→∞an exists.” We leave it to

the reader in the exercises to show the theorem and the above statement areequivalent.

Sequences are a great source of mathema cal inquiry. The On-Line Ency-clopedia of Integer Sequences (http://oeis.org) contains thousands of se-quences and their formulae. (As of this wri ng, there are , sequencesin the database.) Perusing this database quickly demonstrates that a single se-quence can represent several different “real life” phenomena.

Interes ng as this is, our interest actually lies elsewhere. We are more in-terested in the sum of a sequence. That is, given a sequence {an}, we are veryinterested in a +a +a + · · · . Of course, one might immediately counter with“Doesn’t this just add up to ‘infinity’?” Many mes, yes, but there are many im-portant cases where the answer is no. This is the topic of series, which we beginto inves gate in the next sec on.

Notes:

Exercises .Terms and Concepts. Use your own words to define a sequence.

. The domain of a sequence is the numbers.

. Use your own words to describe the range of a sequence.

. Describe what it means for a sequence to be bounded.

ProblemsIn Exercises – , give the first five terms of the given se-quence.

. {an} =

{ n

(n+ )!

}

. {bn} =

{(

−)n}

. {cn} =

{

− nn+

n+

}

. {dn} =

{

√(

(

+√)n

−(

−√)n)}

In Exercises – , determine the nth term of the given se-quence.

. , , , , , . . .

. , − , , − , . . .

. , , , , , . . .

. , , , , , , . . .

In Exercises – , use the following informa on to deter-mine the limit of the given sequences.

• {an} =

{ n −n

}

; limn→∞

an =

• {bn} =

{(

+n

)n}

; limn→∞

bn = e

• {cn} = {sin( /n)}; limn→∞

cn =

. {an} =

{ n −· n

}

. {an} = { bn − an}

. {an} =

{

sin( /n)(

+n

)n}

. {an} =

{

(

+n

) n}

In Exercises – , determine whether the sequence con-verges or diverges. If convergent, give the limit of the se-quence.

. {an} =

{

(− )nn

n+

}

. {an} =

{

n − n+n +

}

. {an} =

{ n

n

}

. {an} =

{

n−n

− nn−

}

, n ≥

. {an} = {ln(n)}

. {an} =

{

n√n +

}

. {an} =

{(

+n

)n}

. {an} =

{

−n

}

. {an} =

{

(− )n+

n

}

. {an} =

{

. n

n

}

. {an} =

{

nn+

}

. {an} =

{

(− )nn

n −

}

In Exercises – , determine whether the sequence isbounded, bounded above, bounded below, or none of theabove.

. {an} = {sin n}

. {an} = {tan n}

. {an} =

{

(− )nn−n

}

. {an} =

{

n −n

}

. {an} = {n cos n}

. {an} = { n − n!}

In Exercises – , determine whether the sequence ismonotonically increasing or decreasing. If it is not, determineif there is anm such that it is monotonic for all n ≥ m.

. {an} =

{

nn+

}

. {an} =

{

n − n+n

}

. {an} =

{

(− )nn

}

. {an} =

{

nn

}

Exercises – explore further the theory of sequences.

. Prove Theorem . . ; that is, use the defini on of the limitof a sequence to show that if lim

n→∞|an| = , then lim

n→∞an =

.

. Let {an} and {bn} be sequences such that limn→∞

an = L andlim

n→∞bn = K.

(a) Show that if an < bn for all n, then L ≤ K.

(b) Give an example where L = K.

. Prove the Squeeze Theorem for sequences: Let {an} and{bn} be such that lim

n→∞an = L and lim

n→∞bn = L, and let

{cn} be such that an ≤ cn ≤ bn for all n. Then limn→∞

cn = L

. Prove the statement “Let {an} be a bounded, monotonicsequence. Then {an} converges; i.e., lim

n→∞an exists.” is

equivalent to Theorem . . . That is,

(a) Show that if Theorem . . is true, then above state-ment is true, and

(b) Show that if the above statement is true, then Theo-rem . . is true.

. Infinite Series

. Infinite SeriesGiven the sequence {an} = { / n} = / , / , / , . . ., consider the follow-ing sums:

a = / = /a + a = / + / = /

a + a + a = / + / + / = /a + a + a + a = / + / + / + / = /

In general, we can show that

a + a + a + · · ·+ an =n −

n = − n .

Let Sn be the sum of the first n terms of the sequence { / n}. From the above,we see that S = / , S = / , etc. Our formula at the end shows that Sn =− / n.Now consider the following limit: lim

n→∞Sn = lim

n→∞

(− / n) = . This limit

can be interpreted as saying something amazing: the sum of all the terms of thesequence { / n} is 1.

This example illustrates some interes ng concepts that we explore in thissec on. We begin this explora on with some defini ons.

Defini on . . Infinite Series, nth Par al Sums, Convergence,Divergence

Let {an} be a sequence.

. The sum∞∑

n=

an is an infinite series (or, simply series).

. Let Sn =n∑

i=

ai ; the sequence {Sn} is the sequence of nth par al

sums of {an}.

. If the sequence {Sn} converges to L, we say the series∞∑

n=

an con-

verges to L, and we write∞∑

n=

an = L.

. If the sequence {Sn} diverges, the series∞∑

n=

an diverges.

Notes:

Chapter Sequences and Series

Using our new terminology, we can state that the series∞∑

n=

/ n converges,

and∞∑

n=

/ n = .

We will explore a variety of series in this sec on. We start with two seriesthat diverge, showing how we might discern divergence.

Example . . Showing series diverge

. Let {an} = {n }. Show∞∑

n=

an diverges.

. Let {bn} = {(− )n+ }. Show∞∑

n=

bn diverges.

S

. Consider Sn, the nth par al sum.

Sn = a + a + a + · · ·+ an= + + · · ·+ n .

By Theorem . . , this is

=n(n+ )( n+ )

.

Since limn→∞

Sn = ∞, we conclude that the series∞∑

n=

n diverges. It is

instruc ve to write∞∑

n=

n = ∞ for this tells us how the series diverges: it

grows without bound.

A sca er plot of the sequences {an} and {Sn} is given in Figure . . (a).The terms of {an} are growing, so the terms of the par al sums {Sn} aregrowing even faster, illustra ng that the series diverges.

Notes:

.....5

..... n.

y

.

..an. Sn

(a)

.....

5

.

10

.−1.

−0.5

.

0.5

.

1

.

n

.

y

...bn. Sn

(b)

Figure . . : Sca er plots rela ng to Ex-ample . . .

. Infinite Series

. The sequence {bn} starts with , − , , − , . . .. Consider some of thepar al sums Sn of {bn}:

S =

S =

S =

S =

This pa ern repeats; we find that Sn =

{n is oddn is even . As {Sn} oscil-

lates, repea ng , , , , . . ., we conclude that limn→∞

Sn does not exist,

hence∞∑

n=

(− )n+ diverges.

A sca er plot of the sequence {bn} and the par al sums {Sn} is given inFigure . . (b). When n is odd, bn = Sn so the marks for bn are drawnoversized to show they coincide.

While it is important to recognize when a series diverges, we are generallymore interested in the series that converge. In this sec on we will demonstratea few general techniques for determining convergence; later sec ons will delvedeeper into this topic.

Geometric Series

One important type of series is a geometric series.

Defini on . . Geometric Series

A geometric series is a series of the form∞∑

n=

rn = + r+ r + r + · · ·+ rn + · · ·

Note that the index starts at n = , not n = .

We started this sec on with a geometric series, although we dropped thefirst term of . One reason geometric series are important is that they have niceconvergence proper es.

Notes:

......4

.6

.8

....

n

.

y

...an. Sn

Figure . . : Sca er plots rela ng to theseries in Example . . .

Chapter Sequences and Series

Theorem . . Geometric Series Test

Consider the geometric series∞∑

n=

rn.

. The nth par al sum is: Sn =− r n+

− r, r ̸= .

. The series converges if, and only if, |r| < . When |r| < ,

∞∑

n=

rn = − r.

According to Theorem . . , the series

∞∑

n=n =

∞∑

n=

( )

= + + + · · ·

converges as r = / , and∞∑

n=n = − /

= . This concurs with our intro-

ductory example; while there we got a sum of , we skipped the first term of .

Example . . Exploring geometric seriesCheck the convergence of the following series. If the series converges, find itssum.

.∞∑

n=

( )n

.∞∑

n=

(− )n

.∞∑

n=

n

S

. Since r = / < , this series converges. By Theorem . . , we have that

∞∑

n=

( )n

= − /= .

However, note the subscript of the summa on in the given series: we areto start with n = . Therefore we subtract off the first two terms, giving:

∞∑

n=

( )n

= − − = .

This is illustrated in Figure . . .

Notes:

........

8

..−.

− .

.

.

..

n

.

y

...an. Sn

(a)

.........

,

. n.

y

.

..an. Sn

(b)

Figure . . : Sca er plots rela ng to theseries in Example . . .

Note: Theorem . . assumes that an +b ̸= for all n. If an + b = forsomen, then of course the series does notconverge regardless of p as not all of theterms of the sequence are defined.

. Infinite Series

. Since |r| = / < , this series converges, and by Theorem . . ,

∞∑

n=

(− )n

= − (− / )= .

The par al sums of this series are plo ed in Figure . . (a). Note howthe par al sums are not purely increasing as some of the terms of thesequence {(− / )n} are nega ve.

. Since r > , the series diverges. (This makes “common sense”; we expectthe sum

+ + + + + + · · ·to diverge.) This is illustrated in Figure . . (b).

p–Series

Another important type of series is the p-series.

Defini on . . p–Series, General p–Series

. A p–series is a series of the form∞∑

n=np

, where p > .

. A general p–series is a series of the form∞∑

n=(an+ b)p

, where p > and a, b are real numbers.

Like geometric series, one of the nice things about p–series is that they haveeasy to determine convergence proper es.

Theorem . . p–Series Test

A general p–series∞∑

n=(an+ b)p

will converge if, and only if, p > .

Notes:

Chapter Sequences and Series

Example . . Determining convergence of seriesDetermine the convergence of the following series.

.∞∑

n=n

.∞∑

n=n

.∞∑

n=

√n

.∞∑

n=

(− )n

n

.∞∑

n= ( n− )

.∞∑

n=n

S

. This is a p–series with p = . By Theorem . . , this series diverges.

This series is a famous series, called the Harmonic Series, so named be-cause of its rela onship to harmonics in the study of music and sound.

. This is a p–series with p = . By Theorem . . , it converges. Note thatthe theorem does not give a formula by which we can determine whatthe series converges to; we just know it converges. A famous, unexpectedresult is that this series converges to π / .

. This is a p–series with p = / ; the theorem states that it diverges.

. This is not a p–series; the defini on does not allow for alterna ng signs.Therefore we cannot apply Theorem . . . (Another famous result statesthat this series, the Alterna ng Harmonic Series, converges to ln .)

. This is a general p–series with p = , therefore it converges.

. This is not a p–series, but a geometric series with r = / . It converges.

Later sec ons will provide tests by which we can determine whether or nota given series converges. This, in general, is much easier than determiningwhata given series converges to. There are many cases, though, where the sum canbe determined.

Example . . Telescoping series

Evaluate the sum∞∑

n=

(

n−

n+

)

.

S It will help to write down some of the first few par al sums

Notes:

........ 8..

.

..n

.

y

...an. Sn

Figure . . : Sca er plots rela ng to theseries of Example . . .

. Infinite Series

of this series.

S = − = −

S =

(

−)

+

(

−)

= −

S =

(

−)

+

(

−)

+

(

−)

= −

S =

(

−)

+

(

−)

+

(

−)

+

(

−)

= −

Note how most of the terms in each par al sum are canceled out! In general,we see that Sn = −

n+. The sequence {Sn} converges, as lim

n→∞Sn =

limn→∞

(

−n+

)

= , and so we conclude that∞∑

n=

(

n−

n+

)

= . Par-

al sums of the series are plo ed in Figure . . .

The series in Example . . is an example of a telescoping series. Informally,a telescoping series is one in which most terms cancel with preceding or follow-ing terms, reducing the number of terms in each par al sum. The par al sum Sndid not contain n terms, but rather just two: and /(n+ ).

When possible, seek away towrite an explicit formula for the nth par al sumSn. This makes evalua ng the limit lim

n→∞Sn much more approachable. We do so

in the next example.

Example . . Evalua ng seriesEvaluate each of the following infinite series.

.∞∑

n=n + n

.∞∑

n=

ln(n+n

)

S

. We can decompose the frac on /(n + n) as

n + n=

n−

n+.

(See Sec on . , Par al Frac onDecomposi on, to recall how this is done,if necessary.)

Notes:

........ 8..

.

..

.

. n.

y

.

..an. Sn

(a)

..... 5...

4

.n

.

y

...an. Sn

(b)

Figure . . : Sca er plots rela ng to theseries in Example . . .

Chapter Sequences and Series

Expressing the terms of {Sn} is now more instruc ve:

S = − = −

S =

(

−)

+

(

−)

= + − −

S =

(

−)

+

(

−)

+

(

−)

= + − −

S =

(

−)

+

(

−)

+

(

−)

+

(

−)

= + − −

S =

(

−)

+

(

−)

+

(

−)

+

(

−)

+

(

−)

= + − −

We again have a telescoping series. In each par al sum, most of the termscancel and we obtain the formula Sn = + −

n+−

n+. Taking

limits allows us to determine the convergence of the series:

limn→∞

Sn = limn→∞

(

+ −n+

−n+

)

= , so∞∑

n=n + n

= .

This is illustrated in Figure . . (a).

. We begin by wri ng the first few par al sums of the series:

S = ln ( )

S = ln ( ) + ln( )

S = ln ( ) + ln( )

+ ln( )

S = ln ( ) + ln( )

+ ln( )

+ ln( )

At first, this does not seem helpful, but recall the logarithmic iden ty:ln x+ ln y = ln(xy). Applying this to S gives:

S = ln ( )+ ln( )

+ ln( )

+ ln( )

= ln(

· · ·)

= ln ( ) .

We can conclude that {Sn} ={ln(n+ )

}. This sequence does not con-

verge, as limn→∞

Sn = ∞. Therefore∞∑

n=

ln(n+n

)

= ∞; the series di-

verges. Note in Figure . . (b) how the sequence of par al sums grows

Notes:

. Infinite Series

slowly; a er terms, it is not yet over . Graphically we may be fooledinto thinking the series converges, but our analysis above shows that itdoes not.

We are learning about a new mathema cal object, the series. As done be-fore, we apply “old” mathema cs to this new topic.

Theorem . . Proper es of Infinite Series

Let∞∑

n=

an = L,∞∑

n=

bn = K, and let c be a constant.

. Constant Mul ple Rule:∞∑

n=

c · an = c ·∞∑

n=

an = c · L.

. Sum/Difference Rule:∞∑

n=

(an ± bn

)=

∞∑

n=

an ±∞∑

n=

bn = L± K.

Before using this theorem, we provide a few “famous” series.

Key Idea . . Important Series

.∞∑

n=n!

= e. (Note that the index starts with n = .)

.∞∑

n=n

=π

.

.∞∑

n=

(− )n+

n=

π.

.∞∑

n=

(− )n

n+=

π.

.∞∑

n=n

diverges. (This is called the Harmonic Series.)

.∞∑

n=

(− )n+

n= ln . (This is called the Alterna ng Harmonic Series.)

Notes:

..........− .

.

.

.

n

.

y

. ..an. Sn

(a)

.....5

..

5

.

,

.

,5

.

,

. n.

y

.

..an. Sn

(b)

Figure . . : Sca er plots rela ng to theseries in Example . . .

Chapter Sequences and Series

Example . . Evalua ng seriesEvaluate the given series.

.∞∑

n=

(− )n+(n − n

)

n.

∞∑

n=n!

. + + + + · · ·

S

. We start by using algebra to break the series apart:

∞∑

n=

(− )n+(n − n

)

n=

∞∑

n=

((− )n+ n

n− (− )n+ n

n

)

=

∞∑

n=

(− )n+

n−

∞∑

n=

(− )n+

n

= ln( )− π ≈ − . .

This is illustrated in Figure . . (a).

. This looks very similar to the series that involves e in Key Idea . . . Note,however, that the series given in this example starts with n = and notn = . The first term of the series in the Key Idea is / ! = , so we willsubtract this from our result below:

∞∑

n=n!

= ·∞∑

n=n!

= · (e− ) ≈ . .

This is illustrated in Figure . . (b). The graph shows how this par cularseries converges very rapidly.

. The denominators in each term are perfect squares; we are adding∞∑

n=n

(note we start with n = , not n = ). This series will converge. Using the

Notes:

. Infinite Series

formula from Key Idea . . , we have the following:∞∑

n=n

=∑

n=n

+

∞∑

n=n

∞∑

n=n

−∑

n=n

=

∞∑

n=n

π −(

+ +

)

=

∞∑

n=n

π − =

∞∑

n=n

. ≈∞∑

n=n

It may take a while before one is comfortable with this statement, whosetruth lies at the heart of the study of infinite series: it is possible that the sum ofan infinite list of nonzero numbers is finite. We have seen this repeatedly in thissec on, yet it s ll may “take some ge ng used to.”

As one contemplates the behavior of series, a few facts become clear.

. In order to add an infinite list of nonzero numbers and get a finite result,“most” of those numbers must be “very near” .

. If a series diverges, it means that the sum of an infinite list of numbers isnot finite (it may approach±∞ or it may oscillate), and:

(a) The series will s ll diverge if the first term is removed.(b) The series will s ll diverge if the first terms are removed.(c) The series will s ll diverge if the first , , terms are removed.(d) The series will s ll diverge if any finite number of terms from any-

where in the series are removed.

These concepts are very important and lie at the heart of the next two the-orems.

Theorem . . nth–Term Test for Divergence

Consider the series∞∑

n=

an. If limn→∞

an ̸= , then∞∑

n=

an diverges.

Notes:

Chapter Sequences and Series

Important! This theorem does not state that if limn→∞

an = then∞∑

n=

an

converges. The standard example of this is the Harmonic Series, as given in KeyIdea . . . TheHarmonic Sequence, { /n}, converges to ; theHarmonic Series,∞∑

n=n, diverges.

Looking back, we can apply this theorem to the series in Example . . . Inthat example, the nth terms of both sequences do not converge to , thereforewe can quickly conclude that each series diverges.

One can rewrite Theorem . . to state “If a series converges, then the un-derlying sequence converges to .” While it is important to understand the truthof this statement, in prac ce it is rarely used. It is generally far easier to provethe convergence of a sequence than the convergence of a series.

Theorem . . Infinite Nature of Series

The convergence or divergence of an infinite series remains unchangedby the addi on or subtrac on of any finite number of terms. That is:

. A divergent series will remain divergent with the addi on or sub-trac on of any finite number of terms.

. A convergent series will remain convergent with the addi on orsubtrac on of any finite number of terms. (Of course, the sumwilllikely change.)

Consider once more the Harmonic Series∞∑

n=nwhich diverges; that is, the

sequence of par al sums {Sn} grows (very, very slowly) without bound. Onemight think that by removing the “large” terms of the sequence that perhapsthe series will converge. This is simply not the case. For instance, the sum of thefirst million terms of the Harmonic Series is about . . Removing the firstmillion terms from the Harmonic Series changes the nth par al sums, effec velysubtrac ng . from the sum. However, a sequence that is growing withoutbound will s ll grow without bound when . is subtracted from it.

The equa ons below illustrate this. The first line shows the infinite sum ofthe Harmonic Series split into the sum of the first million terms plus the sumof “everything else.” The next equa on shows us subtrac ng these first mil-lion terms from both sides. The final equa on employs a bit of “psuedo–math”:

Notes:

. Infinite Series

subtrac ng . from “infinity” s ll leaves one with “infinity.”

∞∑

n=n =

, ,∑

n=n

+

∞∑

n= , ,n

∞∑

n=n −

, ,∑

n=n

=

∞∑

n= , ,n

∞ − . = ∞.

This sec on introduced us to series and defined a few special types of serieswhose convergence proper es are well known: we know when a p-series ora geometric series converges or diverges. Most series that we encounter arenot one of these types, but we are s ll interested in knowing whether or notthey converge. The next three sec ons introduce tests that help us determinewhether or not a given series converges.

Notes:

Exercises .Terms and Concepts

. Use your own words to describe how sequences and seriesare related.

. Use your own words to define a par al sum.

. Given a series∞∑

n=

an, describe the two sequences related

to the series that are important.

. Use your own words to explain what a geometric series is.

. T/F: If {an} is convergent, then∞∑

n=

an is also convergent.

. T/F: If {an} converges to , then∞∑

n=

an converges.

Problems

In Exercises – , a series∞∑

n=

an is given.

(a) Give the first par al sums of the series.

(b) Give a graph of the first terms of an and Sn on thesame axes.

.∞∑

n=

(− )n

n

.∞∑

n=n

.∞∑

n=

cos(πn)

.∞∑

n=

n

.∞∑

n=n!

.∞∑

n=n

.∞∑

n=

(

−)n

.∞∑

n=

( )n

In Exercises – , use Theorem . . to show the given se-ries diverges.

.∞∑

n=

nn(n+ )

.∞∑

n=

n

n

.∞∑

n=

n!n

.∞∑

n=

n − nn + n

.∞∑

n=

n +n+

.∞∑

n=

(

+n

)n

In Exercises – , state whether the given series convergesor diverges.

