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Math 175 Calculus II

Workbook

Math 175 - Calculus II: Concepts and ApplicationsCourse Workbook

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Course Information

Course Webpage: https://hausdorff.boisestate.edu/calculus/

Instructor Name:

Section:

Class Time:

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Exam Information

Exam 1:

Date/Time: Location:

Exam 2:


Final Exam:


Math 175 Calculus II WorkbookJaimos Skriletz, Doug Bullock

We extend our thanks and due credit to the colleagues below, for their contributions to the course materialsand this workbook.

Donna Calhoun, Kathleen Coskey, Brett Crow, Kristen Garcia, Rob Harryman, Tara Sheehan,Shari Ultman, Elena Velasquez

Copyright c©2019 Boise State Mathematics Department

This material is based on work supported by the National Science Foundation under Grant No. #DUE-1347830.

Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s)

and do not necessarily reflect the views of the National Science Foundation.

Math 175 Calculus II: Workbook

This workbook was created for the common Calculus II materials used at Boise State University. Inlearning calculus, students need to practice both how to solve calculus problems and how to communicatesolutions to them. This workbook provides space for students to practice writing problem solutions thatclearly communicate the logic that goes from a problem’s statement to its solution using correct notation.

This workbook is organized into daily lessons. Each daily lesson contains a set of notes and learning goalsalong with a set of worksheets. The notes and learning goals list the main topics and goals of each lessonalong with providing useful formulas and examples. The worksheets provide problem descriptions withvarying levels of guidance for writing down their solutions.

Lessons

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Methods of Integration 5

Calculus I Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Substitution I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Substitution II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

Partial Fractions I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Partial Fractions II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Integration Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Limits and Improper Integrals 57

Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Improper Integrals I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Improper Integrals II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Sequences and Series 75

Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Sigma Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Exam 1 98

Exam 1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

Lessons

Taylor Series 103Taylor Polynomials I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103Taylor Polynomials II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113Binomial Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122Numerics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129Convergence II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

Integration Applications 139Area I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139Area II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156New Slices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1641D Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1702D Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179Volumes I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192Volumes II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202Volumes III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209Centroids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

Exam 2 225Exam 2 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

Vectors and Parametrics 233Vectors I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233Vectors II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243Vector Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252Vector Valued Functions I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261Vector Valued Functions II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269Vector Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278Vector Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296

Final Exam 302Vectors Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302

Appendicies 305Algebra Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305Trigonometric Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306Calculus Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307

Introduction 3

The common materials for Calculus II were created by a group of instructors at Boise State University forour calculus students. These materials are built upon the following three principles:

• Active Learning: Students should spend time discussing calculus concepts, solving problems, andcommunicating their problem strategies and solutions.

• Regular Feedback: Students should receive regular feedback on the correctness of their answers,written solutions, and graphs.

• Collaboration: Students should collaborate with their instructor, learning assistant, and othercalculus students when learning and practicing calculus.

The materials are organized into daily lessons. Each daily lesson comes with a set of notes and learninggoals, one (or more) WebAssign assignment(s), and a set of worksheets. Each lesson’s problem sets arebuilt with varying levels of guidance along with increasing levels of difficulty. The notes and learning goalscan be used both as reference and a guide.

WebAssign

WebAssign is an online platform in which students can submit answers and have them graded. Daily prob-lem sets are written in WebAssign to provide students with regular feedback on their answers. WebAssignproblems come in three basic types

• Multiple choice questions in which correct answer(s) are chosen from a list of possibilities.

• Numerical answers which check if a decimal value is correct within a set tolerance (either exact orrounded to a specified number of decimal digits). Numerical answers may also require correct units.

• Symbolic answers which check if a symbolic formula with variables is correct.

Each daily lesson may come with multiple problem sets. Every lesson comes with a ‘Basic’ problem set,which covers the basic skill set and problems in the lesson. WebAssign ‘Basic’ assignments are generallydue at the start of the next class period.

In addition to the basic problem set, many lessons have an additional assignment that is due another classperiod after its ‘Basic’ assignment. Although the assignment is due on a later date, it is considered partof the daily lesson it is paired with. These additional assignments come in one of two types

• A ‘More’ assignment is an additional WebAssign assignment that contains more practice of the basicskill set. These problems will often be slightly more difficult and contain less guidance.

• An ‘Advanced’ assignment is an additional WebAssign assignment that contains harder or advancedproblems. Advanced problems will use a variety of skills to solve.

WebAssign is good at checking for correctness of answers. WebAssign does not provide feedback ona student’s written solution, graphs, or why an answer is incorrect. In addition to getting graded onWebAssign, students should be writing solutions and graphing on paper. These worksheets provide spacefor the students to produce work and graphs related to a given problem.

