Chapter 5 – Integrals 5.1 Areas and Distances We start with a problem – how can we calculate the area “under” a given function – ie, the area between the function and the x-axis? If the curve happens to be something easy – like a horizontal line - this isn’t too hard, but if it’s a curve, the problem becomes more difficult.

Let’s start by drawing a rectangle that’s f(a) tall by (b – a) wide as an estimate:

Area = f(a) (b – a) How can we get a better estimate? Let’s try 2 rectangles:

Area = f(a) ∆x1 + f(c) ∆x2 How could we get an even better estimate?

Ft.÷HEI÷nk

Ex: Estimate the area under the curve f(x) = x2 between x = 0 and x = 1 by constructing 8 rectangles (the book calls them “strips”) of equal width. This seems like a relatively simple situation, until you realize that you get different answers depending on whether you draw the rectangles from the left or from the right:

Using left endpoints, we get: f(0) !

" + f !"

!" + f #

"!" + f $

"!" + f %

"!" + f &

"!" + f '

"!" + f (

"!"

We can see this will be a lower bound, because all the rectangles fit completely under the curve. Using right endpoints, we get: f !"

!" + f #

"!" + f $

"!" + f %

"!" + f &

"!" + f '

"!" + f (

"!" + f 1 !

" We can see this will be an upper bound, because all the rectangles extend to above the curve. By computing both, we get: 0.2734375 < A < 0.3984375 Hmmm…if only we had a technique for looking at a progression of smaller and smaller rectangles….

ft:-

1

Definition: The Area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles (using right endpoints): A = lim

-→/0- = lim-→/ 3 4! ∆4 + 3 4# ∆4 + ⋯+ 3(4-)∆4

1) It can be proven that this limit always exists. 2) It can also be shown that you get the same limit value A from using left endpoints. 3) In fact, you can choose any x-value in each interval to calculate the height. The values that are chosen - x1

*, x2*, …, xn

* - are called the sample points. Using sigma notation we get: A = lim

-→/3(4:)∆4-

:;! The Distance Problem My odometer is broken, but I’d still like to calculate how far I’ve driven using my speedometer and a stopwatch (held by my passenger, for safety.) I know D = R*T and I have the following data:

Time (s) 0 5 10 15

Speed (ft/s) 0 20 25 10 How can I estimate the distance I’ve travelled?

n = #

rectangle

him I,fcxisn

fcx ,) dx + F Cxddxt . .

tfcxn ) IX

Using Left Endpoints: 0 * 5 + 20 * 5 + 25 * 5 = 225 ft Using Right Endpoints: 20 * 5 + 25 * 5 + 10 * 5 = 275 ft 1) How can we improve our estimate? 2) Do we know whether the true distance lies between the two estimates?

tsp^# seconds

More frequent speed checks

Ex odometer

No

5.2 The Definite Integral The limit of the sum from 5.1: lim-→/

' (! ∆( + ' (# ∆( + ⋯+ '((-)∆( = lim-→/

'((:)∆(-:;!

where ∆( = <=>- , can be written as:

' ( 7(<

>

and is called the definite integral of f from a to b, as long as the limit exits and is the same for all choices of xi. If the limit exists, then f is said to be integrable over [a,b]

Properties of the Integral 1) 87(<

A = c(b – a) where c is any constant 2) ' ( + : ( 7( =<

> ' ( 7(<> + : ( 7(<

> 3) 8' ( 7(<

> = c ' ( 7(<> where c is any constant

4) [' ( − : ( ]7(<

> = ' ( 7(<> - : ( 7(<

> 5) ' ( 7(F

> + ' ( 7(<F = ' ( 7(<

> 6) If f(x) > 0 for a < x < b, then ' ( 7(<

> > 0 7) If f(x) > g(x) for a < x < b, then ' ( 7(<

> > : ( 7(<>

8) If m < f(x) < M for a < x < b, then m (b – a) < ' ( 7(<

> ≤ @(A − B)

rwsum

Boata

Ex: (#33) The graph of f is shown. Evaluate each integral by interpreting it in terms of areas.

a) ' ( 7(#

K d) ' ( 7(L

K

i.ii AN'#(

+0

-

=i+i+i+÷+t5h ,¥'

,

= 4 T

10 + C- 8) = 2

5.3 The Fundamental Theorem of Calculus (Cue exciting music)

The Fundamental Theorem of Calculus, Part I If f is continuous on [a,b], then the function g defined by g(x) = ' E 7EN

> a < x < b is continuous on [a,b] and differentiable on (a,b), and g’(x) = f(x).