.∞∑

n=n

.∞∑

n=n

.∞∑

n=

n

n

.∞∑

n=

n−

.∞∑

n=

√n

.∞∑

n=n!

. T/F: If {an} converges to , then∞∑

n=

an converges.

.∞∑

n=( n+ )

.∞∑

n=n

.∞∑

n=n−

In Exercises – , a series is given.(a) Find a formula for Sn, the nth par al sum of the series.(b) Determine whether the series converges or diverges.

If it converges, state what it converges to.

.∞∑

n=n

.∞∑

n=

. + + + + · · ·

.∞∑

n=

(− )nn

.∞∑

n=n

.∞∑

n=

e−n

. − + − + + · · ·

.∞∑

n=n(n+ )

.∞∑

n=n(n+ )

.∞∑

n=( n− )( n+ )

.∞∑

n=

ln(

nn+

)

.∞∑

n=

n+n (n+ )

. · + · + · + · + · · ·

. +

(

+

)

+

(

+

)

+

(

+

)

+ · · ·

.∞∑

n=n −

.∞∑

n=

(

sin)n

. Break theHarmonic Series into the sumof the odd and eventerms:

∞∑

n=n=

∞∑

n=n− +

∞∑

n=n.

The goal is to show that each of the series on the right di-verge.

(a) Show why∞∑

n=n− >

∞∑

n=n.

(Compare each nth par al sum.)

(b) Show why∞∑

n=n− < +

∞∑

n=n

(c) Explain why (a) and (b) demonstrate that the seriesof odd terms is convergent, if, and only if, the seriesof even terms is also convergent. (That is, show bothconverge or both diverge.)

(d) Explain why knowing the Harmonic Series is diver-gent determines that the even and odd series are alsodivergent.

. Show the series∞∑

n=

n( n− )( n+ )

diverges.

Note: Theorem . . does not state thatthe integral and the summa on have thesame value.

............

y = a(x)

. x.

y

(a)

............

y = a(x)

. x.

y

(b)

Figure . . : Illustra ng the truth of theIntegral Test.

Chapter Sequences and Series

. Integral and Comparison TestsKnowing whether or not a series converges is very important, especially whenwe discuss Power Series in Sec on . . Theorems . . and . . give criteriafor when Geometric and p-series converge, and Theorem . . gives a quick testto determine if a series diverges. There are many important series whose con-vergence cannot be determined by these theorems, though, so we introduce aset of tests that allow us to handle a broad range of series. We start with theIntegral Test.

Integral Test

We stated in Sec on . that a sequence {an} is a func on a(n) whose do-main isN, the set of natural numbers. If we can extend a(n) toR, the real num-bers, and it is both posi ve and decreasing on [ ,∞), then the convergence of∞∑

n=

an is the same as∫ ∞

a(x) dx.

Theorem . . Integral Test

Let a sequence {an} be defined by an = a(n), where a(n) is con nuous,

posi ve and decreasing on [ ,∞). Then∞∑

n=

an converges, if, and only if,∫ ∞

a(x) dx converges.

We can demonstrate the truth of the Integral Test with two simple graphs.In Figure . . (a), the height of each rectangle is a(n) = an for n = , , . . .,and clearly the rectangles enclose more area than the area under y = a(x).Therefore we can conclude that

∫ ∞a(x) dx <

∞∑

n=

an. ( . )

In Figure . . (b), we draw rectangles under y = a(x)with the Right-Hand rule,star ng with n = . This me, the area of the rectangles is less than the area

under y = a(x), so∞∑

n=

an <

∫ ∞a(x) dx. Note how this summa on starts

with n = ; adding a to both sides lets us rewrite the summa on star ng with

Notes:

...... 4. 6. 8... 4. 6. 8..

.

.

.4

.

.6

.

.8

. n.

y

.

..an. Sn

Figure . . : Plo ng the sequence andseries in Example . . .

. Integral and Comparison Tests

n = :∞∑

n=

an < a +

∫ ∞a(x) dx. ( . )

Combining Equa ons ( . ) and ( . ), we have

∞∑

n=

an < a +

∫ ∞a(x) dx < a +

∞∑

n=

an. ( . )

From Equa on ( . ) we can make the following two statements:

. If∞∑

n=

an diverges, so does∫ ∞

a(x)dx (because∞∑

n=

an < a +

∫ ∞a(x)dx)

. If∞∑

n=

an converges, so does∫ ∞

a(x)dx (because∫ ∞

a(x)dx <∞∑

n=

an.)

Therefore the series and integral either both converge or both diverge. Theo-rem . . allows us to extend this theorem to series where a(n) is posi ve anddecreasing on [b,∞) for some b > .

Example . . Using the Integral Test

Determine the convergence of∞∑

n=

ln nn

. (The terms of the sequence {an} =

{ln n/n } and the nth par al sums are given in Figure . . .)

S Figure . . implies that a(n) = (ln n)/n is posi ve anddecreasing on [ ,∞). We can determine this analy cally, too. We know a(n)is posi ve as both ln n and n are posi ve on [ ,∞). To determine that a(n) isdecreasing, consider a ′(n) = ( − ln n)/n , which is nega ve for n ≥ . Sincea ′(n) is nega ve, a(n) is decreasing.

Applying the Integral Test, we test the convergence of∫ ∞ ln x

xdx. Integrat-

ing this improper integral requires the use of Integra on by Parts, with u = ln xand dv = /x dx.

∫ ∞ ln xx

dx = limb→∞

∫ b ln xx

dx

= limb→∞

−xln x∣∣∣

b+

∫ b

xdx

Notes:

Chapter Sequences and Series

= limb→∞

−xln x−

x

∣∣∣

b

= limb→∞

−b− ln b

b. Apply L’Hôpital’s Rule:

= .

Since∫ ∞ ln x

xdx converges, so does

∞∑

n=

ln nn

.

Theorem . . was given without jus fica on, sta ng that the general p-

series∞∑

n=(an+ b)p

converges if, and only if, p > . In the following example,

we prove this to be true by applying the Integral Test.

Example . . Using the Integral Test to establish Theorem . . .

Use the Integral Test to prove that∞∑

n=(an+ b)p

converges if, and only if, p > .

S Consider the integral∫ ∞

(ax+ b)pdx; assuming p ̸= ,

∫ ∞

(ax+ b)pdx = lim

c→∞

∫ c

(ax+ b)pdx

= limc→∞ a( − p)

(ax+ b) −p∣∣∣

c

= limc→∞ a( − p)

((ac+ b) −p − (a+ b) −p).

This limit converges if, and only if, p > . It is easy to show that the integral alsodiverges in the case of p = . (This result is similar to the work preceding KeyIdea . . .)

Therefore∞∑

n=(an+ b)p

converges if, and only if, p > .

We consider two more convergence tests in this sec on, both comparisontests. That is, we determine the convergence of one series by comparing it toanother series with known convergence.

Notes:

Note: A sequence {an} is a posi vesequence if an > for all n.

Because of Theorem . . , any theoremthat relies on a posi ve sequence s llholds true when an > for all but a fi-nite number of values of n.

. Integral and Comparison Tests

Direct Comparison Test

Theorem . . Direct Comparison Test

Let {an} and {bn} be posi ve sequences where an ≤ bn for all n ≥ N,for some N ≥ .

. If∞∑

n=

bn converges, then∞∑

n=

an converges.

. If∞∑

n=

an diverges, then∞∑

n=

bn diverges.

Example . . Applying the Direct Comparison Test

Determine the convergence of∞∑

n=n + n

.

S This series is neither a geometric or p-series, but seems re-lated. We predict it will converge, so we look for a series with larger terms thatconverges. (Note too that the Integral Test seems difficult to apply here.)

Since n < n + n , n > n + nfor all n ≥ . The series

∞∑

n=n is a

convergent geometric series; by Theorem . . ,∞∑

n=n + n

converges.

Example . . Applying the Direct Comparison Test

Determine the convergence of∞∑

n=n− ln n

.

S We know the Harmonic Series∞∑

n=ndiverges, and it seems

that the given series is closely related to it, hence we predict it will diverge.Since n ≥ n− ln n for all n ≥ ,

n≤

n− ln nfor all n ≥ .

The Harmonic Series diverges, so we conclude that∞∑

n=n− ln n

diverges as

well.

Notes:

Chapter Sequences and Series

The concept of direct comparison is powerful and o en rela vely easy toapply. Prac ce helps one develop the necessary intui on to quickly pick a properseries with which to compare. However, it is easy to construct a series for whichit is difficult to apply the Direct Comparison Test.

Consider∞∑

n=n+ ln n

. It is very similar to the divergent series given in Ex-

ample . . . We suspect that it also diverges, asn≈

n+ ln nfor large n. How-

ever, the inequality that we naturally want to use “goes the wrong way”: sincen ≤ n+ ln n for all n ≥ ,

n≥

n+ ln nfor all n ≥ . The given series has terms

less than the terms of a divergent series, and we cannot conclude anything fromthis.

Fortunately, we can apply another test to the given series to determine itsconvergence.

Limit Comparison Test

Theorem . . Limit Comparison Test

Let {an} and {bn} be posi ve sequences.

. If limn→∞

anbn

= L, where L is a posi ve real number, then∞∑

n=

an and

∞∑

n=

bn either both converge or both diverge.

. If limn→∞

anbn

= , then if∞∑

n=

bn converges, then so does∞∑

n=

an.

. If limn→∞

anbn

= ∞, then if∞∑

n=

bn diverges, then so does∞∑

n=

an.

Theorem . . is most useful when the convergence of the series from {bn}is known and we are trying to determine the convergence of the series from{an}.

We use the Limit Comparison Test in the next example to examine the series∞∑

n=n+ ln n

which mo vated this new test.

Notes:

. Integral and Comparison Tests

Example . . Applying the Limit Comparison Test

Determine the convergence of∞∑

n=n+ ln n

using the Limit Comparison Test.

S We compare the terms of∞∑

n=n+ ln n

to the terms of the

Harmonic Sequence∞∑

n=n:

limn→∞

/(n+ ln n)/n

= limn→∞

nn+ ln n

= (a er applying L’Hôpital’s Rule).

Since the Harmonic Series diverges, we conclude that∞∑

n=n+ ln n

diverges as

well.

Example . . Applying the Limit Comparison Test

Determine the convergence of∞∑

n=n − n

S This series is similar to the one in Example . . , but nowweare considering “ n − n ” instead of “ n + n .” This difference makes applyingthe Direct Comparison Test difficult.

Instead, weuse the Limit Comparison Test and comparewith the series∞∑

n=n :

limn→∞

/( n − n )

/ n = limn→∞

n

n − n= (a er applying L’Hôpital’s Rule twice).

We know∞∑

n=n is a convergent geometric series, hence

∞∑

n=n − n

converges

as well.

As men oned before, prac ce helps one develop the intui on to quicklychoose a series with which to compare. A general rule of thumb is to pick aseries based on the dominant term in the expression of {an}. It is also helpfulto note that factorials dominate exponen als, which dominate algebraic func-ons (e.g., polynomials), which dominate logarithms. In the previous example,

Notes:

Chapter Sequences and Series

the dominant term of n − nwas n, so we compared the series to

∞∑

n=n . It is

hard to apply the Limit Comparison Test to series containing factorials, though,as we have not learned how to apply L’Hôpital’s Rule to n!.

Example . . Applying the Limit Comparison Test

Determine the convergence of∞∑

n=

√n+

n − n+.

S We naïvely a empt to apply the rule of thumb given aboveand note that the dominant term in the expression of the series is /n . Knowing

that∞∑

n=n

converges, we a empt to apply the Limit Comparison Test:

limn→∞

(√n+ )/(n − n+ )

/n= lim

n→∞n (

√n+ )

n − n+= ∞ (Apply L’Hôpital’s Rule).

Theorem . . part ( ) only applies when∞∑

n=

bn diverges; in our case, it con-

verges. Ul mately, our test has not revealed anything about the convergenceof our series.

The problem is that we chose a poor series with which to compare. Sincethe numerator and denominator of the terms of the series are both algebraicfunc ons, we should have compared our series to the dominant term of thenumerator divided by the dominant term of the denominator.

The dominant term of the numerator is n / and the dominant term of thedenominator is n . Thus we should compare the terms of the given series ton / /n = /n / :

limn→∞

(√n+ )/(n − n+ )

/n /= lim

n→∞n / (

√n+ )

n − n+= (Apply L’Hôpital’s Rule).

Since the p-series∞∑

n=n /

converges, we conclude that∞∑

n=

√n+

n − n+con-

verges as well.

We men oned earlier that the Integral Test did not work well with seriescontaining factorial terms. The next sec on introduces the Ra o Test, whichdoes handle such series well. We also introduce the Root Test, which is good forseries where each term is raised to a power.

Notes:

Exercises .Terms and Concepts. In order to apply the Integral Test to a sequence {an}, thefunc on a(n) = an must be , and .

. T/F: The Integral Test can be used to determine the sum ofa convergent series.

. What test(s) in this sec on do not work well with factori-als?

. Suppose∞∑

n=

an is convergent, and there are sequences

{bn} and {cn} such that ≤ bn ≤ an ≤ cn for all n. What

can be said about the series∞∑

n=

bn and∞∑

n=

cn?

ProblemsIn Exercises – , use the Integral Test to determine the con-vergence of the given series.

.∞∑

n=n

.∞∑

n=n

.∞∑

n=

nn +

.∞∑

n=n ln n

.∞∑

n=n +

.∞∑

n=n(ln n)

.∞∑

n=

nn

.∞∑

n=

ln nn

In Exercises – , use the Direct Comparison Test to deter-mine the convergence of the given series; state what series isused for comparison.

.∞∑

n=n + n−

.∞∑

n=n + n − n

.∞∑

n=

ln nn

.∞∑

n=n! + n

.∞∑

n=

√n −

.∞∑

n=

√n−

.∞∑

n=

n + n+n −

.∞∑

n=

n

n +

.∞∑

n=

nn −

.∞∑

n=n ln n

In Exercises – , use the Limit Comparison Test to deter-mine the convergence of the given series; state what series isused for comparison.

.∞∑

n=n − n+

.∞∑

n=n − n

.∞∑

n=

ln nn−

.∞∑

n=

√n + n

.∞∑

n= n+√n

.∞∑

n=

n−n + n+

.∞∑

n=

sin(

/n)

.∞∑

n=

n+n −

.∞∑

n=

√n+

n +

.∞∑

n=

√n+

In Exercises – , determine the convergence of the givenseries. State the test used; more than one test may be appro-priate.

.∞∑

n=

nn

.∞∑

n=( n+ )

.∞∑

n=

n!n

.∞∑

n=

ln nn!

.∞∑

n=n + n

.∞∑

n=

n−n+

.∞∑

n=

n

n

.∞∑

n=

cos( /n)√n

. Given that∞∑

n=

an converges, state which of the following

series converges, may converge, or does not converge.

(a)∞∑

n=

ann

(b)∞∑

n=

anan+

(c)∞∑

n=

(an)

(d)∞∑

n=

nan

(e)∞∑

n=an

Note: Theorem . . allows us to applythe Ra o Test to series where {an} is pos-i ve for all but a finite number of terms.

. Ra o and Root Tests

. Ra o and Root Tests

The nth–Term Test of Theorem . . states that in order for a series∞∑

n=

an to

converge, limn→∞

an = . That is, the terms of {an} must get very small. Notonly must the terms approach , they must approach “fast enough”: while

limn→∞

/n = , the Harmonic Series∞∑

n=ndiverges as the terms of { /n} do not

approach “fast enough.”The comparison tests of the previous sec ondetermine convergenceby com-

paring terms of a series to terms of another series whose convergence is known.This sec on introduces the Ra o and Root Tests, which determine convergenceby analyzing the terms of a series to see if they approach “fast enough.”

Ra o Test

Theorem . . Ra o Test

Let {an} be a posi ve sequence where limn→∞

an+an

= L.

. If L < , then∞∑

n=

an converges.

. If L > or L = ∞, then∞∑

n=

an diverges.

. If L = , the Ra o Test is inconclusive.

The principle of the Ra o Test is this: if limn→∞

an+an

= L < , then for large n,

each term of {an} is significantly smaller than its previous term which is enoughto ensure convergence.

Example . . Applying the Ra o TestUse the Ra o Test to determine the convergence of the following series:

.

∞∑

n=

n

n!.

∞∑

n=

n

n.

∞∑

n=n +

.

Notes:

Chapter Sequences and Series

S

.∞∑

n=

n

n!:

limn→∞

n+ /(n+ )!n/n!

= limn→∞

n+ n!n(n+ )!

= limn→∞ n+

= .

Since the limit is < , by the Ra o Test∞∑

n=

n

n!converges.

.∞∑

n=

n

n:

limn→∞

n+ /(n+ )n/n

= limn→∞

n+ nn(n+ )

= limn→∞

n(n+ )

= .

Since the limit is > , by the Ra o Test∞∑

n=

n

ndiverges.

.∞∑

n=n +

:

limn→∞

/((n+ ) +

)

/(n + )= lim

n→∞n +

(n+ ) +

= .

Since the limit is , the Ra o Test is inconclusive. We can easily show thisseries converges using the Direct or Limit Comparison Tests, with each

comparing to the series∞∑

n=n

.

Notes:

. Ra o and Root Tests

The Ra o Test is not effec ve when the terms of a series only contain al-gebraic func ons (e.g., polynomials). It is most effec ve when the terms con-tain some factorials or exponen als. The previous example also reinforces ourdeveloping intui on: factorials dominate exponen als, which dominate alge-braic func ons, which dominate logarithmic func ons. In Part of the example,the factorial in the denominator dominated the exponen al in the numerator,causing the series to converge. In Part , the exponen al in the numerator dom-inated the algebraic func on in the denominator, causing the series to diverge.

While we have used factorials in previous sec ons, we have not exploredthem closely and one is likely to not yet have a strong intui ve sense for howthey behave. The following example gives more prac ce with factorials.

Example . . Applying the Ra o Test

Determine the convergence of∞∑

n=

n!n!( n)!

.

S Before we begin, be sure to note the difference between( n)! and n!. When n = , the former is ! = · · . . . · · = , ,whereas the la er is ( · · · ) = .

Applying the Ra o Test:

limn→∞

(n+ )!(n+ )!/((n+ )

)!

n!n!/( n)!= lim

n→∞(n+ )!(n+ )!( n)!

n!n!( n+ )!

No ng that ( n+ )! = ( n+ ) · ( n+ ) · ( n)!, we have

= limn→∞

(n+ )(n+ )

( n+ )( n+ )

= / .

Since the limit is / < , by the Ra o Test we conclude∞∑

n=

n!n!( n)!

converges.

Root Test

The final test we introduce is the Root Test, which works par cularly well onseries where each term is raised to a power, and does not work well with termscontaining factorials.

Notes:

Note: Theorem . . allows us to applythe Root Test to series where {an} is pos-i ve for all but a finite number of terms.

Chapter Sequences and Series

Theorem . . Root Test

Let {an} be a posi ve sequence, and let limn→∞

(an) /n = L.

. If L < , then∞∑

n=

an converges.

. If L > or L = ∞, then∞∑

n=

an diverges.

. If L = , the Root Test is inconclusive.

Example . . Applying the Root TestDetermine the convergence of the following series using the Root Test:

.

∞∑

n=

(n+n−

)n

.

∞∑

n=

n(ln n)n

.

∞∑

n=

n

n.

S

. limn→∞

((n+n−

)n) /n

= limn→∞

n+n− = .

Since the limit is less than , we conclude the series converges. Note: it isdifficult to apply the Ra o Test to this series.

. limn→∞

(n

(ln n)n

) /n

= limn→∞

(n /n)

ln n.

As n grows, the numerator approaches (apply L’Hôpital’s Rule) and thedenominator grows to infinity. Thus

limn→∞

(n /n)

ln n= .

Since the limit is less than , we conclude the series converges.

. limn→∞

( n

n

) /n

= limn→∞ (

n /n) = .

Since this is greater than , we conclude the series diverges.

Each of the tests we have encountered so far has required that we analyzeseries from posi ve sequences. The next sec on relaxes this restric on by con-sidering alterna ng series, where the underlying sequence has terms that alter-nate between being posi ve and nega ve.

Notes:

Exercises .Terms and Concepts

. The Ra o Test is not effec vewhen the terms of a sequenceonly contain func ons.

. The Ra o Test is most effec ve when the terms of a se-quence contains and/or func ons.

. What three convergence tests do not work well with termscontaining factorials?

. The Root Test works par cularly well on series where eachterm is to a .

Problems

In Exercises – , determine the convergence of the givenseries using the Ra o Test. If the Ra o Test is inconclusive,state so and determine convergence with another test.

.∞∑

n=

nn!

.∞∑

n=

n − nn

.∞∑

n=

n! n

( n)!

.∞∑

n=

n + nn + n

.∞∑

n=n

.∞∑

n=n +

.∞∑

n=

· n

n −

.∞∑

n=

n ·( )n

.∞∑

n=

· · · · · · n· · · · · · n

.∞∑

n=

n!· · · · · ( n)

In Exercises – , determine the convergence of the givenseries using the Root Test. If the Root Test is inconclusive,state so and determine convergence with another test.