Introduction 4

Worksheets

When solving problems, students should practice writing solutions that clearly communicate the logicstarting at the problem’s statement and arriving at its answer using correct mathematical notation. Inaddition to finding answers to problems, students should also reflect on what they have learned after theyhave completed the problem.

Most ‘Basic’ WebAssign problems have a worksheet in this workbook. Each worksheet states the numberof the WebAssign problem it is associated with. The expectation is that students complete the worksheetand discuss both their answer and their written solution with their group, learning assistant, or instructor.After the worksheet is complete, enter the answers into WebAssign to be graded.

• If an answer is correct, students should reflect on what they just did including how to both computeand communicate a problem’s solution. In addition, students should discuss any ideas they may havelearned from that problem. Some worksheets will contain ‘Discussion Points’ to provide studentswith prompts to think about and discuss.

• If an answer is incorrect, students should go back over their worksheet and see if they can spot theirerror. In addition, students should collaborate with other students, the learning assistant, or theirinstructor to help learn the correct process and reasoning to find an answer.

Last, make sure you get feedback on your written solutions on the worksheet to make sure you are usingboth correct logic and notation. You should be getting feedback on at least one worksheet each lessonfrom an instructor or learning assistant.

Best Practices

The following are some best practices for using this workbook along with WebAssign:

• Use the worksheets: Carefully follow the instructions so that you can practice writing solutionsthat are organized, correct, and clearly presented.

• Complete worksheets before entering results into WebAssign: For WebAssign problemswith a worksheet, use the worksheets as a guide to the thought process and the care needed to obtaincorrect results. After the worksheet is complete, enter answers in WebAssign to get feedback to verifyif they are correct.

• Analyze and discuss worksheets with classmates: Ensure that you know why particular stepsare involved in solving any particular problem. Ensure that you know the meaning of the mathe-matical information and notation that you write.

• Get worksheet feedback: Get feedback from your instructor or learning assistant. There is nosubstitute for such direct information about whether you are meeting course expectations.

Calculus I Review Notes and Learning Goals 5

To start with, here is a quick summary of derivatives, antiderivatives, and integrals from Calculus I:

The Derivative

• The derivative of a function f(x) is the functionf ′(x) that gives the slope of the tangent line (orinstantaneous rate of change) at the input value x.

• The derivative is defined as the limit of the slope ofthe secant line (or average rate of change) betweentwo input values x and x+ h as the two points getclose together (h→ 0).

f ′(x) = limh→0

f(x+ h)− f(x)

h

• In practice the derivative can be computed from the derivative rules:

– Basic derivatives:

d

dx

(xn)

= nxn−1

d

dx

(lnx)

=1

xd

dx

(ex)

= ex

d

dx

(sinx

)= cosx

d

dx

(cosx

)= − sinx

– Basic derivatives with linear insides1:

d

dx

((ax+ b)n

)= a · n(ax+ b)n−1

d

dx

(ln(ax+ b)

)= a · 1

(ax+ b)d

dx

(eax+b

)= a · eax+b

d

dx

(sin(ax+ b)

)= a · cos(ax+ b)

d

dx

(cos(ax+ b)

)= −a · sin(ax+ b)

– Linearity rule:d

dx

(af(x) + bg(x)

)= af ′(x) + bg′(x)

– Product and quotient rules:

d

dx

(f(x)g(x)

)= f ′(x)g(x) + f(x)g′(x)

d

dx

(f(x)

g(x)

)=g(x)f ′(x)− f(x)g′(x)(

g(x))2

– Chain rule:

d

dx

(f(g(x)

))= f ′

(g(x)

)g′(x)

• Students are expected to know the derivative rules listed above and to be able to compute derivativesefficiently without the aid of a calculator.

1 This is a special case of the chain rule when the inside is linear (of the form ax+ b).


Derivative Notation

• The derivative operator,d

dx, is the operation of taking the derivative with respect to the variable x.

The derivative operator must operate on (include) the function it is taking the derivative of:

d

dx

(x2e−x

)= 2xe−x − x2e−x

• If y = f(x) is a function, Leibnitz notation is short hand for using the derivative operator on f(x):

dy

dx=df

dx=

d

dx

(f(x)

)• Another common way to write the derivative of f(x) is using prime notation:

f ′(x) =df

dx=

d

dx

(f(x)

)The Antiderivative

• An antiderivative of a function f(x) is any function F (x) whose derivative is f(x), F ′(x) = f(x).