Ex: Find OON sec E7ENS

! (Hint: You’ll need the Chain Rule)

The Fundamental Theorem of Calculus, Part 2 If f is continuous on [a, b], then ' ( 7(<

> = F(b) – F(a) where F is any antiderivative of f. That is, a function s. t. F’ = f.

Why can it be *any* antiderivative of f?

= seek4) . 4×3

The C 'scancel out.

So, differentiation and integration are inverse processes. Ex: Evaluate these integrals 1) 1 + ( + 3(# 7(!

K 2) 4 − E E7E%

K Note: Be careful when rushing in to evaluate integrals. Look at this example:

1(# 7(

$

=!

It seems like it would be easy to find an antiderivative and evaluate it, but first – is this function continuous over [-1, 3]?

= x + I +3¥ ] to

F ( i ) - FO :

= f + ±¥¥j - [ o + It sgj= [ i + E + I] - [D= 2.5

= 64 (4 re . tr ) dt

±I±÷IItI¥± IIIIII:*:*'s¥;**¥= 6¥ - 6¥ = ¥8

⇒ n*±÷¥n⇒K¥¥¥e¥s±÷

No,

not at 0.

5.4 Indefinite Integrals Definite vs. Indefinite Integrals: 2"$"$

J is a definite integral. It’s a string of symbols that represent a number: 2"$"$

J = x2 ]03 = 32 – 02 = 9 2"$"is an indefinite integral. It’s a string of symbols that represent a

function, or family of functions: 2"$"= x2 + C

Table of Indefinite Integrals ' ( " $" =a ((") $"

∫ [f(x) + g(x)] dx = ∫ ( " $" +∫ / " $" ∫ 0$" = kx + C ∫ xndx = M

XYZ

-[! + C (n ≠ -1) ∫ ex dx = ex + C ∫ !

M dx = ln |x| + C ∫ cos x dx = sinx + C ∫ sin x dx = - cos x + C ∫ sec2 x dx = tan x + C ∫ sec x tan x dx = sec x + C

Inktotxe

Net Change Theorem The integral of a rate of change is the net change

8] " $" = 8 : − 8(')<

>

¥Ikarge

5.5 The Substitution Rule Ex: a) Find NNM (%

# + 3%)$ b) Find %# + 3% # 6% + 9 -%

The Substitution Rule If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then . / % /] % -% = . 1 -1

Ex: Evaluate 2% + 1 -% using the Substitution Rule

=3 ( x '+3x} (2×+3)= (x2+3x5( 6×+9 )

*u=×z+s× [s¥tCdu=Rx+})a×

[( ×43×)'+(( 6×+9 )d×=3( 2×+3 )dx

= 3dUSu23du=S}upo=u3+ )

Let

u=2xi⇒Esa .

,⇒s¥¥¥i¥lkxtdx

= Y#z÷

*# . Etc

= ¥= . ± + C

= ÷u÷ . } + C

= stutz + C

= st (2×+1) ÷+ C

Differentiate to check

st . } ( zxtif ! £ (2×+1)

= ÷ . E (2×+1) ±. 2 = Txt

-

Ex: Evaluate 1 + ##!

J #&)# using the Substitution Rule Integrals of Symmetric Functions – If f is even [ f(-x) = f(x) ], then * # )#>

=> = 2 * # )#>J

If f is odd [ f(-x) = -f(x) ], then * # )#>

=> = 0

.

.