.∞∑

n=

(

n+n+

)n

.∞∑

n=

(

. n − n−n + n+

)n

.∞∑

n=

nnn

.∞∑

n=nn

.∞∑

n=

n

n n+

.∞∑

n=

n+

n

.∞∑

n=

(

n − nn + n

)n

.∞∑

n=

(

n−

n

)n

.∞∑

n=

(

ln n)n

.∞∑

n=

n(

ln n)n

In Exercises – , determine the convergence of the givenseries. State the test used; more than one test may be appro-priate.

.∞∑

n=

n + n−n + n − n+

.∞∑

n=

n n

n!

.∞∑

n=

nn + n

.∞∑

n=

n

nn

.∞∑

n=

n√n + n+

.∞∑

n=

n!n!n!( n)!

.∞∑

n=ln n

.∞∑

n=

(

n+n+

)n

.∞∑

n=

n(

ln n)n

.∞∑

n=

(

n−

n+

)

. Alterna ng Series and Absolute Convergence

. Alterna ng Series and Absolute Convergence

All of the series convergence tests we have used require that the underlyingsequence {an} be a posi ve sequence. (We can relax this with Theorem . .and state that there must be an N > such that an > for all n > N; that is,{an} is posi ve for all but a finite number of values of n.)

In this sec on we explore series whose summa on includes nega ve terms.We start with a very specific form of series, where the terms of the summa onalternate between being posi ve and nega ve.

Defini on . . Alterna ng Series

Let {an} be a posi ve sequence. An alterna ng series is a series of eitherthe form ∞∑

n=

(− )nan or∞∑

n=

(− )n+ an.

Recall the termsofHarmonic Series come from theHarmonic Sequence {an} ={ /n}. An important alterna ng series is the Alterna ng Harmonic Series:

∞∑

n=

(− )n+n= − + − + − + · · ·

Geometric Series can also be alterna ng series when r < . For instance, ifr = − / , the geometric series is

∞∑

n=

(− )n

= − + − + − + · · ·

Theorem . . states that geometric series converge when |r| < and gives

the sum:∞∑

n=

rn = − r. When r = − / as above, we find

∞∑

n=

(− )n

= − (− / )=

/= .

Apowerful convergence theoremexists for other alterna ng series thatmeeta few condi ons.

Notes:

.....

L

....8

..

.

..

n

.

y

...an. Sn

Figure . . : Illustra ng convergencewith the Alterna ng Series Test.

Chapter Sequences and Series

Theorem . . Alterna ng Series Test

Let {an} be a posi ve, decreasing sequence where limn→∞

an = . Then

∞∑

n=

(− )nan and∞∑

n=

(− )n+ an

converge.

The basic idea behind Theorem . . is illustrated in Figure . . . A posi ve,decreasing sequence {an} is shown along with the par al sums

Sn =n∑

i=

(− )i+ ai = a − a + a − a + · · ·+ (− )n+ an.

Because{an} is decreasing, the amount bywhich Sn bounces up/downdecreases.Moreover, the odd terms of Sn form a decreasing, bounded sequence, while theeven terms of Sn form an increasing, bounded sequence. Since bounded, mono-tonic sequences converge (see Theorem . . ) and the terms of {an} approach, one can show the odd and even terms of Sn converge to the same common

limit L, the sum of the series.

Example . . Applying the Alterna ng Series TestDetermine if the Alterna ng Series Test applies to each of the following series.

.

∞∑

n=

(− )n+n

.

∞∑

n=

(− )nln nn

.

∞∑

n=

(− )n+| sin n|n

S

. This is the Alterna ng Harmonic Series as seen previously. The underlyingsequence is {an} = { /n}, which is posi ve, decreasing, and approachesas n → ∞. Therefore we can apply the Alterna ng Series Test and

conclude this series converges.While the test does not state what the series converges to, we will see

later that∞∑

n=

(− )n+n= ln .

. The underlying sequence is {an} = {ln n/n}. This is posi ve and ap-proaches as n → ∞ (use L’Hôpital’s Rule). However, the sequence is notdecreasing for all n. It is straigh orward to compute a = , a ≈ . ,

Notes:

. Alterna ng Series and Absolute Convergence

a ≈ . , and a ≈ . : the sequence is increasing for at least thefirst terms.

We do not immediately conclude that we cannot apply the Alterna ngSeries Test. Rather, consider the long–term behavior of {an}. Trea ngan = a(n) as a con nuous func on of n defined on [ ,∞), we can takeits deriva ve:

a ′(n) =− ln nn

.

The deriva ve is nega ve for all n ≥ (actually, for all n > e), mean-ing a(n) = an is decreasing on [ ,∞). We can apply the Alterna ngSeries Test to the series when we start with n = and conclude that∞∑

n=

(− )nln nn

converges; adding the terms with n = and n = do not

change the convergence (i.e., we apply Theorem . . ).

The important lesson here is that as before, if a series fails to meet thecriteria of the Alterna ng Series Test on only a finite number of terms, wecan s ll apply the test.

. The underlying sequence is {an} = | sin n|/n . This sequence is posi veand approaches as n → ∞. However, it is not a decreasing sequence;the value of | sin n| oscillates between and as n → ∞. We cannotremove a finite number of terms to make {an} decreasing, therefore wecannot apply the Alterna ng Series Test.

Keep in mind that this does not mean we conclude the series diverges;in fact, it does converge. We are just unable to conclude this based onTheorem . . .

Key Idea . . gives the sum of some important series. Two of these are

∞∑

n=n

=π ≈ . and

∞∑

n=

(− )n+

n=

π ≈ . .

These two series converge to their sums at different rates. To be accurate totwo places a er the decimal, we need terms of the first series though only

of the second. To get places of accuracy, we need terms of the firstseries though only of the second. Why is it that the second series convergesso much faster than the first?

While there are many factors involved when studying rates of convergence,the alterna ng structure of an alterna ng series gives us a powerful tool whenapproxima ng the sum of a convergent series.

Notes:

Chapter Sequences and Series

Theorem . . The Alterna ng Series Approxima on Theorem

Let {an} be a sequence that sa sfies the hypotheses of the Alterna ngSeries Test, and let Sn and L be the nth par al sums and sum, respec vely,

of either∞∑

n=

(− )nan or∞∑

n=

(− )n+ an. Then

. |Sn − L| < an+ , and

. L is between Sn and Sn+ .

Part of Theorem . . states that the nth par al sum of a convergent al-terna ng series will be within an+ of its total sum. Consider the alterna ng

series we looked at before the statement of the theorem,∞∑

n=

(− )n+

n. Since

a = / ≈ . , we know that S is within . of the total sum.Moreover, Part of the theorem states that since S ≈ . and S ≈

. , we know the sum L lies between . and . . One use of this isthe knowledge that S is accurate to two places a er the decimal.

Some alterna ng series converge slowly. In Example . . we determined

the series∞∑

n=

(− )n+ln nn

converged. With n = , we find ln n/n ≈ . ,

meaning that S ≈ . is accurate to one, maybe two, places a er thedecimal. Since S ≈ . , we know the sum L is . ≤ L ≤ . .

Example . . Approxima ng the sum of convergent alterna ng seriesApproximate the sum of the following series, accurate to within . .

.

∞∑

n=

(− )n+n

.

∞∑

n=

(− )n+ln nn

.

S

. Using Theorem . . , we want to find n where /n ≤ . :

n≤ . =

n ≥n ≥

√

n ≥ .

Notes:

. Alterna ng Series and Absolute Convergence

Let L be the sum of this series. By Part of the theorem, |S − L| < a =/ . We can compute S = . , which our theorem states is

within . of the total sum.We can use Part of the theorem to obtain an even more accurate result.Aswe know the th termof the series is− / , we can easily computeS = . . Part of the theorem states that L is between S and S ,so . < L < . .

. We want to find n where ln(n)/n < . . We start by solving ln(n)/n =. for n. This cannot be solved algebraically, so we will use Newton’s

Method to approximate a solu on.Let f(x) = ln(x)/x− . ; we want to know where f(x) = . We make aguess that xmust be “large,” so our ini al guess will be x = . Recallhow Newton’s Method works: given an approximate solu on xn, our nextapproxima on xn+ is given by

xn+ = xn −f(xn)f ′(xn)

.

We find f ′(x) =(

− ln(x))/x . This gives

x = − ln( )/ − .(

− ln( ))/

= .

Using a computer, we find that Newton’s Method seems to converge to asolu on x = . a er itera ons. Taking the next integer higher, wehave n = , where ln( )/ = . < . .Again using a computer, we find S = − . . Part of the theo-rem states that this is within . of the actual sum L. Already knowingthe , th term,we can compute S = − . , meaning− . <L < − . .

No ce how the first series converged quite quickly, where we needed onlyterms to reach the desired accuracy, whereas the second series took over ,terms.

One of the famous results of mathema cs is that the Harmonic Series,∞∑

n=n

diverges, yet the Alterna ng Harmonic Series,∞∑

n=

(− )n+n, converges. The

Notes:

Note: In Defini on . . ,∞∑

n=

an is not

necessarily an alterna ng series; it justmay have some nega ve terms.

Chapter Sequences and Series

no on that alterna ng the signs of the terms in a series can make a series con-verge leads us to the following defini ons.

Defini on . . Absolute and Condi onal Convergence

. A series∞∑

n=

an converges absolutely if∞∑

n=

|an| converges.

. A series∞∑

n=

an converges condi onally if∞∑

n=

an converges but

∞∑

n=

|an| diverges.

Thus we say the Alterna ng Harmonic Series converges condi onally.

Example . . Determining absolute and condi onal convergence.Determine if the following series converge absolutely, condi onally, or diverge.

.

∞∑

n=

(− )nn+

n + n+.

∞∑

n=

(− )nn + n+

n .

∞∑

n=

(− )nn−n−

S

. We can show the series∞∑

n=

∣∣∣∣(− )n

n+n + n+

∣∣∣∣=

∞∑

n=

n+n + n+

diverges using the Limit Comparison Test, comparing with /n.

The series∞∑

n=

(− )nn+

n + n+converges using the Alterna ng Series

Test; we conclude it converges condi onally.

. We can show the series∞∑

n=

∣∣∣∣(− )n

n + n+n

∣∣∣∣=

∞∑

n=

n + n+n

converges using the Ra o Test.

Notes:

. Alterna ng Series and Absolute Convergence

Therefore we conclude∞∑

n=

(− )nn + n+

n converges absolutely.

. The series∞∑

n=

∣∣∣∣(− )n

n−n−

∣∣∣∣=

∞∑

n=

n−n−

diverges using the nth Term Test, so it does not converge absolutely.

The series∞∑

n=

(− )nn−n− fails the condi ons of the Alterna ng Series

Test as ( n− )/( n− ) does not approach as n → ∞. We can statefurther that this series diverges; as n → ∞, the series effec vely adds andsubtracts / over and over. This causes the sequence of par al sums tooscillate and not converge.

Therefore the series∞∑

n=

(− )nn−n− diverges.

Knowing that a series converges absolutely allows us to make two impor-tant statements, given in the following theorem. The first is that absolute con-

vergence is “stronger” than regular convergence. That is, just because∞∑

n=

an

converges, we cannot conclude that∞∑

n=

|an| will converge, but knowing a series

converges absolutely tells us that∞∑

n=

an will converge.

One reason this is important is that our convergence tests all require that theunderlying sequence of terms be posi ve. By taking the absolute value of theterms of a series where not all terms are posi ve, we are o en able to apply anappropriate test and determine absolute convergence. This, in turn, determinesthat the series we are given also converges.

The second statement relates to rearrangements of series. When dealingwith a finite set of numbers, the sum of the numbers does not depend on theorder which they are added. (So + + = + + .) Onemay be surprised tofind out that when dealing with an infinite set of numbers, the same statementdoes not always hold true: some infinite lists of numbers may be rearranged indifferent orders to achieve different sums. The theorem states that the terms ofan absolutely convergent series can be rearranged in any way without affec ngthe sum.

Notes:

Chapter Sequences and Series

Theorem . . Absolute Convergence Theorem

Let∞∑

n=

an be a series that converges absolutely.

.∞∑

n=

an converges.

. Let {bn} be any rearrangement of the sequence {an}. Then∞∑

n=

bn =∞∑

n=

an.

In Example . . , we determined the series in part converges absolutely.Theorem . . tells us the series converges (which we could also determine us-ing the Alterna ng Series Test).

The theorem states that rearranging the terms of an absolutely convergentseries does not affect its sum. This implies that perhaps the sum of a condi on-ally convergent series can change based on the arrangement of terms. Indeed,it can. The Riemann Rearrangement Theorem (named a er Bernhard Riemann)states that any condi onally convergent series can have its terms rearranged sothat the sum is any desired value, including∞!

As an example, consider the Alterna ng Harmonic Series once more. Wehave stated that

∞∑

n=

(− )n+n= − + − + − + · · · = ln ,

(see Key Idea . . or Example . . ).

Consider the rearrangement where every posi ve term is followed by twonega ve terms:

− − + − − + − − · · ·

(Convince yourself that these are exactly the same numbers as appear in theAlterna ng Harmonic Series, just in a different order.) Now group some terms

Notes:

. Alterna ng Series and Absolute Convergence

and simplify:(

−)

− +

(

−)

− +

(

−)

− + · · · =

− + − + − + · · · =(

− + − + − + · · ·)

= ln .

By rearranging the terms of the series, we have arrived at a different sum!(One could try to argue that the Alterna ng Harmonic Series does not actuallyconverge to ln , because rearranging the terms of the series shouldn’t changethe sum. However, the Alterna ng Series Test proves this series converges toL, for some number L, and if the rearrangement does not change the sum, thenL = L/ , implying L = . But the Alterna ng Series Approxima on Theoremquickly shows that L > . The only conclusion is that the rearrangement didchange the sum.) This is an incredible result.

We end here our study of tests to determine convergence. The end of thistext contains a table summarizing the tests that one may find useful.

While series are worthy of study in and of themselves, our ul mate goalwithin calculus is the study of Power Series, which we will consider in the nextsec on. We will use power series to create func ons where the output is theresult of an infinite summa on.

Notes:

Exercises .Terms and Concepts

. Why is∞∑

n=

sin n not an alterna ng series?

. A series∞∑

n=

(− )nan converges when {an} is ,

and limn→∞

an = .

. Give an example of a series where∞∑

n=

an converges but

∞∑

n=

|an| does not.

. The sum of a convergent series can be changed byrearranging the order of its terms.

Problems

In Exercises – , an alterna ng series∞∑

n=i

an is given.

(a) Determine if the series converges or diverges.

(b) Determine if∞∑

n=

|an| converges or diverges.

(c) If∞∑

n=

an converges, determine if the convergence is

condi onal or absolute.

.∞∑

n=

(− )n+

n

.∞∑

n=

(− )n+√n!

.∞∑

n=

(− )nn+n−

.∞∑

n=

(− )nn

n

.∞∑

n=

(− )n+n+

n − n+

.∞∑

n=

(− )n

ln n+

.∞∑

n=

(− )nnln n

.∞∑

n=

(− )n+

+ + + · · ·+ ( n− )

.∞∑

n=

cos(

πn)

.∞∑

n=

sin(

(n+ / )π)

n ln n

.∞∑

n=

(

−)n

.∞∑

n=

(−e)−n

.∞∑

n=

(− )nnn!

.∞∑

n=

(− )n −n

.∞∑

n=

(− )n√n

.∞∑

n=

(− )n

n!

Let Sn be the nth par al sum of a series. In Exercises – , aconvergent alterna ng series is given and a value of n. Com-pute Sn and Sn+ and use these values to find bounds on thesum of the series.

.∞∑

n=

(− )n

ln(n+ ), n =

.∞∑

n=

(− )n+

n, n =

.∞∑

n=

(− )n

n!, n =

.∞∑

n=

(

−)n

, n =

In Exercises – , a convergent alterna ng series is givenalongwith its sum and a value of ε. Use Theorem . . to findn such that the nth par al sum of the series is within ε of thesum of the series.

.∞∑

n=

(− )n+

n=

π , ε = .

.∞∑

n=

(− )n

n!=

e, ε = .

.∞∑

n=

(− )n

n+=

π , ε = .

.∞∑

n=

(− )n

( n)!= cos , ε = −

Chapter Sequences and Series

. Power SeriesSo far, our study of series has examined the ques on of “Is the sum of theseinfinite terms finite?,” i.e., “Does the series converge?” We now approach seriesfrom a different perspec ve: as a func on. Given a value of x, we evaluate f(x)by finding the sum of a par cular series that depends on x (assuming the seriesconverges). We start this new approach to series with a defini on.

Defini on . . Power Series

Let {an} be a sequence, let x be a variable, and let c be a real number.

. The power series in x is the series

∞∑

n=

anxn = a + a x+ a x + a x + . . .

. The power series in x centered at c is the series∞∑

n=

an(x− c)n = a + a (x− c) + a (x− c) + a (x− c) + . . .

Example . . Examples of power seriesWrite out the first five terms of the following power series:

.

∞∑

n=

xn .

∞∑

n=

(− )n+(x+ )n

n.

∞∑

n=

(− )n+(x− π) n

( n)!.

S

. One of the conven ons we adopt is that x = regardless of the value ofx. Therefore ∞∑

n=

xn = + x+ x + x + x + . . .

This is a geometric series in x.

. This series is centered at c = − . Note how this series starts with n = .We could rewrite this series star ng at n = with the understanding thata = , and hence the first term is .∞∑

n=

(− )n+(x+ )n

n= (x+ )− (x+ )

+(x+ ) − (x+ )

+(x+ )

. . .

Notes:

. Power Series

. This series is centered at c = π. Recall that ! = .

∞∑

n=

(− )n+(x− π) n

( n)!= − +

(x− π) − (x− π)+(x− π)

!− (x− π)

!. . .

We introduced power series as a type of func on, where a value of x is givenand the sum of a series is returned. Of course, not every series converges. For

instance, in part of Example . . , we recognized the series∞∑

n=

xn as a geo-

metric series in x. Theorem . . states that this series converges only when|x| < .

This raises the ques on: “For what values of xwill a given power series con-verge?,” which leads us to a theorem and defini on.

Theorem . . Convergence of Power Series

Let a power series∞∑

n=

an(x− c)n be given. Then one of the following is

true:

. The series converges only at x = c.

. There is an R > such that the series converges for all x in(c− R, c+ R) and diverges for all x < c− R and x > c+ R.

. The series converges for all x.

The value of R is important when understanding a power series, hence it isgiven a name in the following defini on. Also, note that part of Theorem . .makes a statement about the interval (c− R, c+ R), but the not the endpointsof that interval. A series may/may not converge at these endpoints.

Notes:

Chapter Sequences and Series

Defini on . . Radius and Interval of Convergence

. The number R given in Theorem . . is the radius of convergenceof a given series. When a series converges for only x = c, we saythe radius of convergence is , i.e., R = . When a series convergesfor all x, we say the series has an infinite radius of convergence, i.e.,R = ∞.

. The interval of convergence is the set of all values of x for whichthe series converges.

To find the values of x for which a given series converges, wewill use the con-vergence tests we studied previously (especially the Ra o Test). However, thetests all required that the terms of a series be posi ve. The following theoremgives us a work–around to this problem.

Theorem . . The Radius of Convergence of a Series and AbsoluteConvergence

The series∞∑

n=

an(x − c)n and∞∑

n=

∣∣an(x − c)n

∣∣ have the same radius of

convergence R.

Theorem . . allows us to find the radius of convergence R of a series byapplying the Ra o Test (or any applicable test) to the absolute value of the termsof the series. We prac ce this in the following example.

Example . . Determining the radius and interval of convergence.Find the radius and interval of convergence for each of the following series:

.

∞∑

n=

xn

n!.

∞∑

n=

(− )n+xn

n.

∞∑

n=

n(x− )n .

∞∑

n=

n!xn

S

Notes:

. Power Series

. We apply the Ra o Test to the series∞∑

n=

∣∣∣∣

xn

n!

∣∣∣∣:

limn→∞

∣∣xn+ /(n+ )!

∣∣

∣∣xn/n!

∣∣

= limn→∞

∣∣∣∣

xn+

xn· n!(n+ )!

∣∣∣∣

= limn→∞

∣∣∣∣

xn+

∣∣∣∣

= for all x.

The Ra o Test shows us that regardless of the choice of x, the series con-verges. Therefore the radius of convergence is R = ∞, and the interval ofconvergence is (−∞,∞).

. We apply the Ra o Test to the series∞∑

n=

∣∣∣∣(− )n+

xn

n

∣∣∣∣=

∞∑

n=

∣∣∣∣

xn

n

∣∣∣∣:

limn→∞

∣∣xn+ /(n+ )

∣∣

∣∣xn/n

∣∣

= limn→∞

∣∣∣∣

xn+

xn· nn+

∣∣∣∣

= limn→∞

|x| nn+

= |x|.

The Ra o Test states a series converges if the limit of |an+ /an| = L < .We found the limit above to be |x|; therefore, the power series convergeswhen |x| < , or when x is in (− , ). Thus the radius of convergence isR = .

To determine the interval of convergence, we need to check the endpointsof (− , ). When x = − , we have the opposite of the Harmonic Series:

∞∑

n=

(− )n+(− )n

n=

∞∑

n=

−n

= −∞.