• Since the derivative of a constant is zero, ddx

(C)

= 0, there are multiple possible antiderivatives.Additionally antiderivatives differ by at most a constant.

• An antiderivative problem (indefinite integral) asks for all possible antiderivatives of f(x)∫f(x) dx = F (x) + C

where F (x) is any antiderivative and C is an unknown constant (called the constant of integration).

• Basic antiderivatives:∫xn dx =

1

n+ 1xn+1 + C; n 6= −1∫

1

xdx = ln |x|+ C∫

ex dx = ex + C∫sinx dx = − cosx+ C∫cosx dx = sinx+ C

• Basic antiderivatives with linear insides2:∫(ax+ b)n dx =

1

a· 1

n+ 1(ax+ b)n+1 + C∫

1

ax+ bdx =

1

aln |ax+ b|+ C∫

eax+b dx =1

aeax+b + C∫

sin(ax+ b) dx = −1

acos(ax+ b) + C∫

cos(ax+ b) dx =1

asin(ax+ b) + C

• Linearity rule for antiderivatives:∫ (af(x) + bg(x)

)dx = a

∫f(x)dx+ b

∫g(x)dx

• Students are expected to know the basic antiderivative rules above and be able to efficiently computethem without the aid of a calculator.

• Antiderivatives do not have product, quotient, or chain rules like derivatives do.

2 Linear insides are of the form ax+ b. Unlike derivatives there is no rule for antiderivatives with complicated insides.


Integrals

• A Riemann Integral computes the signed area betweenthe curve y = f(x) and the x-axis on the interval [a, b].

• A Riemann Sum approximates the area by covering itwith tiny rectangles, and then summing their areas.

• The Riemann Integral is defined to be the limit of thearea approximations as the number of rectangles be-comes infinite (or their width approaches 0).

• Integrals can be computed using the Fundamental Theorem of Calculus.

– Here’s the formal theorem:

If f(x) is continuous on the interval [a, b] and F (x) is any antiderivative of f(x) then∫ b

a

f(x) dx = F (x)∣∣∣ba

= F (b)− F (a)

– Here’s how you use it to compute integrals:∫ b

a

formula with x dx

↓

find antiderivative∣∣∣ba

↓

Answer: Evaluate antiderivative at b − Evaluate antiderivative at a

Integral Notation

• Integral notation is used for both antiderivatives (without bounds) and integrals (with bounds).

– An antiderivative problem asks for all possible antiderivatives:∫f(x)dx = F (x) + C

– An integral problem asks for a number, which gives the signed area under the curve∫ b

a

f(x)dx = F (x)∣∣∣ba

= F (b)− F (a)

• Note: Integrals and antiderivatives are two different concepts, and need to be treated as such. Oneis the inverse of the derivative while the other computes the signed area of a region by slicing it upinto tiny rectangles, and summing the area of all the rectangles.

Due to the Fundamental Theorem of Calculus, integrals can be computed using antiderivatives. Thisis the motivation of why the integral symbol is used in antiderivative problems. But it is importantto keep track of the distinction between antiderivatives and integrals.

Substitution I Notes and Learning Goals 8

Integration by substitution is the process of transforming an antiderivative or integral problem in termsof the variable x into a new antiderivative or integral problem in terms of a new variable u. Then use theanswer to the new problem to find an answer to the original problem.

Transform an Antiderivative Problem using u-Substitution

Here is the process to transform then answer an antiderivative problem,

∫f(x) dx, via u-substitution.

The process begins with a formula: u = formula with x . You are then locked into the following steps.

Step 1. Compute du. Write it as du = u′ dx.

Step 2. Use the equations for u and du to substitute all the x’s and the dx in the original problem. Youwill have to recognize the following types of substitution opportunities:

• Things that look like the formula for u. Replace with u.

• Things that look like the formula for du. Replace with du.

Note: Sometimes you have to do a bit of algebra to get the du to work out.

• If the previous steps leave some x’s in the problem, solve u = formula with x for the variable

x. Use your answer to replace remaining x’s.

The result is a transformed antiderivative problem3 only in terms of u.

Step 3. Find an answer to the new antiderivative problem. Your answer will have u’s in it.

Step 4. Change all the u’s back into formula with x to answer original problem.

Example: Answer the following antiderivative problem using the substitution u = 1 + x2.∫2x√

1 + x2 dx

Step 1. Start with u = 1 + x2 and find du = u′ dx = 2x dx.

Step 2. Substitute using the equations u = 1 + x2 and du = 2xdx:∫2x√

1 + x2 dx =

∫ √1 + x2

(2x dx

)=

∫ √u du

At this stage we have only transformed the antiderivative problem.

Step 3. Answer the new antiderivative problem:∫ √u du =

∫u1/2 du =

2

3u3/2 + C

Step 4. Use the answer to the new problem and the substitution u = 1+x2 to answer the original problem:∫2x√

1 + x2 dx =2

3

(1 + x2

)3/2+ C

3 Know that substitution does not solve the original problem; it just replaces the original antiderivative problem with anew one. Expect homework and exam questions to ask only for the transformed problem.

Substitution I Notes and Learning Goals 9

Transform an Integral using u-Substitution

The process to transform an integral,

∫ b

a

f(x) dx, via substitution is similar, but you also have to

transform the limits of integration.

The process starts with a formula: u = formula with x .

Step 1. Find du = u′dx.

Step 2. Use the equations for u and du to rewrite the integrand only in terms of u.

In addition transform the bounds: The bounds x = a and x = b are values of the original variable x.

Use the formula with x to get the new limits of integration in terms of the new variable u.

The integral is transformed when both the integrand and bounds are in terms of u.

Step 3. Answer the new integral problem using the Fundamental Theorem of Calculus4.

Step 4. The answer to the new integral problem (in above step) also answers the original integral problem.

Example: Answer the following integral problem using the substitution u = 1 + x2.∫ 2

1

2x√

1 + x2 dx

Step 1. Start with u = 1 + x2 and find du = u′ dx = 2x dx.

Step 2. Find the new bounds:

• Lower Bound: If x = 1 then u = 1 + 12 = 2.

• Upper Bound: If x = 2 then u = 1 + 22 = 5.

Substitute using the equations u = 1 + x2, du = 2xdx, and the new bounds5:∫ 2

1

2x√

1 + x2 dx =

∫ 2

1

√1 + x2

(2x dx

)=

∫ 5

2

√u du

This substitution shows that the area represented by the originalintegral and the area represented by the transformed integral arethe same (as shown at right). Thus the answer to the new problemwill also answer the original problem.

Step 3. Answer the new problem using the FTC:

∫ 5

2

u1/2 du =2

3u3/2

∣∣∣52

=2

3

(53/2 − 23/2

)Step 4. This is also the answer to the original integral.

4 Since both the integrand and bounds are written in terms of u, there is no need to revert back to the old variable. Justplug in the u bounds into the antiderivative in terms of u using the FTC.

5 The bounds must match the differential, du or dx. Integrals whose bounds are not translated to the new variable willgive an incorrect value.

Substitution I Worksheet 1 10

WebAssign #1: Find the following antiderivative using the substitution u = x4 + 10.∫4x3(x4 + 10)7 dx

Start by writing the substitution equation u = x4 + 10.

Step 1 Find du.

Step 2 Use the equation for u and du to rewrite the antiderivative problem in terms of u and du.∫4x3(x4 + 10)7 dx =

Step 3 Find an antiderivative for the new problem in terms of u (include +C).

Step 4 Use the substitution u = x4 + 10 to transform the answer in step 3 back to the original variable x.

WebAssign #2: Find the following antiderivative using the substitution u = sin(x).∫sin3(x) cos(x) dx

Follow the above steps.


Transform each of the following antiderivative problems into a new antiderivative problem using the givenu-substitution. Your answer will be a new antiderivative problem in terms of u and du.

• WebAssign #3: Use u = −x2∫−2xe−x

2

dx

• WebAssign #4: Use u = ln(x)∫ln(x)

xdx

• WebAssign #5: Use u = 2x3 − 4∫x2(2x3 − 4)9 dx

• WebAssign #6: Use u = x2 − 4∫x√x2 − 4 dx


WebAssign #9: Compute the following integral using the substitution u = cos(x).∫ π/3

0

cos3(x) sin(x) dx

Start by writing the substitution equation u = cos(x).

• Find du.

• Use the substitution equation, u = cos(x), to transform the x-bounds into u-bounds.

• Use the equation for u and du and the new bounds to rewrite the integral in terms of u and du.∫ π/3

0

cos3(x) sin(x) dx =

• Use the Fundamental Theorem of Calculus to compute the new integral.

WebAssign #10: Compute the following integral using the substitution u = ln(x).∫ e

1

(ln(x)

)2x

dx

Follow the above steps.


Compute each of the following integrals using the given u-substitution. Show all work as described below.

• Statement of u. Even if this is given, you have to write it on your paper.

• Statement of du. Must include the other differential (usually dx).

• Transformed integral, written entirely in terms of u.

• Computation of the new integral using the Fundamental Theorem of Calculus.