The series diverges when x = − .

When x = , we have the series∞∑

n=

(− )n+( )n

n, which is the Alterna ng

Harmonic Series, which converges. Therefore the interval of convergenceis (− , ].

Notes:

Chapter Sequences and Series

. We apply the Ra o Test to the series∞∑

n=

∣∣ n(x− )n

∣∣:

limn→∞

∣∣ n+ (x− )n+

∣∣

∣∣ n(x− )n

∣∣

= limn→∞

∣∣∣∣

n+

n · (x− )n+

(x− )n

∣∣∣∣

= limn→∞

∣∣ (x− )

∣∣.

According to the Ra o Test, the series convergeswhen∣∣ (x− )

∣∣ < =⇒

∣∣x−

∣∣ < / . The series is centered at , and xmust be within / of

in order for the series to converge. Therefore the radius of convergenceis R = / , and we know that the series converges absolutely for all x in( − / , + / ) = ( . , . ).We check for convergence at the endpoints to find the interval of conver-gence. When x = . , we have:

∞∑

n=

n( . − )n =

∞∑

n=

n(− / )n

=∞∑

n=

(− )n,

which diverges. A similar process shows that the series also diverges atx = . . Therefore the interval of convergence is ( . , . ).

. We apply the Ra o Test to∞∑

n=

∣∣n!xn

∣∣:

limn→∞

∣∣(n+ )!xn+

∣∣

∣∣n!xn

∣∣

= limn→∞

∣∣(n+ )x

∣∣

= ∞ for all x, except x = .

The Ra o Test shows that the series diverges for all x except x = . There-fore the radius of convergence is R = .

We can use a power series to define a func on:

f(x) =∞∑

n=

anxn

where the domain of f is a subset of the interval of convergence of the powerseries. One can apply calculus techniques to such func ons; in par cular, wecan find deriva ves and an deriva ves.

Notes:

. Power Series

Theorem . . Deriva ves and Indefinite Integrals of Power SeriesFunc ons

Let f(x) =∞∑

n=

an(x − c)n be a func on defined by a power series, with

radius of convergence R.

. f(x) is con nuous and differen able on (c− R, c+ R).

. f ′(x) =∞∑

n=

an · n · (x− c)n− , with radius of convergence R.

.∫

f(x) dx = C+∞∑

n=

an(x− c)n+

n+, with radius of convergence R.

A few notes about Theorem . . :

. The theorem states that differen a on and integra on do not change theradius of convergence. It does not state anything about the interval ofconvergence. They are not always the same.

. No ce how the summa on for f ′(x) starts with n = . This is because theconstant term a of f(x) goes to .

. Differen a on and integra on are simply calculated term–by–term usingthe Power Rules.

Example . . Deriva ves and indefinite integrals of power series

Let f(x) =∞∑

n=

xn. Find f ′(x) and F(x) =∫

f(x) dx, along with their respec ve

intervals of convergence.

S We find the deriva ve and indefinite integral of f(x), follow-ing Theorem . . .

. f ′(x) =∞∑

n=

nxn− = + x+ x + x + · · · .

In Example . . , we recognized that∞∑

n=

xn is a geometric series in x. We

know that such a geometric series converges when |x| < ; that is, theinterval of convergence is (− , ).

Notes:

Chapter Sequences and Series

To determine the interval of convergence of f ′(x), we consider the end-points of (− , ):

f ′(− ) = − + − + · · · , which diverges.

f ′( ) = + + + + · · · , which diverges.

Therefore, the interval of convergence of f ′(x) is (− , ).

. F(x) =∫

f(x) dx = C+∞∑

n=

xn+

n+= C+ x+

x+

x+ · · ·

To find the interval of convergence of F(x), we again consider the end-points of (− , ):

F(− ) = C− + / − / + / + · · ·

The value of C is irrelevant; no ce that the rest of the series is an Alter-na ng Series that whose terms converge to . By the Alterna ng SeriesTest, this series converges. (In fact, we can recognize that the terms of theseries a er C are the opposite of the Alterna ng Harmonic Series. We canthus say that F(− ) = C− ln .)

F( ) = C+ + / + / + / + · · ·

No ce that this summa on is C + the Harmonic Series, which diverges.Since F converges for x = − and diverges for x = , the interval ofconvergence of F(x) is [− , ).

The previous example showed how to take the deriva ve and indefinite in-tegral of a power series without mo va on for why we care about such opera-ons. Wemay care for the sheer mathema cal enjoyment “that we can”, which

is mo va on enough for many. However, we would be remiss to not recognizethat we can learn a great deal from taking deriva ves and indefinite integrals.

Recall that f(x) =

∞∑

n=

xn in Example . . is a geometric series. According

to Theorem . . , this series converges to /( − x) when |x| < . Thus we cansay

f(x) =∞∑

n=

xn = − x, on (− , ).

Integra ng the power series, (as done in Example . . ,) we find

F(x) = C +

∞∑

n=

xn+

n+, ( . )

Notes:

. Power Series

while integra ng the func on f(x) = /( − x) gives

F(x) = − ln | − x|+ C . ( . )

Equa ng Equa ons ( . ) and ( . ), we have

F(x) = C +

∞∑

n=

xn+

n+= − ln | − x|+ C .

Le ng x = , we have F( ) = C = C . This implies that we can drop theconstants and conclude

∞∑

n=

xn+

n+= − ln | − x|.

We established in Example . . that the series on the le converges at x = − ;subs tu ng x = − on both sides of the above equality gives

− + − + − + · · · = − ln .

On the le we have the opposite of the Alterna ng Harmonic Series; on theright, we have− ln . We conclude that

− + − + · · · = ln .

Important: We stated in Key Idea . . (in Sec on . ) that the Alterna ng Har-monic Series converges to ln , and referred to this fact again in Example . .of Sec on . . However, we never gave an argument for why this was the case.The work above finally shows how we conclude that the Alterna ng HarmonicSeries converges to ln .

We use this type of analysis in the next example.

Example . . Analyzing power series func ons

Let f(x) =∞∑

n=

xn

n!. Find f ′(x) and

∫

f(x) dx, and use these to analyze the behav-

ior of f(x).

S We start by making two notes: first, in Example . . , wefound the interval of convergence of this power series is (−∞,∞). Second, wewill find it useful later to have a few terms of the series wri en out:

∞∑

n=

xn

n!= + x+

x+

x+

x+ · · · ( . )

Notes:

Chapter Sequences and Series

We now find the deriva ve:

f ′(x) =∞∑

n=

nxn−

n!

=

∞∑

n=

xn−

(n− )!= + x+

x!+ · · · .

Since the series starts at n = and each term refers to (n− ), we can re-indexthe series star ng with n = :

=

∞∑

n=

xn

n!

= f(x).

We found the deriva ve of f(x) is f(x). The only func ons for which this is trueare of the form y = cex for some constant c. As f( ) = (see Equa on ( . )), cmust be . Therefore we conclude that

f(x) =∞∑

n=

xn

n!= ex

for all x.We can also find

∫

f(x) dx:

∫

f(x) dx = C+∞∑

n=

xn+

n!(n+ )

= C+∞∑

n=

xn+

(n+ )!

We write out a few terms of this last series:

C+∞∑

n=

xn+

(n+ )!= C+ x+

x+

x+

x+ · · ·

The integral of f(x) differs from f(x) only by a constant, again indica ng thatf(x) = ex.

Example . . and the work following Example . . established rela on-ships between a power series func on and “regular” func ons that we havedealt with in the past. In general, given a power series func on, it is difficult (if

Notes:

. Power Series

not impossible) to express the func on in terms of elementary func ons. Wechose examples where things worked out nicely.

In this sec on’s last example, we show how to solve a simple differen alequa on with a power series.

Example . . Solving a differen al equa on with a power series.Give the first terms of the power series solu on to y ′ = y, where y( ) = .

S The differen al equa on y ′ = y describes a func on y =f(x) where the deriva ve of y is twice y and y( ) = . This is a rather simpledifferen al equa on; with a bit of thought one should realize that if y = Ce x,then y ′ = Ce x, and hence y ′ = y. By le ng C = we sa sfy the ini alcondi on of y( ) = .

Let’s ignore the fact that we already know the solu on and find a powerseries func on that sa sfies the equa on. The solu on we seek will have theform

f(x) =∞∑

n=

anxn = a + a x+ a x + a x + · · ·

for unknown coefficients an. We can find f ′(x) using Theorem . . :

f ′(x) =∞∑

n=

an · n · xn− = a + a x+ a x + a x · · · .

Since f ′(x) = f(x), we have

a + a x+ a x + a x · · · =(a + a x+ a x + a x + · · ·

)

= a + a x+ a x + a x + · · ·

The coefficients of like powers of xmust be equal, so we find that

a = a , a = a , a = a , a = a , etc.

The ini al condi on y( ) = f( ) = indicates that a = ; with this, we canfind the values of the other coefficients:

a = and a = a ⇒ a = ;

a = and a = a ⇒ a = / = ;

a = and a = a ⇒ a = /( · ) = / ;

a = / and a = a ⇒ a = /( · · ) = / .

Thus the first terms of the power series solu on to the differen al equa ony ′ = y is

f(x) = + x+ x + x + x + · · ·

Notes:

Chapter Sequences and Series

In Sec on . , as we study Taylor Series, we will learn how to recognize this se-ries as describing y = e x.

Our last example illustrates that it can be difficult to recognize an elementaryfunc on by its power series expansion. It is far easier to start with a known func-on, expressed in terms of elementary func ons, and represent it as a power

series func on. One may wonder why we would bother doing so, as the la erfunc on probably seems more complicated. In the next two sec ons, we showboth how to do this and why such a process can be beneficial.

Notes:

Exercises .Terms and Concepts. We adopt the conven on that x = , regardless of thevalue of x.

. What is the difference between the radius of convergenceand the interval of convergence?

. If the radius of convergence of∞∑

n=

anxn is , what is the ra-

dius of convergence of∞∑

n=

n · anxn− ?

. If the radius of convergence of∞∑

n=

anxn is , what is the ra-

dius of convergence of∞∑

n=

(− )nanxn?

ProblemsIn Exercises – , write out the sum of the first terms of thegiven power series.

.∞∑

n=

nxn

.∞∑

n=n

xn

.∞∑

n=n!xn

.∞∑

n=

(− )n

( n)!x n

In Exercises – , a power series is given.(a) Find the radius of convergence.(b) Find the interval of convergence.

.∞∑

n=

(− )n+

n!xn

.∞∑

n=

nxn

.∞∑

n=

(− )n(x− )n

n

.∞∑

n=

(x+ )n

n!

.∞∑

n=

xnn

.∞∑

n=

(− )n(x− )n

n

.∞∑

n=

n(x− )n

.∞∑

n=

(− )nxn

.∞∑

n=

√nxn

.∞∑

n=

nn x

n

.∞∑

n=

n

n!(x− )n

.∞∑

n=

(− )nn!(x− )n

.∞∑

n=

xn

n

.∞∑

n=

(x+ )n

n

.∞∑

n=

n!( x )n

.∞∑

n=

n(

x+)n

In Exercises – , a func on f(x) =∞∑

n=

anxn is given.

(a) Give a power series for f ′(x) and its interval of conver-gence.

(b) Give a power series for∫

f(x) dx and its interval of con-vergence.

.∞∑

n=

nxn

.∞∑

n=

xn

n

.∞∑

n=

( x)n

.∞∑

n=

(− x)n

.∞∑

n=

(− )nx n

( n)!

.∞∑

n=

(− )nxn

n!

In Exercises – , give the first terms of the series that isa solu on to the given differen al equa on.

. y ′ = y, y( ) =

. y ′ = y, y( ) =

. y ′ = y , y( ) =

. y ′ = y+ , y( ) =

. y ′′ = −y, y( ) = , y ′( ) =

. y ′′ = y, y( ) = , y ′( ) =

.....

y = f(x)

.

y = p1(x)

.

−4

.

−2

.

2

.

4

. −5.

5

.

x

.

y

f( ) = f ′′′( ) = −f ′( ) = f ( )( ) = −f ′′( ) = f ( )( ) = −

Figure . . : Plo ng y = f(x) and a tableof deriva ves of f evaluated at .

...

..

y = p2(x)

.

y = p4(x)

.

−4

.

−2

.

2

.

4

.

−5

.

5

.

x

.

y

Figure . . : Plo ng f, p and p .

...

..

y = p (x)

.

−

.

−

...−

.

.

x

.

y

Figure . . : Plo ng f and p .

. Taylor Polynomials

. Taylor PolynomialsConsider a func on y = f(x) and a point

(c, f(c)

). The deriva ve, f ′(c), gives

the instantaneous rate of change of f at x = c. Of all lines that pass through thepoint

(c, f(c)

), the line that best approximates f at this point is the tangent line;

that is, the line whose slope (rate of change) is f ′(c).In Figure . . , we see a func on y = f(x) graphed. The table below the

graph shows that f( ) = and f ′( ) = ; therefore, the tangent line to f atx = is p (x) = (x− )+ = x+ . The tangent line is also given in the figure.Note that “near” x = , p (x) ≈ f(x); that is, the tangent line approximates fwell.

One shortcoming of this approxima on is that the tangent line only matchesthe slope of f; it does not, for instance, match the concavity of f. We can find apolynomial, p (x), that doesmatch the concavitywithoutmuchdifficulty, though.The table in Figure . . gives the following informa on:

f( ) = f ′( ) = f ′′( ) = .

Therefore, we want our polynomial p (x) to have these same proper es. Thatis, we need

p ( ) = p′ ( ) = p′′( ) = .

This is simply an ini al–value problem. We can solve this using the tech-niques first described in Sec on . . To keep p (x) as simple as possible, we’llassume that not only p′′( ) = , but that p′′(x) = . That is, the second deriva-ve of p is constant.If p′′(x) = , then p′ (x) = x + C for some constant C. Since we have

determined that p′ ( ) = , we find that C = and so p′ (x) = x + . Finally,we can compute p (x) = x +x+C. Using our ini al values, we know p ( ) =so C = .We conclude that p (x) = x + x+ . This func on is plo ed with f inFigure . . .

We can repeat this approxima on process by crea ng polynomials of higherdegree that matchmore of the deriva ves of f at x = . In general, a polynomialof degree n can be created to match the first n deriva ves of f. Figure . . alsoshows p (x) = −x / −x / +x +x+ , whose first four deriva ves at matchthose of f. (Using the table in Figure . . , start with p( )

(x) = − and solvethe related ini al–value problem.)

As we use more and more deriva ves, our polynomial approxima on to fgets be er and be er. In this example, the interval on which the approxima onis “good” gets bigger and bigger. Figure . . shows p (x); we can visually affirmthat this polynomial approximates f very well on [− , ]. (The polynomial p (x)is not par cularly “nice”. It is

x+

x − x − x − x+

x+

x+

x − x − x − x+x +x+ .)

Notes:

f(x) = ex ⇒ f( ) =f ′(x) = ex ⇒ f ′( ) =f ′′(x) = ex ⇒ f ′′( ) =...

...f (n)(x) = ex ⇒ f (n)( ) =

Figure . . : The deriva ves of f(x) = ex

evaluated at x = .

Chapter Sequences and Series

Thepolynomialswehave created are examples of Taylor polynomials, nameda er the Bri sh mathema cian Brook Taylor who made important discoveriesabout such func ons. While we created the above Taylor polynomials by solvingini al–value problems, it can be shown that Taylor polynomials follow a generalpa ern that make their forma on much more direct. This is described in thefollowing defini on.

Defini on . . Taylor Polynomial, Maclaurin Polynomial

Let f be a func on whose first n deriva ves exist at x = c.

. The Taylor polynomial of degree n of f at x = c is

pn(x) = f(c)+f ′(c)(x−c)+f ′′(c)

!(x−c) +

f ′′′(c)!

(x−c) +· · ·+ f (n)(c)n!

(x−c)n.

. A special case of the Taylor polynomial is theMaclaurin polynomial, where c =. That is, theMaclaurin polynomial of degree n of f is

pn(x) = f( ) + f ′( )x+f ′′( )

!x +

f ′′′( )

!x + · · ·+ f (n)( )

n!xn.

We will prac ce crea ng Taylor and Maclaurin polynomials in the followingexamples.

Example . . Finding and using Maclaurin polynomials

. Find the nth Maclaurin polynomial for f(x) = ex.

. Use p (x) to approximate the value of e.

S

. We start with crea ng a table of the deriva ves of ex evaluated at x = .In this par cular case, this is rela vely simple, as shown in Figure . . .By the defini on of the Maclaurin series, we have

pn(x) = f( ) + f ′( )x+f ′′( )

!x +

f ′′′( )

!x + · · ·+ f (n)( )

n!xn

= + x+ x + x + x + · · ·+n!xn.

Notes:

.....y = p5(x).−

..

5

..

x

.

y

Figure . . : A plot of f(x) = ex and itsth degree Maclaurin polynomial p (x).

f(x) = ln x ⇒ f( ) =f ′(x) = /x ⇒ f ′( ) =f ′′(x) = − /x ⇒ f ′′( ) = −f ′′′(x) = /x ⇒ f ′′′( ) =

f ( )(x) = − /x ⇒ f ( )( ) = −...

...f (n)(x) = ⇒ f (n)( ) =

(− )n+ (n− )!

xn(− )n+ (n− )!

Figure . . : Deriva ves of ln x evaluatedat x = .

. Taylor Polynomials

. Using our answer from part , we have

p = + x+ x + x + x + x .

To approximate the value of e, note that e = e = f( ) ≈ p ( ). It is verystraigh orward to evaluate p ( ):

p ( ) = + + + + + = ≈ . .

A plot of f(x) = ex and p (x) is given in Figure . . .

Example . . Finding and using Taylor polynomials

. Find the nth Taylor polynomial of y = ln x at x = .

. Use p (x) to approximate the value of ln . .

. Use p (x) to approximate the value of ln .

S

. We begin by crea ng a table of deriva ves of ln x evaluated at x = .While this is not as straigh orward as it was in the previous example, apa ern does emerge, as shown in Figure . . .Using Defini on . . , we have

pn(x) = f(c) + f ′(c)(x− c) + f ′′(c)!

(x− c) +f ′′′(c)

!(x− c) + · · ·+ f (n)(c)

n!(x− c)n

= + (x− )− (x− ) + (x− ) − (x− ) + · · ·+ (− )n+

n(x− )n.

Note how the coefficients of the (x− ) terms turn out to be “nice.”

. We can compute p (x) using our work above:

p (x) = (x− )− (x− ) + (x− ) − (x− ) + (x− ) − (x− ) .

Since p (x) approximates ln x well near x = , we approximate ln . ≈p ( . ):

Notes:

.....

y = ln x

.

y = p6(x)

....−

.

−

..

x

.

y

Figure . . : A plot of y = ln x and its th

degree Taylor polynomial at x = .

...

..

y = ln x

.

y = p (x)

....

−

.

−

.

.

x

.

y

Figure . . : A plot of y = ln x and its th

degree Taylor polynomial at x = .

Chapter Sequences and Series

p ( . ) = ( . − )− ( . − ) + ( . − ) − ( . − ) + · · ·

· · ·+ ( . − ) − ( . − )

=

≈ . .

This is a good approxima on as a calculator shows that ln . ≈ . .Figure . . plots y = ln x with y = p (x). We can see that ln . ≈p ( . ).

. We approximate ln with p ( ):

p ( ) = ( − )− ( − ) + ( − ) − ( − ) + · · ·

· · ·+ ( − ) − ( − )

= − + − + −

=

≈ . .

This approxima on is not terribly impressive: a handheld calculator showsthat ln ≈ . . The graph in Figure . . shows that p (x) providesless accurate approxima ons of ln x as x gets close to or .Surprisingly enough, even the th degree Taylor polynomial fails to ap-proximate ln x for x > , as shown in Figure . . . We’ll soon discuss whythis is.

Taylor polynomials are used to approximate func ons f(x) in mainly two sit-ua ons:

. When f(x) is known, but perhaps “hard” to compute directly. For instance,we can define y = cos x as either the ra o of sides of a right triangle(“adjacent over hypotenuse”) or with the unit circle. However, neither ofthese provides a convenient way of compu ng cos . A Taylor polynomialof sufficiently high degree can provide a reasonablemethod of compu ngsuch values using only opera ons usually hard–wired into a computer (+,−,× and÷).

Notes:

Note: Even though Taylor polynomialscould be used in calculators and com-puters to calculate values of trigonomet-ric func ons, in prac ce they generallyaren’t. Other more efficient and accuratemethods have been developed, such asthe CORDIC algorithm.

. Taylor Polynomials

. When f(x) is not known, but informa on about its deriva ves is known.This occurs more o en than one might think, especially in the study ofdifferen al equa ons.

In both situa ons, a cri cal piece of informa on to have is “How good is myapproxima on?” If we use a Taylor polynomial to compute cos , how do weknow how accurate the approxima on is?

We had the same problem when studying Numerical Integra on. Theorem. . provided bounds on the error when using, say, Simpson’s Rule to approx-

imate a definite integral. These bounds allowed us to determine that, for in-stance, using subintervals provided an approxima on within ±. of the ex-act value. The following theorem gives similar bounds for Taylor (and henceMaclaurin) polynomials.

Theorem . . Taylor’s Theorem

. Let f be a func on whose n+ th deriva ve exists on an interval I and let c be in I.Then, for each x in I, there exists zx between x and c such that

f(x) = f(c) + f ′(c)(x− c) +f ′′(c)

!(x− c) + · · ·+ f (n)(c)

n!(x− c)n + Rn(x),

where Rn(x) =f (n+ )(zx)(n+ )!

(x− c)(n+ ).

.∣∣Rn(x)

∣∣ ≤ max

∣∣ f (n+ )(z)

∣∣

(n+ )!

∣∣(x− c)(n+ )

∣∣, where z is in I.

The first part of Taylor’s Theorem states that f(x) = pn(x) + Rn(x), wherepn(x) is the nth order Taylor polynomial and Rn(x) is the remainder, or error, inthe Taylor approxima on. The second part gives bounds on how big that errorcan be. If the (n + )th deriva ve is large on I, the error may be large; if x is farfrom c, the error may also be large. However, the (n+ )! term in the denomi-nator tends to ensure that the error gets smaller as n increases.

The following example computes error es mates for the approxima ons ofln . and ln made in Example . . .

Example . . Finding error bounds of a Taylor polynomialUse Theorem . . to find error bounds when approxima ng ln . and ln withp (x), the Taylor polynomial of degree of f(x) = ln x at x = , as calculated inExample . . .

Notes:

Chapter Sequences and Series

S

. We start with the approxima on of ln . with p ( . ). The theorem ref-erences an open interval I that contains both x and c. The smaller theinterval we use the be er; it will give us a more accurate (and smaller!)approxima on of the error. We let I = ( . , . ), as this interval containsboth c = and x = . .The theorem references max

∣∣f (n+ )(z)

∣∣. In our situa on, this is asking

“How big can the th deriva ve of y = ln x be on the interval ( . , . )?”The seventh deriva ve is y = − !/x . The largest value it a ains on I isabout . Thus we can bound the error as:

∣∣R ( . )

∣∣ ≤ max

∣∣f ( )(z)

∣∣

!

∣∣( . − )

∣∣

≤ ·

≈ . .

We computed p ( . ) = . ; using a calculator, we find ln . ≈. , so the actual error is about . , which is less than our

bound of . . This affirms Taylor’s Theorem; the theorem states thatour approxima on would be within about thousandths of the actualvalue, whereas the approxima on was actually closer.

. We again find an interval I that contains both c = and x = ; we chooseI = ( . , . ). The maximum value of the seventh deriva ve of f on thisinterval is again about (as the largest values come near x = . ).Thus

∣∣R ( )

∣∣ ≤ max

∣∣f ( )(z)

∣∣

!

∣∣( − )

∣∣

≤ ·

≈ . .

This bound is not as nearly as good as before. Using the degree Taylorpolynomial at x = will bring us within . of the correct answer. Asp ( ) ≈ . , our error es mate guarantees that the actual value ofln is somewhere between . and . . These bounds are notpar cularly useful.In reality, our approxima on was only off by about . . However, weare approxima ng ostensibly because we do not know the real answer. Inorder to be assured that we have a good approxima on, we would haveto resort to using a polynomial of higher degree.

Notes:

f(x) = cos x ⇒ f( ) =f ′(x) = − sin x ⇒ f ′( ) =f ′′(x) = − cos x ⇒ f ′′( ) = −f ′′′(x) = sin x ⇒ f ′′′( ) =

f ( )(x) = cos x ⇒ f ( )( ) =

f ( )(x) = − sin x ⇒ f ( )( ) =

f ( )(x) = − cos x ⇒ f ( )( ) = −f ( )(x) = sin x ⇒ f ( )( ) =

f ( )(x) = cos x ⇒ f ( )( ) =

f ( )(x) = − sin x ⇒ f ( )( ) =

Figure . . : A table of the deriva ves off(x) = cos x evaluated at x = .

.....

y = p8(x)

.

−

.

−

.

−

.

−

.

−

...... −..

x

.

y

... f(x) = cos x

Figure . . : A graph of f(x) = cos x andits degree Maclaurin polynomial.

. Taylor Polynomials

We prac ce again. This me, we use Taylor’s theorem to find n that guaran-tees our approxima on is within a certain amount.

Example . . Finding sufficiently accurate Taylor polynomialsFind n such that the nth Taylor polynomial of f(x) = cos x at x = approximatescos to within . of the actual answer. What is pn( )?

S Following Taylor’s theorem, we need bounds on the size ofthe deriva ves of f(x) = cos x. In the case of this trigonometric func on, this iseasy. All deriva ves of cosine are± sin x or± cos x. In all cases, these func onsare never greater than in absolute value. We want the error to be less than. . To find the appropriate n, consider the following inequali es:

max∣∣f (n+ )(z)

∣∣

(n+ )!

∣∣( − )(n+ )

∣∣ ≤ .

(n+ )!· (n+ ) ≤ .

We find an n that sa sfies this last inequality with trial–and–error. When n = ,

we have+

( + )!≈ . ; when n = , we have

+

( + )!≈ . <

. . Thus we want to approximate cos with p ( ).

We now set out to compute p (x). We again need a table of the deriva vesof f(x) = cos x evaluated at x = . A table of these values is given in Figure . . .No ce how the deriva ves, evaluated at x = , follow a certain pa ern. All theodd powers of x in the Taylor polynomial will disappear as their coefficient is .While our error bounds state that we need p (x), our work shows that this willbe the same as p (x).

Since we are forming our polynomial at x = , we are crea ng a Maclaurinpolynomial, and:

p (x) = f( ) + f ′( )x+f ′′( )

!x +

f ′′′( )

!x + · · ·+ f ( )( )

!x

= −!x +

!x −

!x +

!x

We finally approximate cos :

cos ≈ p ( ) = − ≈ − . .

Our error bound guarantee that this approxima on is within . of the correctanswer. Technology shows us that our approxima on is actually within about. of the correct answer.

Figure . . shows a graph of y = p (x) and y = cos x. Note how well thetwo func ons agree on about (−π, π).

Notes:

f(x) =√x ⇒ f( ) =

f ′(x) = √x

⇒ f ′( ) =

f ′′(x) = −x /

⇒ f ′′( ) =−

f ′′′(x) =x /

⇒ f ′′′( ) =

f ( )(x) = −x /

⇒ f ( )( ) =−

Figure . . : A table of the deriva ves off(x) =

√x evaluated at x = .

.....

.. y =√

x.y = p (x)

...... x.

y

Figure . . : A graph of f(x) =√x and

its degree Taylor polynomial at x = .

Chapter Sequences and Series

Example . . Finding and using Taylor polynomials

. Find the degree Taylor polynomial, p (x), for f(x) =√x at x = .

. Use p (x) to approximate√

.

. Find bounds on the error when approxima ng√

with p ( ).

S

. We begin by evalua ng the deriva ves of f at x = . This is done in Figure. . . These values allow us to form the Taylor polynomial p (x):

p (x) = + (x− )+− /

!(x− ) +

/

!(x− ) +

− /

!(x− ) .

. As p (x) ≈ √x near x = , we approximate

√with p ( ) = . .

. To find a bound on the error, we need an open interval that contains x =and x = . We set I = ( . , . ). The largest value the fi h deriva ve off(x) =

√x takes on this interval is near x = . , at about . . Thus

∣∣R ( )

∣∣ ≤ .

!

∣∣( − )

∣∣ ≈ . .

This shows our approxima on is accurate to at least the first places a erthe decimal. (It turns out that our approxima on is actually accurate toplaces a er the decimal.) A graph of f(x) =

√x and p (x) is given in

Figure . . . Note how the two func ons are nearly indis nguishable on( , ).

Our final example gives a brief introduc on to using Taylor polynomials tosolve differen al equa ons.

Example . . Approxima ng an unknown func onA func on y = f(x) is unknown save for the following two facts.

. y( ) = f( ) = , and

. y ′ = y

(This second fact says that amazingly, the deriva ve of the func on is actuallythe func on squared!)

Find the degree Maclaurin polynomial p (x) of y = f(x).

Notes:

.....

.. y =− x

.

y = p (x)

.−

.− .5

..5

..... x.

y

Figure . . : A graph of y = − /(x− )and y = p (x) from Example . . .

. Taylor Polynomials

S Onemight ini ally think that not enough informa on is givento find p (x). However, note how the second fact above actually lets us knowwhat y ′( ) is:

y ′ = y ⇒ y ′( ) = y ( ).

Since y( ) = , we conclude that y ′( ) = .Nowwe find informa on about y ′′. Star ng with y ′ = y , take deriva ves of

both sides, with respect to x. That means we must use implicit differen a on.

y ′ = yddx(y ′)=

ddx(y)

y ′′ = y · y ′.

Now evaluate both sides at x = :

y ′′( ) = y( ) · y ′( )

y ′′( ) =

We repeat this once more to find y ′′′( ). We again use implicit differen a on;this me the Product Rule is also required.

ddx(y ′′)=

ddx(yy ′)

y ′′′ = y ′ · y ′ + y · y ′′.

Now evaluate both sides at x = :

y ′′′( ) = y ′( ) + y( )y ′′( )

y ′′′( ) = + =

In summary, we have:

y( ) = y ′( ) = y ′′( ) = y ′′′( ) = .

We can now form p (x):

p (x) = + x+!x +

!x

= + x+ x + x .

It turns out that the differen al equa on we started with, y ′ = y , wherey( ) = , can be solved without too much difficulty: y = − x

. Figure . .shows this func on plo ed with p (x). Note how similar they are near x = .

Notes:

Chapter Sequences and Series

It is beyond the scope of this text to pursue error analysis when using Tay-lor polynomials to approximate solu ons to differen al equa ons. This topic iso en broached in introductory Differen al Equa ons courses and usually cov-ered in depth in Numerical Analysis courses. Such an analysis is very important;one needs to know how good their approxima on is. We explored this examplesimply to demonstrate the usefulness of Taylor polynomials.

Most of this chapter has been devoted to the study of infinite series. Thissec on has taken a step back from this study, focusing instead on finite summa-on of terms. In the next sec on, we explore Taylor Series, where we represent

a func on with an infinite series.

Notes:

Exercises .Terms and Concepts

. What is the difference between a Taylor polynomial and aMaclaurin polynomial?

. T/F: In general, pn(x) approximates f(x) be er and be eras n gets larger.

. For some func on f(x), theMaclaurin polynomial of degreeis p (x) = + x− x + x − x . What is p (x)?

. For some func on f(x), theMaclaurin polynomial of degreeis p (x) = + x− x + x − x . What is f ′′′( )?

ProblemsIn Exercises – , find the Maclaurin polynomial of degreen for the given func on.

. f(x) = e−x, n =

. f(x) = sin x, n =

. f(x) = x · ex, n =

. f(x) = tan x, n =

. f(x) = e x, n =

. f(x) = − x, n =

. f(x) =+ x

, n =

. f(x) =+ x

, n =

In Exercises – , find the Taylor polynomial of degree n,at x = c, for the given func on.

. f(x) =√x, n = , c =

. f(x) = ln(x+ ), n = , c =

. f(x) = cos x, n = , c = π/

. f(x) = sin x, n = , c = π/

. f(x) =x, n = , c =

. f(x) =x, n = , c =

. f(x) =x +

, n = , c = −

. f(x) = x cos x, n = , c = π

In Exercises – , approximate the func on value with theindicated Taylor polynomial and give approximate bounds onthe error.

. Approximate sin . with the Maclaurin polynomial of de-gree .

. Approximate cos with the Maclaurin polynomial of de-gree .

. Approximate√

with the Taylor polynomial of degreecentered at x = .

. Approximate ln . with the Taylor polynomial of degreecentered at x = .

Exercises – ask for an n to be found such that pn(x) ap-proximates f(x) within a certain bound of accuracy.

. Find n such that the Maclaurin polynomial of degree n off(x) = ex approximates ewithin . of the actual value.

. Find n such that the Taylor polynomial of degree n of f(x) =√x, centered at x = , approximates

√within . of

the actual value.

. Find n such that the Maclaurin polynomial of degree n off(x) = cos x approximates cos π/ within . of the ac-tual value.

. Find n such that the Maclaurin polynomial of degree n off(x) = sin x approximates cos π within . of the actualvalue.

In Exercises – , find the nth term of the indicated Taylorpolynomial.

. Find a formula for the nth term of theMaclaurin polynomialfor f(x) = ex.

. Find a formula for the nth term of theMaclaurin polynomialfor f(x) = cos x.

. Find a formula for the nth term of theMaclaurin polynomialfor f(x) = sin x.

. Find a formula for the nth term of theMaclaurin polynomialfor f(x) = − x

.

. Find a formula for the nth term of theMaclaurin polynomialfor f(x) =

+ x.

. Find a formula for the nth term of the Taylor polynomial forf(x) = ln x centered at x = .

In Exercises – , approximate the solu on to the givendifferen al equa on with a degree Maclaurin polynomial.

. y′ = y, y( ) =

. y′ = y, y( ) =

. y′ =y, y( ) =

f(x) = cos x ⇒ f( ) =f ′(x) = − sin x ⇒ f ′( ) =f ′′(x) = − cos x ⇒ f ′′( ) = −f ′′′(x) = sin x ⇒ f ′′′( ) =

f ( )(x) = cos x ⇒ f ( )( ) =

f ( )(x) = − sin x ⇒ f ( )( ) =

f ( )(x) = − cos x ⇒ f ( )( ) = −f ( )(x) = sin x ⇒ f ( )( ) =

f ( )(x) = cos x ⇒ f ( )( ) =

f ( )(x) = − sin x ⇒ f ( )( ) =

Figure . . : A table of the deriva ves off(x) = cos x evaluated at x = .

. Taylor Series

. Taylor Series

In Sec on . , we showed how certain func ons can be represented by a powerseries func on. In Sec on . , we showed how we can approximate func onswith polynomials, given that enough deriva ve informa on is available. In thissec on we combine these concepts: if a func on f(x) is infinitely differen able,we show how to represent it with a power series func on.

Defini on . . Taylor and Maclaurin Series

Let f(x) have deriva ves of all orders at x = c.

. The Taylor Series of f(x), centered at c is

∞∑

n=

f (n)(c)n!

(x− c)n.

. Se ng c = gives theMaclaurin Series of f(x):

∞∑

n=

f (n)( )

n!xn.

If pn(x) is the nth degree Taylor polynomial for f(x) centered at x = c, we sawhow f(x) is approximately equal to pn(x) near x = c. We also saw how increasingthe degree of the polynomial generally reduced the error.

We are now considering series, where we sum an infinite set of terms. Ourul mate hope is to see the error vanish and claim a func on is equal to its Taylorseries.

When crea ng the Taylor polynomial of degree n for a func on f(x) at x = c,we needed to evaluate f, and the first n deriva ves of f, at x = c. When crea ngthe Taylor series of f, it helps to find a pa ern that describes the nth deriva veof f at x = c. We demonstrate this in the next two examples.

Example . . The Maclaurin series of f(x) = cos xFind the Maclaurin series of f(x) = cos x.

S In Example . . we found the th degree Maclaurin poly-nomial of cos x. In doing so, we created the table shown in Figure . . . No-ce how f (n)( ) = when n is odd, f (n)( ) = when n is divisible by , and

f (n)( ) = − when n is even but not divisible by . Thus the Maclaurin series

Notes:

f(x) = ln x ⇒ f( ) =f ′(x) = /x ⇒ f ′( ) =f ′′(x) = − /x ⇒ f ′′( ) = −f ′′′(x) = /x ⇒ f ′′′( ) =

f ( )(x) = − /x ⇒ f ( )( ) = −f ( )(x) = /x ⇒ f ( )( ) =...

...f (n)(x) = ⇒ f (n)( ) =

(− )n+ (n− )!

xn(− )n+ (n− )!

Figure . . : Deriva ves of ln x evaluatedat x = .

Chapter Sequences and Series

of cos x is− x

+x!− x

!+

x!− · · ·

We can go further and write this as a summa on. Since we only need the termswhere the power of x is even, we write the power series in terms of x n:

∞∑

n=

(− )nx n

( n)!.

This Maclaurin series is a special type of power series. As such, we should de-termine its interval of convergence. Applying the Ra o Test, we have

limn→∞

∣∣∣∣∣(− )n+

x (n+ )

((n+ )

)!

∣∣∣∣∣

/∣∣∣∣(− )n

x n

( n)!

∣∣∣∣= lim

n→∞

∣∣∣∣

x n+

x n

∣∣∣∣

( n)!( n+ )!

= limn→∞

x( n+ )( n+ )

.

For any fixed x, this limit is . Therefore this power series has an infinite radiusof convergence, converging for all x. It is important to note what we have, andhave not, determined: we have determined the Maclaurin series for cos x alongwith its interval of convergence. We have not shown that cos x is equal to thispower series.

Example . . The Taylor series of f(x) = ln x at x =Find the Taylor series of f(x) = ln x centered at x = .

S Figure . . shows the nth deriva ve of ln x evaluated at x =for n = , . . . , , along with an expression for the nth term:

f (n)( ) = (− )n+ (n− )! for n ≥ .

Remember that this is what dis nguishes Taylor series from Taylor polynomials;we are very interested in finding a pa ern for the nth term, not just finding afinite set of coefficients for a polynomial. Since f( ) = ln = , we skip thefirst term and start the summa on with n = , giving the Taylor series for ln x,centered at x = , as

∞∑

n=

(− )n+ (n− )!n!(x− )n =

∞∑

n=

(− )n+(x− )n

n.

We now determine the interval of convergence, using the Ra o Test.

limn→∞

∣∣∣∣(− )n+

(x− )n+

n+

∣∣∣∣

/∣∣∣∣(− )n+

(x− )n

n

∣∣∣∣= lim

n→∞

∣∣∣∣

(x− )n+

(x− )n

∣∣∣∣

nn+

=∣∣(x− )

∣∣.

Notes:

Note: It can be shown that ln x is equal tothis Taylor series on ( , ]. From the workin Example . . , this jus fies our previ-ous declara on that the Alterna ng Har-monic Series converges to ln .

. Taylor Series

By the Ra o Test, we have convergence when∣∣(x − )

∣∣ < : the radius of con-

vergence is , and we have convergence on ( , ). We now check the endpoints.At x = , the series is

∞∑

n=

(− )n+(− )n

n= −

∞∑

n=n,

which diverges (it is the Harmonic Series mes (− ).)At x = , the series is

∞∑

n=

(− )n+( )n

n=

∞∑

n=

(− )n+n,

the Alterna ng Harmonic Series, which converges.We have found the Taylor series of ln x centered at x = , and have deter-

mined the series converges on ( , ]. We cannot (yet) say that ln x is equal tothis Taylor series on ( , ].

It is important to note that Defini on . . defines a Taylor series given afunc on f(x), but makes no claim about their equality. We will find that “mostof the me” they are equal, but we need to consider the condi ons that allowus to conclude this.

Theorem . . states that the error between a func on f(x) and its nth–degree Taylor polynomial pn(x) is Rn(x), where

∣∣Rn(x)

∣∣ ≤ max

∣∣ f (n+ )(z)

∣∣

(n+ )!

∣∣(x− c)(n+ )

∣∣.

If Rn(x) goes to for each x in an interval I as n approaches infinity, we con-clude that the func on is equal to its Taylor series expansion.

Theorem . . Func on and Taylor Series Equality

Let f(x) have deriva ves of all orders at x = c, let Rn(x) be as stated inTheorem . . , and let I be an interval on which the Taylor series of f(x)converges. If lim

n→∞Rn(x) = for all x in I, then

f(x) =∞∑

n=

f (n)(c)n!

(x− c)n on I.

Notes:

Chapter Sequences and Series

We demonstrate the use of this theorem in an example.

Example . . Establishing equality of a func on and its Taylor seriesShow that f(x) = cos x is equal to itsMaclaurin series, as found in Example . . ,for all x.

S Given a value x, the magnitude of the error term Rn(x) isbounded by

∣∣Rn(x)

∣∣ ≤ max

∣∣ f (n+ )(z)

∣∣

(n+ )!

∣∣xn+

∣∣.

Since all deriva ves of cos x are± sin xor± cos x, whosemagnitudes are boundedby , we can state

∣∣Rn(x)

∣∣ ≤

(n+ )!

∣∣xn+

∣∣

which implies

− |xn+ |(n+ )!

≤ Rn(x) ≤|xn+ |(n+ )!

. ( . )

For any x, limn→∞

xn+

(n+ )!= . Applying the Squeeze Theorem to Equa on ( . ),

we conclude that limn→∞

Rn(x) = for all x, and hence

cos x =∞∑

n=

(− )nx n

( n)!for all x.

It is natural to assume that a func on is equal to its Taylor series on the series’interval of convergence, but this is not always the case. In order to properlyestablish equality, one must use Theorem . . . This is a bit disappoin ng, aswe developed beau ful techniques for determining the interval of convergenceof a power series, and proving that Rn(x) → can be difficult. For instance, itis not a simple task to show that ln x equals its Taylor series on ( , ] as foundin Example . . ; in the Exercises, the reader is only asked to show equality on( , ), which is simpler.

There is good news. A func on f(x) that is equal to its Taylor series, centeredat any point the domain of f(x), is said to be an analy c func on, and most, ifnot all, func ons that we encounter within this course are analy c func ons.Generally speaking, any func on that one creates with elementary func ons(polynomials, exponen als, trigonometric func ons, etc.) that is not piecewisedefined is probably analy c. Formost func ons, we assume the func on is equalto its Taylor series on the series’ interval of convergence and only use Theorem. . when we suspect something may not work as expected.

Notes:

. Taylor Series

We develop the Taylor series for one more important func on, then give atable of the Taylor series for a number of common func ons.

Example . . The Binomial SeriesFind the Maclaurin series of f(x) = ( + x)k, k ̸= .

S When k is a posi ve integer, the Maclaurin series is finite.For instance, when k = , we have

f(x) = ( + x) = + x+ x + x + x .

The coefficients of x when k is a posi ve integer are known as the binomial co-efficients, giving the series we are developing its name.

When k = / , we have f(x) =√

+ x. Knowing a series representa on ofthis func on would give a useful way of approxima ng

√. , for instance.

To develop the Maclaurin series for f(x) = ( + x)k for any value of k ̸= ,we consider the deriva ves of f evaluated at x = :

f(x) = ( + x)k f( ) =

f ′(x) = k( + x)k− f ′( ) = k

f ′′(x) = k(k− )( + x)k− f ′′( ) = k(k− )

f ′′′(x) = k(k− )(k− )( + x)k− f ′′′( ) = k(k− )(k− )

......

f (n)(x) = k(k− ) · · ·(

k− (n− ))

( + x)k−n f (n)( ) = k(k− ) · · ·(

k− (n− ))

Thus the Maclaurin series for f(x) = ( + x)k is

+kx+k(k− )

!x +

k(k− )(k− )

!x +. . .+

k(k− ) · · ·(k− (n− )

)

n!xn+. . .

It is important to determine the interval of convergence of this series. With

an =k(k− ) · · ·

(k− (n− )

)

n!xn,

we apply the Ra o Test:

limn→∞

|an+ ||an|

= limn→∞

∣∣∣∣

k(k− ) · · · (k− n)(n+ )!

xn+∣∣∣∣

/∣∣∣∣∣

k(k− ) · · ·(k− (n− )

)

n!xn∣∣∣∣∣

= limn→∞

∣∣∣∣

k− nn+

x∣∣∣∣

= |x|.

Notes:

Chapter Sequences and Series

The series converges absolutely when the limit of the Ra o Test is less than; therefore, we have absolute convergence when |x| < .

While outside the scope of this text, the interval of convergence dependson the value of k. When k > , the interval of convergence is [− , ]. When− < k < , the interval of convergence is [− , ). If k ≤ − , the interval ofconvergence is (− , ).

We learned that Taylor polynomials offer a way of approxima ng a “difficultto compute” func on with a polynomial. Taylor series offer a way of exactlyrepresen ng a func on with a series. One probably can see the use of a goodapproxima on; is there any use of represen ng a func on exactly as a series?

Whilewe should not overlook themathema cal beauty of Taylor series (whichis reason enough to study them), there are prac cal uses as well. They providea valuable tool for solving a variety of problems, including problems rela ng tointegra on and differen al equa ons.

In Key Idea . . (on the following page) we give a table of the Taylor seriesof a number of common func ons. We then give a theorem about the “algebraof power series,” that is, how we can combine power series to create powerseries of new func ons. This allows us to find the Taylor series of func ons likef(x) = ex cos x by knowing the Taylor series of ex and cos x.

Before we inves gate combining func ons, consider the Taylor series for thearctangent func on (see Key Idea . . ). Knowing that tan− ( ) = π/ , we canuse this series to approximate the value of π:

π= tan− ( ) = − + − + − · · ·

π =

(

− + − + − · · ·)

Unfortunately, this par cular expansion of π converges very slowly. The firstterms approximate π as . , which is not par cularly good.

Notes:

. Taylor Series

Key Idea . . Important Taylor Series Expansions

Func on and Series First Few Terms Interval ofConvergence

ex =∞∑

n=

xn

n!+ x+

x!+

x!+ · · · (−∞,∞)

sin x =∞∑

n=

(− )nx n+

( n+ )!x− x

!+

x!− x

!+ · · · (−∞,∞)

cos x =∞∑

n=

(− )nx n

( n)!− x

!+

x!− x

!+ · · · (−∞,∞)

ln x =∞∑

n=

(− )n+(x− )n

n(x− )− (x− )

+(x− ) − · · · ( , ]

− x=

∞∑

n=

xn + x+ x + x + · · · (− , )

( + x)k =∞∑

n=

k(k− ) · · ·(

k− (n− ))

n!xn + kx+

k(k− )

!x + · · · (− , )a

tan− x =∞∑

n=

(− )nx n+

n+x− x

+x − x

+ · · · [− , ]

aConvergence at x = ± depends on the value of k.

Theorem . . Algebra of Power Series

Let f(x) =∞∑

n=

anxn and g(x) =∞∑

n=

bnxn converge absolutely for |x| < R, and let h(x) be con nuous.

. f(x)± g(x) =∞∑

n=

(an ± bn)xn for |x| < R.

. f(x)g(x) =

( ∞∑

n=

anxn)( ∞∑

n=

bnxn)

=

∞∑

n=

(a bn + a bn− + . . . anb

)xn for |x| < R.

. f(h(x)

)=

∞∑

n=

an(h(x)

)n for |h(x)| < R.

Notes:

Chapter Sequences and Series

Example . . Combining Taylor seriesWrite out the first terms of the Taylor Series for f(x) = ex cos x using Key Idea. . and Theorem . . .

S Key Idea . . informs us that

ex = + x+x!+

x!+ · · · and cos x = − x

!+

x!+ · · · .

Applying Theorem . . , we find that

ex cos x =(

+ x+x!+

x!+ · · ·

)(

− x!+

x!+ · · ·

)

.

Distribute the right hand expression across the le :

=

(

− x!+

x!+ · · ·

)

+ x(

− x!+

x!+ · · ·

)

+x!

(

− x!+

x!+ · · ·

)

+x!

(

− x!+

x!+ · · ·

)

+x!

(

− x!+

x!+ · · ·

)

+ · · ·

Distribute again and collect like terms.

= + x− x − x − x+

x+ · · ·

While this process is a bit tedious, it is much faster than evalua ng all the nec-essary deriva ves of ex cos x and compu ng the Taylor series directly.

Because the series for ex and cos x both converge on (−∞,∞), so does theseries expansion for ex cos x.

Example . . Crea ng new Taylor seriesUse Theorem . . to create series for y = sin(x ) and y = ln(

√x).

S Given that

sin x =∞∑

n=

(− )nx n+

( n+ )!= x− x

!+

x!− x

!+ · · · ,

we simply subs tute x for x in the series, giving

sin(x ) =

∞∑

n=

(− )n(x ) n+

( n+ )!= x − x

!+

x!− x

!· · · .

Notes:

Note: In Example . . , one could cre-ate a series for ln(

√x) by simply recogniz-

ing that ln(√x) = ln(x / ) = / ln x,

and hence mul plying the Taylor seriesfor ln x by / . This example was cho-sen to demonstrate other aspects of se-ries, such as the fact that the interval ofconvergence changes.

. Taylor Series

Since the Taylor series for sin x has an infinite radius of convergence, so does theTaylor series for sin(x ).

The Taylor expansion for ln x given in Key Idea . . is centered at x = , sowe will center the series for ln(

√x) at x = as well. With

ln x =∞∑

n=

(− )n+(x− )n

n= (x− )− (x− )

+(x− ) − · · · ,

we subs tute√x for x to obtain

ln(√x) =

∞∑

n=

(− )n+(√x− )n

n= (

√x− )− (

√x− )

+(√x− ) −· · · .

While this is not strictly a power series, it is a series that allows us to study thefunc on ln(

√x). Since the interval of convergence of ln x is ( , ], and the range

of√x on ( , ] is ( , ], the interval of convergence of this series expansion of

ln(√x) is ( , ].

Example . . Using Taylor series to evaluate definite integrals

Use the Taylor series of e−x to evaluate∫

e−x dx.

S We learned, when studying Numerical Integra on, that e−x

does not have an an deriva ve expressible in terms of elementary func ons.This means any definite integral of this func on must have its value approxi-mated, and not computed exactly.

We can quickly write out the Taylor series for e−x using the Taylor series ofex:

ex =∞∑

n=

xn

n!= + x+

x!+

x!+ · · ·

and so

e−x =

∞∑

n=

(−x )n

n!

=

∞∑

n=

(− )nx n

n!

= − x +x!− x

!+ · · · .

Notes:

Chapter Sequences and Series

We use Theorem . . to integrate:∫

e−x dx = C+ x− x+

x· !

− x· !

+ · · ·+ (− )nx n+

( n+ )n!+ · · ·

This is the an deriva ve of e−x ; while we can write it out as a series, we can-not write it out in terms of elementary func ons. We can evaluate the definite

integral∫

e−x dx using this an deriva ve; subs tu ng and for x and sub-trac ng gives

∫

e−x dx = − + · !− · !

+ · !· · · .

Summing the terms shown above give the approxima on of . . Sincethis is an alterna ng series, we can use the Alterna ng Series Approxima onTheorem, (Theorem . . ), to determine how accurate this approxima on is.The next term of the series is /( · !) ≈ . . Thus we know ourapproxima on is within . of the actual value of the integral. This isarguably much less work than using Simpson’s Rule to approximate the value ofthe integral.

Example . . Using Taylor series to solve differen al equa onsSolve the differen al equa on y ′ = y in terms of a power series, and use thetheory of Taylor series to recognize the solu on in terms of an elementary func-on.

S We found the first terms of the power series solu on tothis differen al equa on in Example . . in Sec on . . These are:

a = , a = , a = = , a = · = , a = · · = .

We include the “unsimplified” expressions for the coefficients found in Example. . as we are looking for a pa ern. It can be shown that an = n/n!. Thus the

solu on, wri en as a power series, is

y =∞∑

n=

n

n!xn =

∞∑

n=

( x)n

n!.

Using Key Idea . . and Theorem . . , we recognize f(x) = e x:

ex =∞∑

n=

xn

n!⇒ e x =

∞∑

n=

( x)n

n!.

Notes:

. Taylor Series

Finding a pa ern in the coefficients that match the series expansion of aknown func on, such as those shown in Key Idea . . , can be difficult. What ifthe coefficients in the previous example were given in their reduced form; howcould we s ll recover the func on y = e x?

Suppose that all we know is that

a = , a = , a = , a = , a = .

Defini on . . states that each term of the Taylor expansion of a func on in-cludes an n!. This allows us to say that

a = =b!, a = =

b!, and a = =

b!

for some values b , b and b . Solving for these values, we see that b = ,b = and b = . That is, we are recovering the pa ern we had previouslyseen, allowing us to write

f(x) =∞∑

n=

anxn =∞∑

n=

bnn!xn

= + x+!x +

!x +

!x + · · ·

From here it is easier to recognize that the series is describing an exponen alfunc on.

There are simpler, more direct ways of solving the differen al equa on y ′ =y. We applied power series techniques to this equa on to demonstrate its u l-

ity, and went on to show how some mes we are able to recover the solu on interms of elementary func ons using the theory of Taylor series. Most differen-al equa ons faced in real scien fic and engineering situa ons are much more

complicated than this one, but power series can offer a valuable tool in finding,or at least approxima ng, the solu on.

This chapter introduced sequences, which are ordered lists of numbers, fol-lowed by series, wherein we add up the terms of a sequence. We quickly sawthat such sums do not always add up to “infinity,” but rather converge. We stud-ied tests for convergence, then ended the chapter with a formal way of definingfunc ons based on series. Such “series–defined func ons” are a valuable toolin solving a number of different problems throughout science and engineering.

Coming in the next chapters are new ways of defining curves in the planeapart from using func ons of the form y = f(x). Curves created by these newmethods can be beau ful, useful, and important.

Notes:

Exercises .Terms and Concepts

. What is the difference between a Taylor polynomial and aTaylor series?

. What theoremmustwe use to show that a func on is equalto its Taylor series?

ProblemsKey Idea . . gives the nth term of the Taylor series of com-mon func ons. In Exercises – , verify the formula given inthe Key Idea by finding the first few terms of the Taylor seriesof the given func on and iden fying a pa ern.

. f(x) = ex; c =

. f(x) = sin x; c =

. f(x) = /( − x); c =

. f(x) = tan− x; c =

In Exercises – , find a formula for the nth term of the Tay-lor series of f(x), centered at c, by finding the coefficients ofthe first few powers of x and looking for a pa ern. (The for-mulas for several of these are found in Key Idea . . ; showwork verifying these formula.)

. f(x) = cos x; c = π/

. f(x) = /x; c =

. f(x) = e−x; c =

. f(x) = ln( + x); c =

. f(x) = x/(x+ ); c =

. f(x) = sin x; c = π/

In Exercises – , show that the Taylor series for f(x), asgiven in Key Idea . . , is equal to f(x) by applying Theorem. . ; that is, show lim

n→∞Rn(x) = .

. f(x) = ex

. f(x) = sin x

. f(x) = ln x (show equality only on ( , ))

. f(x) = /( − x) (show equality only on (− , ))

In Exercises – , use the Taylor series given in Key Idea. . to verify the given iden ty.

. cos(−x) = cos x

. sin(−x) = − sin x

. ddx

(

sin x)

= cos x

. ddx

(

cos x)

= − sin x

In Exercises – , write out the first terms of the Binomialseries with the given k-value.

. k = /

. k = − /

. k = /

. k =

In Exercises – , use the Taylor series given in Key Idea. . to create the Taylor series of the given func ons.

. f(x) = cos(

x)

. f(x) = e−x

. f(x) = sin(

x+)

. f(x) = tan−(

x/)

. f(x) = ex sin x (only find the first terms)

. f(x) = ( + x) / cos x (only find the first terms)

In Exercises – , approximate the value of the given def-inite integral by using the first nonzero terms of the inte-grand’s Taylor series.

.∫

√π

sin(

x)

dx

.∫ π /

cos(√

x)

dx

A: S T S PChapterSec on .

. Answers will vary.

. Answers will vary.

. Answers will vary.

. velocity

. / x + C

. / x − x+ C

. s+ C

. − /(t) + C

. tan θ + C

. sec x− csc x+ C

. t/ ln + C

. / t + t + t+ C

. x / + C

. ax+ C

. − cos x+

. x − x +

. x/ ln + − / ln

. x − x +

. θ − sin(θ)− π +

. x−

. dy = ( xex cos x+ x ex cos x− x ex sin x)dx

Sec on .

. Answers will vary.

.

. (a)(b)(c)(d)(e) −(f)

. (a)(b)(c)(d)(e)(f)

. (a) π

(b) π

(c) π

(d) π

. (a) −(b) −

(c) −(d) −

. (a)(b)(c) −(d) −

. (a) /s(b)(c) .

. (a) /s(b)(c) t =(d) t = +

√≈ . seconds

.

.

.

. −

. / x − / x + / x − x+ C

. / t / − /t+ t/ ln + C

Sec on .

. limits

. Rectangles.

. + + =

. − + + + =

. + / + / + / + / = /

. / + / + / + / = /

. Answers may vary;∑

i= i

. Answers may vary;∑

i=i

i+

. · =

.

. −

.

.

.

.

. π/ + π/(√

) ≈ .

. .

. (a) Exact expressions will vary; ( +n)n .

(b) / , / , /

(c) /

. (a) .(b) , ,(c)

. (a) Exact expressions will vary; − /n.(b) , , /

(c)

. F(x) = tan x+

. G(t) = / t − / t + t+

. G(t) = sin t− cos t−

Sec on .

. Answers will vary.

. T

.

.

.

. ( − / )/ ln

. −

. /

. /

. /

. /

. /

.

.

. Explana ons will vary. A sketch will help.

. c = /√

. c = ln(e− ) ≈ .

. /π

.

.

. −

.

. −

. − /s

. /s

. /

. /

. F′(x) = ( x + ) x +x

. F′(x) = x(x + )− (x+ )

Sec on .

. F

. They are superseded by the Trapezoidal Rule; it takes an equalamount of work and is generally more accurate.

. (a) /

(b) /

(c) /

. (a) ( +√

)π ≈ .

(b) ( +√

)π ≈ .

(c)

. (a) .

(b) / ≈ .

(c) / ≈ .

. (a)(b)(c)

. Trapezoidal Rule: .

Simpson’s Rule: .

. Trapezoidal Rule: .

Simpson’s Rule: .

. Trapezoidal Rule: .

Simpson’s Rule: .

. Trapezoidal Rule: .

Simpson’s Rule: .

. (a) n = (using max(

f ′′(x))

= )

(b) n = (using max(

f ( )(x))

= )

. (a) n = (using max(

f ′′(x))

= )

(b) n = (using max(

f ( )(x))

= )

. (a) Area is . cm .(b) Area is , yd .

ChapterSec on .

. Chain Rule.

. (x − ) + C

.(

x +)

+ C

. ln | x+ |+ C

. (x+ ) / − (x+ ) / + C = (x− )√x+ + C

. e√

x + C

. − x − x + C

. sin (x)+ C

. − sin( − x) + C

. ln | sec( x) + tan( x)|+ C

. sin(x )+ C

. The key is to rewrite cot x as cos x/ sin x, and let u = sin x.

. e x− + C

. e(x− ) + C

. ln(

ex +)

+ C

.x

ln + C

. ln (x) + C

. (ln x) + C

. x + x+ ln |x|+ C

. x − x + x− ln |x+ |+ C

. x − x+ ln |x+ |+ C

.√

tan−(

x√)

+ C

. sin−(

x√)

+ C

. sec− (|x|/ ) + C

.tan−

(

x−√)

√ + C

A.

. sin−( x− )

+ C

. −(x + )

+ C

. −√

− x + C

. − cos (x) + C

. ln |x− |+ C

. x + ln∣

∣x + x+∣

∣− x+ C

. ln∣

∣ x + x+∣

∣+ C

. tan−(

x)

+ C

. sec− (| x|) + C

. ln∣

∣x − x+∣

∣+ tan−( x− )

+ C

. ln∣

∣x − x+∣

∣+ x+tan−

(

x−√)

√ + C

. x + ln∣

∣x + x+∣

∣− x+tan−

(

x+√)

√ + C

. tan− (sin(x)) + C

.√x − x− + C

. − ln

. /

. ( − e)/

. π/

Sec on .

. T

. Determining which func ons in the integrand to set equal to “u”and which to set equal to “dv”.

. sin x− x cos x+ C

. −x cos x+ x sin x+ cos x+ C

. / ex + C

. − xe− x − e− x+ C

. / e x(sin x+ cos x) + C

. / e x(sin( x) + cos( x)) + C

.√

− x + x sin− (x) + C

. x tan− (x)− x + tan− (x) + C

. x ln |x| − x + C

. − x + x ln |x− | − x − ln |x− |+ C

. x ln |x| − x + C

. (x+ ) + (x+ ) (ln(x+ )) − (x+ ) ln(x+ ) + C

. ln | sin(x)| − x cot(x) + C

. (x − ) / + C

. x sec x− ln | sec x+ tan x|+ C

. / x(

sin(ln x)− cos(ln x))

+ C

. sin(√

x)

− √x cos

(√x)

+ C

.√xe

√x − e

√x + C

. π

.

. /

. e − e

. /(

eπ + e−π)

Sec on .

. F

. F

. − cos (x) + C

. cos x− cos x+ C

. sin x− sin x+ sin x+ C

. x − sin( x) + C

.(

− cos( x)− cos( x))

+ C

.(

sin( x)− sin( x))

+ C

.(

sin(x) + sin( x))

+ C

. tan (x)+ C

. tan (x)+

tan (x)+ C

. sec (x) − sec (x)+ C

. tan x− tan x+ x+ C

. (sec x tan x− ln | sec x+ tan x|) + C

.

. /

. /

. /

Sec on .

. backwards

. (a) tan θ + = sec θ

(b) sec θ.

.(

x√x + + ln |

√x + + x|

)

+ C

.(

sin− x+ x√

− x)

+ C

. x√x − − ln |x+

√x − |+ C

. x√

x + / + ln |√

x + / + x|+ C =

x√

x + + ln |√

x + + x|+ C

.(

x√

x − / − ln | x+√

x − / |)

+ C =

x√

x − − ln | x+√

x − |+ C

. sin−(

x√)

+ C (Trig. Subst. is not needed)

.√x − −

√sec− (x/

√) + C

.√x − + C (Trig. Subst. is not needed)

. −√x +

+ C (Trig. Subst. is not needed)

. x+x + x+ + tan−

( x+ )

+ C

.(

−√

−xx − sin− (x/

√)

)

+ C

. π/

.√

+ ln( +√

)

. sin− ( / ) +√

Note: the new lower bound isθ = sin− (− / ) and the new upper bound is θ = sin− ( / ).The final answer comes with recognizing thatsin− (− / ) = − sin− ( / ) and thatcos(

sin− ( / ))

= cos(

sin− (− / ))

=√

/ . A.

Sec on .

. ra onal

. Ax + B

x−

. Ax−

√ + Bx+

√

. ln |x− |+ ln |x+ |+ C

. (ln |x+ | − ln |x− |) + C

. ln |x+ | − x+ + C

. x+ + ln |x|+ ln |x+ |+ C

. − ln | x− |+ ln | x− |+ ln | x+ |+ C

. x + x+ ln |x− |+ ln |x+ | − + C

.(

− ln∣

∣x + x+∣

∣+ ln |x| −√

tan−(

x+√))

+ C

. ln∣

∣ x + x−∣

∣+ ln |x+ |+ C

. ln∣

∣x +∣

∣+ ln |x+ | − tan−( x )+ C

.(

ln∣

∣x − x+∣

∣+ ln |x− |)

+√

tan−(

x−√)

+ C

. ln( / ) ≈ .

. −π/ + tan− − ln( / ) ≈ .

Sec on .

. Because cosh x is always posi ve.

. coth x− csch x =(

ex + e−x

ex − e−x

)

−(

ex − e−x

)

=(e x + + e− x)− ( )

e x − + e− x

=e x − + e− x

e x − + e− x

=

. cosh x =(

ex + e−x)

=e x + + e− x

=(e x + e− x) +

=

(

e x + e− x+

)

=cosh x+

.

.ddx

[sech x] =ddx

[

ex + e−x

]

=− (ex − e−x)

(ex + e−x)

= − (ex − e−x)

(ex + e−x)(ex + e−x)

= −ex + e−x · e

x − e−x

ex + e−x

= − sech x tanh x

.∫

tanh x dx =∫ sinh x

cosh xdx

Let u = cosh x; du = (sinh x)dx

=

∫

udu

= ln |u|+ C= ln(cosh x) + C.

. cosh x

. x sec (x )

. sinh x+ cosh x

. − x(x )

√−x

. x√x −

. − csc x

. y = x

. y = (x+ ln )−

. y = x

. / ln(cosh( x)) + C

. / sinh x+ C or / cosh x+ C

. x cosh(x)− sinh(x) + C

. cosh− x/ + C = ln(

x+√x −

)

+ C

. cosh− (x / ) + C = ln(x +√x − ) + C

. tan− (x/ ) + ln |x− |+ ln |x+ |+ C

. tan− (ex) + C

. x tanh− x+ / ln |x − |+ C

.

.

Sec on .

. / ,∞/∞, · ∞,∞−∞, , ∞,∞

. F

. deriva ves; limits

. Answers will vary.

.

. −

.

. /

. ∞

.

.

. ∞

.

. −

.

.

. ∞

. ∞

.

.

.

.

.

.

.

A.

. −∞

.

Sec on .

. The interval of integra on is finite, and the integrand iscon nuous on that interval.

. converges; could also state< .

. p >

. e /

. /

. / ln

. diverges

.

. diverges

. diverges

. diverges

.

.

. − /

. diverges

.

. /

. diverges; Limit Comparison Test with /x.

. diverges; Limit Comparison Test with /x.

. converges; Direct Comparison Test with e−x.

. converges; Direct Comparison Test with /(x − ).

. converges; Direct Comparison Test with /ex.

ChapterSec on .

. T

. Answers will vary.

. π + π ≈ .

. π

. /

. / ln

. .

. − π/

. /

. All enclosed regions have the same area, with regions being thereflec on of adjacent regions. One region is formed on[π/ , π/ ], with area

√.

.

. /

. / ( −√

) ≈ .

.

.

. ,

Sec on .

. T

. Recall that “dx” does not just “sit there;” it is mul plied by A(x)and represents the thickness of a small slice of the solid.Therefore dx has units of in, giving A(x) dx the units of in .

. π√

/ units

. π / units

. π/ units

. π − π units

. (a) π/

(b) π/

(c) π/

(d) π/

. (a) π/

(b) π/

(c) π/

(d) π/

. (a) π /

(b) π / − π sinh− ( )

(c) π / + π sinh− ( )

. Placing the p of the cone at the origin such that the x-axis runsthrough the center of the circular base, we have A(x) = πx / .Thus the volume is π/ units .

. Orient the cone such that the p is at the origin and the x-axis isperpendicular to the base. The cross–sec ons of this cone areright, isosceles triangles with side length x/ ; thus thecross–sec onal areas are A(x) = x / , giving a volume of /units .

Sec on .

. T

. F

. π/ units

. π − π units

. π√

/ units

. π / units

. (a) π/

(b) π/

(c) π/

(d) π/

. (a) π/

(b) π/

(c) π/

(d) π/

. (a) π(√

− )

(b) π( −√

+ sinh− ( ))

Sec on .

. T

.√

. /

. /

A.

. /

. − ln( −√

) ≈ .

.∫ √

+ x dx

.∫

√

+ x dx

.∫

−

√

+ x−x dx

.∫

√

+ x dx

. .

. Simpson’s Rule fails, as it requires one to divide by . However,recognize the answer should be the same as for y = x ; why?

. Simpson’s Rule fails.

. .

. π∫

x√

dx = π√

. π∫

x√

+ x dx = π/ (√

− )

. π∫ √

− x√

+ x/( − x ) dx = π

Sec on .

. In SI units, it is one joule, i.e., one Newton–meter, or kg·m/s ·m.In Imperial Units, it is –lb.

. Smaller.

. (a) –lb(b) −

√≈ . –lb

. (a) · d · l –lb(b) %(c) ℓ( −

√/ ) ≈ . ℓ

. (a) –lb(b) , –lb(c) Yes, for the cable accounts for about % of the total work.

. –lb

. . J

. / –lb

. f · d/ J

. –lb

. (a) , . –lb(b) , . –lb(c) When . of water have been pumped from the tank,

leaving about . in the tank.

. , –lb

. , –lb

. , , J

Sec on .

. Answers will vary.

. . lb

. . lb

. . lb

. lb

. . lb

. (a) lb

(b) lb

. (a) . lb(b) lb

. (a) . lb(b) . lb

. .

ChapterSec on .

. Answers will vary.

. Answers will vary.

. , , , ,

. − ,− ,− ,− ,−

. an = n+

. an = · n−

. /

.

. diverges

. converges to

. diverges

. converges to e

. converges to

. converges to

. bounded

. bounded

. neither bounded above or below

. monotonically increasing

. never monotonic

. Let {an} be given such that limn→∞

|an| = . By the defini on ofthe limit of a sequence, given any ε > , there is am such that forall n > m,

∣

∣ |an| −∣

∣ < ε. Since∣

∣ |an| −∣

∣ = |an − |, thisdirectly implies that for all n > m, |an − | < ε, meaning thatlim

n→∞an = .

. A sketch of one proof method:Let any ε > be given. Since {an} and {bn} converge, thereexists an N > such that for all n ≥ N, both an and bn are withinε/ of L; we can conclude that they are at most ε apart from eachother. Since an ≤ cn ≤ bn, one can show that cn is within ε of L,showing that {cn} also converges to L.

Sec on .

. Answers will vary.

. One sequence is the sequence of terms {an}. The other is thesequence of nth par al sums, {Sn} = {

∑ni= ai}.

. F

. (a) − ,− ,− ,− ,−(b) Plot omi ed

. (a) − , ,− , ,−(b) Plot omi ed

. (a) , , , ,

(b) Plot omi ed

. (a) − . ,− . ,− . ,− . ,− .

A.

(b) Plot omi ed

. limn→∞

an = ; by Theorem . . the series diverges.

. limn→∞

an = ∞; by Theorem . . the series diverges.

. limn→∞

an = / ; by Theorem . . the series diverges.

. Converges; p-series with p = .

. Diverges; geometric series with r = / .

. Diverges; fails nth term test

. F

. Diverges; by Theorem . . this is half the Harmonic Series, whichdiverges by growing without bound. “Half of growing withoutbound” is s ll growing without bound.

. (a) Sn =−( / )n

/

(b) Converges to / .

. (a) Sn =(

n(n+ ))

(b) Diverges

. (a) Sn =− / n

/

(b) Converges to .

. (a) Sn =−(− / )n

/

(b) Converges to / .

. (a) With par al frac ons, an =(

n − n+

)

. Thus

Sn =(

− n+ − n+

)

.

(b) Converges to /

. (a) Sn = ln(

/(n+ ))

(b) Diverges (to−∞).

. (a) an = n(n+ ); using par al frac ons, the resul ng

telescoping sum reduces toSn =

(

+ + − n+ − n+ − n+

)

(b) Converges to / .

. (a) With par al frac ons, an =(

n− − n+

)

. Thus

Sn =(

/ − n − n+

)

.

(b) Converges to / .

. (a) The nth par al sum of the odd series is+ + + · · ·+ n− . The nth par al sum of the even

series is + + + · · ·+ n . Each term of the evenseries is less than the corresponding term of the oddseries, giving us our result.

(b) The nth par al sum of the odd series is+ + + · · ·+ n− . The nth par al sum of plus the

even series is + + + · · ·+(n− )

. Each term of theeven series is now greater than or equal to thecorresponding term of the odd series, with equality only onthe first term. This gives us the result.

(c) If the odd series converges, the work done in (a) shows theeven series converges also. (The sequence of the nthpar al sum of the even series is bounded andmonotonically increasing.) Likewise, (b) shows that if theeven series converges, the odd series will, too. Thus ifeither series converges, the other does.Similarly, (a) and (b) can be used to show that if eitherseries diverges, the other does, too.

(d) If both the even and odd series converge, then their sumwould be a convergent series. This would imply that theHarmonic Series, their sum, is convergent. It is not. Henceeach series diverges.

Sec on .

. con nuous, posi ve and decreasing

. The Integral Test (we do not have a con nuous defini on of n!yet) and the Limit Comparison Test (same as above, hence wecannot take its deriva ve).

. Converges

. Diverges

. Converges

. Converges

. Converges; compare to∞∑

n= n, as /(n + n− ) ≤ /n for

all n > .

. Diverges; compare to∞∑

n= n, as /n ≤ ln n/n for all n ≥ .

. Diverges; compare to∞∑

n= n. Since n =

√n >

√n − ,

/n ≤ /√n − for all n ≥ .

. Diverges; compare to∞∑

n= n:

n=

nn

<n + n+

n<

n + n+

n − ,

for all n ≥ .

. Diverges; compare to∞∑

n= n. Note that

nn − =

nn − ·

n>

n,

as nn −

> , for all n ≥ .

. Converges; compare to∞∑

n= n.

. Diverges; compare to∞∑

n=

ln nn

.

. Diverges; compare to∞∑

n= n.

. Diverges; compare to∞∑

n= n. Just as lim

n→

sin nn

= ,

limn→∞

sin( /n)/n

= .

. Converges; compare to∞∑

n= n /.

. Converges; Integral Test

. Diverges; the nth Term Test and Direct Comparison Test can beused.

. Converges; the Direct Comparison Test can be used with sequence/ n.

. Diverges; the nth Term Test can be used, along with the IntegralTest.

. (a) Converges; use Direct Comparison Test as ann < n.

A.

(b) Converges; since original series converges, we knowlimn→∞ an = . Thus for large n, anan+ < an.

(c) Converges; similar logic to part (b) so (an) < an.

(d) May converge; certainly nan > an but that does not meanit does not converge.

(e) Does not converge, using logic from (b) and nth Term Test.

Sec on .

. algebraic, or polynomial.

. Integral Test, Limit Comparison Test, and Root Test

. Converges

. Converges

. The Ra o Test is inconclusive; the p-Series Test states it diverges.

. Converges

. Converges; note the summa on can be rewri en as∞∑

n=

nn!nn!

, to

which the Ra o Test or Geometric Series Test can be applied.

. Converges

. Converges

. Diverges

. Diverges. The Root Test is inconclusive, but the nth-Term Testshows divergence. (The terms of the sequence approach e , not, as n → ∞.)

. Converges

. Diverges; Limit Comparison Test with /n.

. Converges; Ra o Test or Limit Comparison Test with / n.

. Diverges; nth-Term Test or Limit Comparison Test with .

. Diverges; Direct Comparison Test with /n

. Converges; Root Test

Sec on .

. The signs of the terms do not alternate; in the given series, someterms are nega ve and the others posi ve, but they do notnecessarily alternate.

. Many examples exist; one common example is an = (− )n/n.

. (a) converges

(b) converges (p-Series)

(c) absolute

. (a) diverges (limit of terms is not )

(b) diverges

(c) n/a; diverges

. (a) converges

(b) diverges (Limit Comparison Test with /n)

(c) condi onal

. (a) diverges (limit of terms is not )

(b) diverges

(c) n/a; diverges

. (a) diverges (terms oscillate between± )

(b) diverges

(c) n/a; diverges

. (a) converges

(b) converges (Geometric Series with r = / )(c) absolute

. (a) converges(b) converges (Ra o Test)(c) absolute

. (a) converges(b) diverges (p-Series Test with p = / )(c) condi onal

. S = − . ; S = − . ;

− . ≤∞∑

n=

(− )n

ln(n+ )≤ − .

. S = . ; S = . ;

. ≤∞∑

n=

(− )n

n!≤ .

. n =

. Using the theorem, we find n = guarantees the sum is within. of π/ . (Convergence is actually faster, as the sum is withinε of π/ when n ≥ .)

Sec on .

.

.

. + x+ x + x + x

. + x+ x + x + x

. (a) R = ∞(b) (−∞,∞)

. (a) R =

(b) ( , ]

. (a) R =

(b) (− , )

. (a) R = /

(b) ( / , / )

. (a) R =

(b) (− , )

. (a) R = ∞(b) (−∞,∞)

. (a) R =

(b) [− , ]

. (a) R =

(b) x =

. (a) f ′(x) =∞∑

n=n xn− ; (− , )

(b)∫

f(x) dx = C+

∞∑

n=

nn+

xn+ ; (− , )

. (a) f ′(x) =∞∑

n=

nn x

n− ; (− , )

(b)∫

f(x) dx = C+∞∑

n= (n+ ) n xn+ ; [− , )

. (a) f ′(x) =∞∑

n=

(− )nx n−

( n− )!=

∞∑

n=

(− )n+ x n+

( n+ )!;

(−∞,∞)

A.

(b)∫

f(x) dx = C+∞∑

n=

(− )nx n+

( n+ )!; (−∞,∞)

. + x+ x + x + x

. + x+ x + x + x

. + x+ x − x + x

Sec on .

. The Maclaurin polynomial is a special case of Taylor polynomials.Taylor polynomials are centered at a specific x-value; when thatx-value is , it is a Maclauring polynomial.

. p (x) = + x− x .

. p (x) = − x+ x − x

. p (x) = x+ x + x + x + x

. p (x) = x + x + x + x+

. p (x) = x − x + x − x+

. p (x) = + (− +x)− (− +x) + (− +x) − (− +x)

. p (x) = √ − − π +x√ − (− π +x)

√ +(− π +x)

√ +(− π +x)

√ −(− π +x)

√ − (− π +x)√

. p (x) = − x− + (x− ) − (x− ) + (x− ) − (x− )

. p (x) = + +x + ( + x)

. p (x) = x− x ; p ( . ) = . . Error is bounded by±

!· . ≈ ± . .

. p (x) = + (− + x)− (− + x) ; p ( ) = . .The third deriva ve of f(x) =

√x is bounded on ( , ) by . .

Error is bounded by± .!

· = ± . .

. The nth deriva ve of f(x) = ex is bounded by on intervalscontaining and . Thus |Rn( )| ≤

(n+ )!(n+ ). When n = ,

this is less than . .

. The nth deriva ve of f(x) = cos x is bounded by on intervalscontaining and π/ . Thus |Rn(π/ )| ≤

(n+ )!(π/ )(n+ ).

When n = , this is less than . . Since the Maclaurinpolynomial of cos x only uses even powers, we can actually justuse n = .

. The nth term is n! xn.

. The nth term is: when n even, ; when n is odd, (− )(n− )/

n! xn.

. The nth term is (− )nxn.

. + x+ x + x + x

. + x− x + x − x

Sec on .

. A Taylor polynomial is a polynomial, containing a finite number ofterms. A Taylor series is a series, the summa on of an infinitenumber of terms.

. All deriva ves of ex are ex which evaluate to at x = .The Taylor series starts + x+ x +

!x +

!x + · · · ;

the Taylor series is∞∑

n=

xn

n!

. The nth deriva ve of /( − x) is f (n)(x) = (n)!/( − x)n+ ,which evaluates to n! at x = .The Taylor series starts + x+ x + x + · · · ;

the Taylor series is∞∑

n=xn

. The Taylor series starts− (x− π/ ) + x + (x− π/ ) + x − (x− π/ ) ;

the Taylor series is∞∑

n=(− )n+

(x− π/ ) n+

( n+ )!

. f (n)(x) = (− )ne−x; at x = , f (n)( ) = − when n is odd andf (n)( ) = when n is even.The Taylor series starts − x+ x −

!x + · · · ;

the Taylor series is∞∑

n=(− )n

xn

n!.

. f (n)(x) = (− )n+ n!(x+ )n+

; at x = , f (n)( ) = (− )n+ n!n+

The Taylor series starts+ (x− )− (x− ) + (x− ) · · · ;

the Taylor series is∞∑

n=(− )n+

(x− )n

n+ .

. Given a value x, the magnitude of the error term Rn(x) is boundedby

∣

∣Rn(x)∣

∣ ≤max

∣

∣ f (n+ )(z)∣

∣

(n+ )!

∣

∣x(n+ )∣

∣,

where z is between and x.If x > , then z < x and f (n+ )(z) = ez < ex. If x < , thenx < z < and f (n+ )(z) = ez < . So given a fixed x value, letM = max{ex, }; f (n)(z) < M. This allows us to state

∣

∣Rn(x)∣

∣ ≤ M(n+ )!

∣

∣x(n+ )∣

∣.

For any x, limn→∞

M(n+ )!

∣

∣x(n+ )∣

∣ = . Thus by the Squeeze

Theorem, we conclude that limn→∞

Rn(x) = for all x, and hence

ex =∞∑

n=

xn

n!for all x.

. Given a value x, the magnitude of the error term Rn(x) is boundedby

∣

∣Rn(x)∣

∣ ≤max

∣

∣ f (n+ )(z)∣

∣

(n+ )!

∣

∣(x− )(n+ )∣

∣,

where z is between and x.Note that

∣

∣f (n+ )(x)∣

∣ = n!xn+ .

Per the statement of the problem, we only consider the case< x < .

If < x < , then < z < x and f (n+ )(z) = n!zn+ < n!. Thus

∣

∣Rn(x)∣

∣ ≤ n!(n+ )!

∣

∣(x− )(n+ )∣

∣ =(x− )n+

n+<

n+.

Thuslim

n→∞

∣

∣Rn(x)∣

∣ < limn→∞ n+

= ,

hence

ln x =∞∑

n=(− )n+

(x− )n

non ( , ).

. Given cos x =∞∑

n=(− )n

x n

( n)!,

cos(−x) =∞∑

n=(− )n

(−x) n

( n)!=

∞∑

n=(− )n

x n

( n)!= cos x, as all

powers in the series are even.

A.

. Given sin x =∞∑

n=(− )n

x n+

( n+ )!,

ddx(

sin x)

=ddx

(

∞∑

n=(− )n

x n+

( n+ )!

)

=

∞∑

n=(− )n

( n+ )x n

( n+ )!=

∞∑

n=(− )n

x n

( n)!= cos x. (The

summa on s ll starts at n = as there was no constant term inthe expansion of sin x).

. +x − x

+x − x

. +x − x

+x − x

.∞∑

n=(− )n

(x ) n

( n)!=

∞∑

n=(− )n

x n

( n)!.

.∞∑

n=(− )n

( x+ ) n+

( n+ )!.

. x+ x +x − x

.∫

√π

sin(

x)

dx ≈∫

√π (

x − x+

x − x)

dx =

.

A.

Index

!,Absolute Convergence Theorem,absolute maximum,absolute minimum,Absolute Value Theorem,accelera on, ,Alterna ng Harmonic Series, , ,Alterna ng Series Test,aN, ,analy c func on,angle of eleva on,an deriva ve,

of vector–valued func on,arc length, , , , ,arc length parameter, ,asymptote

horizontal,ver cal,

aT, ,average rate of change,average value of a func on,average value of func on,

Binomial Series,Bisec on Method,boundary point,bounded sequence,

convergence,bounded set,

center of mass, – , ,Chain Rule,

mul variable, ,nota on,

circle of curvature,circula on,closed,closed disk,concave down,concave up,concavity, ,

inflec on point,test for,

conic sec ons,degenerate,ellipse,hyperbola,parabola,

connected,simply,

conserva ve field, , ,

Constant Mul ple Ruleof deriva ves,of integra on,of series,

constrained op miza on,con nuous func on, ,

proper es, ,vector–valued,

contour lines,convergence

absolute, ,Alterna ng Series Test,condi onal,Direct Comparison Test,

for integra on,Integral Test,interval of,Limit Comparison Test,

for integra on,nth–term test,of geometric series,of improper int., , ,of monotonic sequences,of p-series,of power series,of sequence, ,of series,radius of,Ra o Comparison Test,Root Comparison Test,

coordinatescylindrical,polar,spherical,

cri cal number,cri cal point, , –cross product

and deriva ves,applica ons,

area of parallelogram,torque,volume of parallelepiped,

defini on,proper es, ,

curl,of conserva ve fields,

curvature,and mo on,equa ons for,of circle, ,radius of,

A.

curveparametrically defined,rectangular equa on,smooth,

curve sketching,cusp,cycloid,cylinder,cylindrical coordinates,

decreasing func on,finding intervals,

definite integral,and subs tu on,of vector–valued func on,proper es,

del operator,deriva ve

accelera on,as a func on,at a point,basic rules,Chain Rule, , , ,Constant Mul ple Rule,Constant Rule,differen al,direc onal, , , , ,exponen al func ons,First Deriv. Test,Generalized Power Rule,higher order,

interpreta on,hyperbolic funct.,implicit, ,interpreta on,inverse func on,inverse hyper.,inverse trig.,logarithmic differen a on,Mean Value Theorem,mixed par al,mo on,mul variable differen ability, ,normal line,nota on, ,parametric equa ons,par al, ,Power Rule, , ,power series,Product Rule,Quo ent Rule,second,Second Deriv. Test,Sum/Difference Rule,tangent line,third,trigonometric func ons,vector–valued func ons, , ,velocity,

differen able, , ,

differen al,nota on,

Direct Comparison Testfor integra on,for series,

direc onal deriva ve, , , , ,directrix, ,Disk Method,displacement, , ,distance

between lines,between point and line,between point and plane,between points in space,traveled,

divergence, ,Alterna ng Series Test,Direct Comparison Test,

for integra on,Integral Test,Limit Comparison Test,

for integra on,nth–term test,of geometric series,of improper int., , ,of p-series,of sequence,of series,Ra o Comparison Test,Root Comparison Test,

Divergence Theoremin space,in the plane,

dot productand deriva ves,defini on,proper es, ,

double integral, ,in polar,proper es,

eccentricity, ,elementary func on,ellipse

defini on,eccentricity,parametric equa ons,reflec ve property,standard equa on,

extremaabsolute, ,and First Deriv. Test,and Second Deriv. Test,finding,rela ve, , ,

Extreme Value Theorem, ,extreme values,

factorial,First Deriva ve Test,

first octant,floor func on,flow, ,fluid pressure/force, ,flux, , , ,focus, , ,Fubini’s Theorem,func on

of three variables,of two variables,vector–valued,

Fundamental Theorem of Calculus, ,and Chain Rule,

Fundamental Theorem of Line Integrals, ,

Gabriel’s Horn,Gauss’s Law,Generalized Power Rule,geometric series, ,gradient, , , , ,

and level curves,and level surfaces,

Green’s Theorem,

Harmonic Series,Head To Tail Rule,Hooke’s Law,hyperbola

defini on,eccentricity,parametric equa ons,reflec ve property,standard equa on,

hyperbolic func ondefini on,deriva ves,iden es,integrals,inverse,

deriva ve,integra on,logarithmic def.,

implicit differen a on, ,improper integra on, ,incompressible vector field,increasing func on,

finding intervals,indefinite integral,

of vector–valued func on,indeterminate form, , , ,inflec on point,ini al point,ini al value problem,Integral Test,integra on

arc length,area, , ,area between curves, ,average value,by parts,

by subs tu on,definite,

and subs tu on,proper es,Riemann Sums,

displacement,distance traveled,double,fluid force, ,Fun. Thm. of Calc., ,general applica on technique,hyperbolic funct.,improper, , , ,indefinite,inverse hyper.,iterated,Mean Value Theorem,mul ple,nota on, , , ,numerical,

Le /Right Hand Rule, ,Simpson’s Rule, , ,Trapezoidal Rule, , ,

of mul variable func ons,of power series,of trig. func ons,of trig. powers, ,of vector–valued func on,of vector–valued func ons,par al frac on decomp.,Power Rule,Sum/Difference Rule,surface area, , ,trig. subst.,triple, , –volume

cross-sec onal area,Disk Method,Shell Method, ,Washer Method, ,

with cylindrical coordinates,with spherical coordinates,work,

interior point,Intermediate Value Theorem,interval of convergence,iterated integra on, , , , , –

changing order,proper es, ,

L’Hôpital’s Rule, ,lamina,Le Hand Rule, , ,Le /Right Hand Rule,level curves, ,level surface, ,limit

Absolute Value Theorem,at infinity,defini on,

difference quo ent,does not exist, ,indeterminate form, , , ,L’Hôpital’s Rule, ,le handed,of infinity,of mul variable func on, , ,of sequence,of vector–valued func ons,one sided,proper es, ,pseudo-defini on,right handed,Squeeze Theorem,

Limit Comparison Testfor integra on,for series,

line integralFundamental Theorem, ,over scalar field, , ,over vector field,path independent, ,proper es over a scalar field,proper es over a vector field,

lines,distances between,equa ons for,intersec ng,parallel,skew,

logarithmic differen a on,

Möbius band,Maclaurin Polynomial, see Taylor Polynomial

defini on,Maclaurin Series, see Taylor Series

defini on,magnitude of vector,mass, , , ,

center of, ,maximum

absolute, ,and First Deriv. Test,and Second Deriv. Test,rela ve/local, , ,

Mean Value Theoremof differen a on,of integra on,

Midpoint Rule, ,minimum

absolute, ,and First Deriv. Test, ,rela ve/local, , ,

moment, , ,monotonic sequence,mul ple integra on, see iterated integra onmul variable func on, ,

con nuity, – , ,differen ability, , , ,domain, ,

level curves,level surface,limit, , ,range, ,

Newton’s Method,norm,normal line, , ,normal vector,nth–term test,numerical integra on,

Le /Right Hand Rule, ,Simpson’s Rule, ,

error bounds,Trapezoidal Rule, ,

error bounds,

octantfirst,

one to one,open,open ball,open disk,op miza on,

constrained,orientable,orthogonal, ,

decomposi on,orthogonal decomposi on of vectors,orthogonal projec on,oscula ng circle,outer unit normal vector,

p-series,parabola

defini on,general equa on,reflec ve property,

parallel vectors,Parallelogram Law,parametric equa ons

arc length,concavity,defini on,finding d y

dx ,finding dy

dx ,normal line,of a surface,surface area,tangent line,

parametrized surface,par al deriva ve, ,

high order,meaning,mixed,second deriva ve,total differen al, ,

par on,size of,

path independent, ,

perpendicular, see orthogonalpiecewise smooth curve,planes

coordinate plane,distance between point and plane,equa ons of,introduc on,normal vector,tangent,

point of inflec on,polar

coordinates,func on

arc length,gallery of graphs,surface area,

func ons,area,area between curves,finding dy

dx ,graphing,

polar coordinates,plo ng points,

poten al func on, ,Power Rule

differen a on, , , ,integra on,

power series,algebra of,convergence,deriva ves and integrals,

projec le mo on, , ,

quadric surfacedefini on,ellipsoid,ellip c cone,ellip c paraboloid,gallery, –hyperbolic paraboloid,hyperboloid of one sheet,hyperboloid of two sheets,sphere,trace,

Quo ent Rule,

R,radius of convergence,radius of curvature,Ra o Comparison Test

for series,rearrangements of series, ,related rates,Riemann Sum, , ,

and definite integral,Right Hand Rule, , ,right hand rule

of Cartesian coordinates,of the cross product,

Rolle’s Theorem,

Root Comparison Testfor series,

saddle point, ,Second Deriva ve Test, ,sensi vity analysis,sequence

Absolute Value Theorem,posi ve,

sequencesboundedness,convergent, , ,defini on,divergent,limit,limit proper es,monotonic,

seriesabsolute convergence,Absolute Convergence Theorem,alterna ng,

Approxima on Theorem,Alterna ng Series Test,Binomial,condi onal convergence,convergent,defini on,Direct Comparison Test,divergent,geometric, ,Integral Test,interval of convergence,Limit Comparison Test,Maclaurin,nth–term test,p-series,par al sums,power, ,

deriva ves and integrals,proper es,radius of convergence,Ra o Comparison Test,rearrangements, ,Root Comparison Test,Taylor,telescoping, ,

Shell Method, ,signed area,signed volume, ,simple curve,simply connected,Simpson’s Rule, ,

error bounds,smooth,

curve,surface,

smooth curvepiecewise,

speed,sphere,

spherical coordinates,Squeeze Theorem,Stokes’ Theorem,Sum/Difference Rule

of deriva ves,of integra on,of series,

summa onnota on,proper es,

surface,smooth,

surface area,of parametrized surface, ,solid of revolu on, , ,

surface integral,surface of revolu on, ,

tangent line, , , ,direc onal,

tangent plane,Taylor Polynomial

defini on,Taylor’s Theorem,

Taylor Seriescommon series,defini on,equality with genera ng func on,

Taylor’s Theorem,telescoping series, ,terminal point,torque,total differen al, ,

sensi vity analysis,total signed area,trace,Trapezoidal Rule, ,

error bounds,triple integral, , –

proper es,

unbounded sequence,unbounded set,unit normal vector

aN,and accelera on, ,and curvature,defini on,in R ,

unit tangent vectorand accelera on, ,and curvature, ,aT,defini on,in R ,

unit vector,proper es,standard unit vector,unit normal vector,unit tangent vector,

vector field,conserva ve, ,curl of,divergence of, ,over vector field,poten al func on of, ,

vector–valued func onalgebra of,arc length,average rate of change,con nuity,defini on,deriva ves, , ,describing mo on,displacement,distance traveled,graphing,integra on,limits,of constant length, , , ,projec le mo on, ,smooth,tangent line,

vectors,algebra of,algebraic proper es,component form,cross product, , ,defini on,dot product, –Head To Tail Rule,magnitude,norm,normal vector,orthogonal,orthogonal decomposi on,orthogonal projec on,parallel,Parallelogram Law,resultant,standard unit vector,unit vector, ,zero vector,

velocity, ,volume, , ,

Washer Method, ,work, ,

Differen a on Rules

.ddx

(cx) = c

.ddx

(u± v) = u′ ± v′

.ddx

(u · v) = uv′ + u′v

.ddx

(

uv

)

=vu′ − uv′

v

.ddx

(u(v)) = u′(v)v′

.ddx

(c) =

.ddx

(x) =

.ddx

(xn) = nxn−

.ddx

(ex) = ex

.ddx

(ax) = ln a · ax

.ddx

(ln x) =x

.ddx

(loga x) = ln a·x

.ddx

(sin x) = cos x

.ddx

(cos x) = − sin x

.ddx

(csc x) = − csc x cot x

.ddx

(sec x) = sec x tan x

.ddx

(tan x) = sec x

.ddx

(cot x) = − csc x

.ddx(

sin− x)

= √− x

.ddx(

cos− x)

=−√− x

.ddx(

csc− x)

=−

|x|√x −

.ddx(

sec− x)

=|x|

√x −

.ddx(

tan− x)

=+ x

.ddx(

cot− x)

=−+ x

.ddx

(cosh x) = sinh x

.ddx

(sinh x) = cosh x

.ddx

(tanh x) = sech x

.ddx

(sech x) = − sech x tanh x

.ddx

(csch x) = − csch x coth x

.ddx

(coth x) = − csch x

.ddx(

cosh− x)

= √x −

.ddx(

sinh− x)

= √x +

.ddx(

sech− x)

=−

x√

− x

.ddx(

csch− x)

=−

|x|√

+ x

.ddx(

tanh− x)

= − x

.ddx(

coth− x)

= − x

Integra on Rules

.∫

c · f(x) dx = c∫

f(x) dx

.∫

f(x)± g(x) dx =

∫

f(x) dx±∫

g(x) dx

.∫

dx = C

.∫

dx = x+ C

.∫

xn dx =n+

xn+ + C, n ̸= −

.∫

ex dx = ex + C

.∫

ln x dx = x ln x− x+ C

.∫

ax dx =ln a

· ax + C

.∫

xdx = ln |x|+ C

.∫

cos x dx = sin x+ C

.∫

sin x dx =− cos x+ C

.∫

tan x dx =− ln | cos x|+ C

.∫

sec x dx = ln | sec x+ tan x|+ C

.∫

csc x dx =− ln | csc x+ cot x|+ C

.∫

cot x dx = ln | sin x|+ C

.∫

sec x dx = tan x+ C

.∫

csc x dx =− cot x+ C

.∫

sec x tan x dx = sec x+ C

.∫

csc x cot x dx =− csc x+ C

.∫

cos x dx = x+ sin(

x)

+ C

.∫

sin x dx = x− sin(

x)

+ C

.∫

x + adx =

atan−

(

xa

)

+ C

.∫

√a − x

dx = sin−(

xa

)

+ C

.∫

x√x − a

dx =asec−

( |x|a

)

+ C

.∫

cosh x dx = sinh x+ C

.∫

sinh x dx = cosh x+ C

.∫

tanh x dx = ln(cosh x) + C

.∫

coth x dx = ln | sinh x|+ C

.∫

√x − a

dx = ln∣

∣x+√

x − a∣

∣+ C

.∫

√x + a

dx = ln∣

∣x+√

x + a∣

∣+ C

.∫

a − xdx =

aln∣

∣

∣

∣

a+ xa− x

∣

∣

∣

∣

+ C

.∫

x√a − x

dx =aln(

xa+

√a − x

)

+ C

.∫

x√x + a

dx =aln∣

∣

∣

∣

xa+

√x + a

∣

∣

∣

∣

+ C

The Unit Circle

x

y

◦ ( , )

◦

π/

(√

,)

◦

π/

(√

,√ )

◦

π/

(

,√ )

◦

π/

( , )

◦

π/

(

− ,√ )

◦

π/

(

−√

,√ )

◦

π/

(

−√

,)

◦π(− , )

◦

π/(

−√

,−)

◦

π/

(

−√

,−√ )

◦

π/

(

− ,−√ )

◦

π/

( ,− )

◦

π/

(

,−√ )

◦

π/

(√

,−√ )

◦

π/(√

,−)

Defini ons of the Trigonometric Func ons

Unit Circle Defini on

x

y

(x, y)

y

x

θ

sin θ = y cos θ = x

csc θ =y

sec θ =x

tan θ =yx

cot θ =xy

Right Triangle Defini on

Adjacent

OppositeHypot

enuse

θ

sin θ =OH

csc θ =HO

cos θ =AH

sec θ =HA

tan θ =OA

cot θ =AO

Common Trigonometric Iden es

Pythagorean Iden essin x+ cos x =

tan x+ = sec x

+ cot x = csc x

Cofunc on Iden essin(π − x

)

= cos x

cos(π − x

)

= sin x

tan(π − x

)

= cot x

csc(π − x

)

= sec x

sec(π − x

)

= csc x

cot(π − x

)

= tan x

Double Angle Formulassin x = sin x cos x

cos x = cos x− sin x

= cos x−= − sin x

tan x =tan x

− tan x

Sum to Product Formulas

sin x+ sin y = sin(x+ y

)

cos(x− y

)

sin x− sin y = sin(x− y

)

cos(x+ y

)

cos x+ cos y = cos(x+ y

)

cos(x− y

)

cos x− cos y = − sin(x+ y

)

sin(x− y

)

Power–Reducing Formulas

sin x =− cos x

cos x =+ cos x

tan x =− cos x+ cos x

Even/Odd Iden essin(−x) = − sin x

cos(−x) = cos x

tan(−x) = − tan x

csc(−x) = − csc x

sec(−x) = sec x

cot(−x) = − cot x

Product to Sum Formulas

sin x sin y =(cos(x− y)− cos(x+ y)

)

cos x cos y =(cos(x− y) + cos(x+ y)

)

sin x cos y =(sin(x+ y) + sin(x− y)

)

Angle Sum/Difference Formulassin(x± y) = sin x cos y± cos x sin y

cos(x± y) = cos x cos y∓ sin x sin y

tan(x± y) =tan x± tan y∓ tan x tan y

Areas and Volumes

Trianglesh = a sin θ

Area = bh

Law of Cosines:c = a + b − ab cos θ

bθ

ach

Right Circular ConeVolume = πr h

Surface Area =πr√r + h + πr

h

r

ParallelogramsArea = bh

b

h

Right Circular CylinderVolume = πr h

Surface Area =πrh+ πr

h

r

TrapezoidsArea = (a+ b)h

b

a

h

SphereVolume = πr

Surface Area = πrr

CirclesArea = πr

Circumference = πrr

General ConeArea of Base = A

Volume = Ahh

A

Sectors of Circlesθ in radians

Area = θr

s = rθ r

s

θ

General Right CylinderArea of Base = A

Volume = Ahh

A

Algebra

Factors and Zeros of PolynomialsLet p(x) = anxn + an− xn− + · · ·+ a x+ a be a polynomial. If p(a) = , then a is a zero of the polynomial and a solu on ofthe equa on p(x) = . Furthermore, (x− a) is a factor of the polynomial.

Fundamental Theorem of AlgebraAn nth degree polynomial has n (not necessarily dis nct) zeros. Although all of these zeros may be imaginary, a realpolynomial of odd degree must have at least one real zero.

Quadra c FormulaIf p(x) = ax + bx+ c, and ≤ b − ac, then the real zeros of p are x = (−b±

√b − ac)/ a

Special Factorsx − a = (x− a)(x+ a) x − a = (x− a)(x + ax+ a )x + a = (x+ a)(x − ax+ a ) x − a = (x − a )(x + a )

(x+ y)n = xn + nxn− y+ n(n− )! xn− y + · · ·+ nxyn− + yn

(x− y)n = xn − nxn− y+ n(n− )! xn− y − · · · ± nxyn− ∓ yn

Binomial Theorem(x+ y) = x + xy+ y (x− y) = x − xy+ y(x+ y) = x + x y+ xy + y (x− y) = x − x y+ xy − y(x+ y) = x + x y+ x y + xy + y (x− y) = x − x y+ x y − xy + y

Ra onal Zero TheoremIf p(x) = anxn + an− xn− + · · ·+ a x+ a has integer coefficients, then every rational zero of p is of the form x = r/s,where r is a factor of a and s is a factor of an.

Factoring by Groupingacx + adx + bcx+ bd = ax (cs+ d) + b(cx+ d) = (ax + b)(cx+ d)

Arithme c Opera onsab+ ac = a(b+ c)

ab+

cd=

ad+ bcbd

a+ bc

=ac+

bc

(ab

)

( cd

) =(ab

)(dc

)

=adbc

(ab

)

c=

abc

a(bc

) =acb

a(bc

)

=abc

a− bc− d

=b− ad− c

ab+ aca

= b+ c

Exponents and Radicals

a = , a ̸= (ab)x = axbx axay = ax+y √a = a / ax

ay= ax−y n

√a = a /n

(ab

)x=

ax

bxn√am = am/n a−x =

axn√ab = n

√a n√b (ax)y = axy n

√ab=

n√a

n√b

Addi onal Formulas

Summa on Formulas:n∑

i=

c = cnn∑

i=

i =n(n+ )

n∑

i=

i =n(n+ )( n+ )

n∑

i=

i =

(n(n+ )

)

Trapezoidal Rule:∫ b

af(x) dx ≈ ∆x[

f(x ) + f(x ) + f(x ) + ...+ f(xn) + f(xn+ )]

with Error ≤ (b− a)n

[max

∣∣f ′′(x)

∣∣]

Simpson’s Rule:∫ b

af(x) dx ≈ ∆x[

f(x ) + f(x ) + f(x ) + f(x ) + ...+ f(xn− ) + f(xn) + f(xn+ )]

with Error ≤ (b− a)n[max

∣∣f ( )(x)

∣∣]

Arc Length:

L =∫ b

a

√

+ f ′(x) dx

Surface of Revolu on:

S = π

∫ b

af(x)√

+ f ′(x) dx

(where f(x) ≥ )

S = π

∫ b

ax√

+ f ′(x) dx

(where a, b ≥ )

Work Done by a Variable Force:

W =

∫ b

aF(x) dx

Force Exerted by a Fluid:

F =∫ b

awd(y) ℓ(y) dy

Taylor Series Expansion for f(x):

pn(x) = f(c) + f ′(c)(x− c) +f ′′(c)

!(x− c) +

f ′′′(c)!

(x− c) + ...+f (n)(c)n!

(x− c)n

Maclaurin Series Expansion for f(x), where c = :

pn(x) = f( ) + f ′( )x+f ′′( )

!x +

f ′′′( )

!x + ...+

f (n)( )

n!xn

Summary of Tests for Series:

Test Series Condi on(s) ofConvergence

Condi on(s) ofDivergence Comment

nth-Term∞∑

n=

an limn→∞

an ̸= This test cannot be used toshow convergence.

Geometric Series∞∑

n=

rn |r| < |r| ≥ Sum = − r

Telescoping Series∞∑

n=

(bn − bn+a) limn→∞

bn = L Sum =

(a∑

n=

bn

)

− L

p-Series∞∑

n=(an+ b)p

p > p ≤

Integral Test∞∑

n=

an

∫ ∞a(n) dn

is convergent

∫ ∞a(n) dn

is divergent

an = a(n) must becon nuous

Direct Comparison∞∑

n=

an

∞∑

n=

bn

converges and≤ an ≤ bn

∞∑

n=

bn

diverges and≤ bn ≤ an

Limit Comparison∞∑

n=

an

∞∑

n=

bn

converges andlim

n→∞an/bn ≥

∞∑

n=

bn

diverges andlim

n→∞an/bn >

Also diverges iflim

n→∞an/bn = ∞

Ra o Test∞∑

n=

an limn→∞

an+an

< limn→∞

an+an

>

{an}must be posi veAlso diverges iflim

n→∞an+ /an = ∞

Root Test∞∑

n=

an limn→∞

(an) /n

< limn→∞

(an) /n

>

{an}must be posi veAlso diverges if

limn→∞

(an) /n

= ∞