• WebAssign #11: Use u = −x2/4∫ 1

0

3xe−x2/4 dx

• WebAssign #12: Use u = 1− x2∫ 1

0

x√

1− x2 dx

Substitution II Notes and Learning Goals 14

In Substitution I every problem you were given started with:

u = formula with x

In today’s lesson you will choose the substitution formula used.

1. Learn to make good choices for u = formula with x

2. Know that any choice is allowed. But not all choices are useful.

Here are some suggestions for how to make useful choices.

• If any part of the problem is a function with insides, consider:

u = insides

Example:

∫x√

4− x2 dx The insides are u = 4− x2.

• If there is a denominator more complicated than x or xn, consider:

u = denominator

Example:

∫x2

3x+ 4dx The denominator is u = 3x+ 4.

• If some stuff looks like the derivative of some other stuff, consider:

u = the other stuff

Example:

∫1

xlnx dx The derivative of lnx is

1

x, so try u = lnx.

3. Know that these are not rules. They are suggestions for what works. The only way to know if asubstitution formula is useful is to use it to transform the integral and see if the new integrand isnicer than the original.

Substitution II Worksheet 1 15

WebAssign #1: Consider the antiderivative problem

∫x2(x3 + 5)9 dx.

• Transform this antiderivative problem using the substitution u = x3 + 5. Your answer will be a newantiderivative problem written in terms of u and du

• Transform this antiderivative problem using the substitution u = x2. Your answer will be a newantiderivative problem written in terms of u and du

• Choose the above substitution which was useful in transforming the antiderivative problem into anelementary antiderivative. Use that substitution to finish the problem and find the antiderivative ofthe original problem in terms of x. Include +C.


WebAssign #2: Consider the antiderivative problem

∫x2√x3 + 1 dx.



• Choose the above substitution which was useful in transforming the antiderivative problem into anelementary antiderivative. Use that substitution to finish the problem and find the antiderivative ofthe original problem in terms of x. Include +C.


Compute each of the following antiderivatives using substitution. Show all work as described below.

• Statement of u. Determine du.

• Transformed antiderivative problem, written entirely in terms of u.

• Antiderivative of new problem followed by answer to original problem (Include +C).

• WebAssign #7:

∫3t√

7− 3t2 dt

• WebAssign #8:

∫6x2 sin(x3) dx

• WebAssign #9:

∫4x

(16− x2)3/2dx






• WebAssign #11:

∫ 1

0

x3(1 + x4)3 dx

• WebAssign #12:

∫ √30

4x√x2 + 1

dx






• WebAssign #13:

∫ π/2

0

3 cos2(x) sin(x)dx

• WebAssign #14:

∫ 4

1

1

2√x(1 +√x)2 dx

Integration by Parts Notes and Learning Goals 20

1. The new skill is a technique called integration by parts. Here’s how it works on an example:∫3x cosx dx

• Visualize the integrand as a product. ∫3x cosx dx

• Name the two pieces u and dv. The dv piece must contain the original differential, dx.

u = 3x

dv = cosx dx

• Differentiate u, and antidifferentiate dv. Note that du picks up the dx, while v loses it.

u = 3x −→ du = 3 dx

dv = cosx dx −→ v = sinx

• Use the Integration by Parts Rule:∫u dv = uv −

∫v du

The original integral is transformed into

3x sinx −∫

sinx 3 dx

2. The rule also works on definite integrals. This will be explored in the WebAssign exercises.

3. If you are interested in where the rule comes from:

• Start with product rule: (uv)′ = vu′ + uv′

• Solve for uv′ = (uv)′ − vu′

• Integrate both sides∫uv′ =

∫(uv)′ −

∫vu′ = uv −

∫vu′

Integration by Parts Worksheet 1 21

WebAssign #1: Find the following antiderivative using integration by parts:∫2xe−x dx

Start by writing down the two parts: u = 2x and dv = e−xdx

Step 1 Find du and v.

Step 2 Use the integration by parts formula to rewrite the antiderivative problem:∫u dv = uv −

∫v du∫

2xe−x dx =

Step 3 Find an antiderivative for the new problem (include +C).

WebAssign #2: Find the following antiderivative using integration by parts.∫12t sin(2t) dt

Use the parts u = 12t and dv = sin(2t) dt. Follow the above steps.


WebAssign #3: Find the following antiderivative using integration by parts.∫t2 ln(t) dt

Use the parts u = ln(t) and dv = t2 dt.

WebAssign #4: Find the following antiderivative using integration by parts.∫(2x+ 7)e2x dx

Use the parts u = 2x+ 7 and dv = e2x dx.


WebAssign #5: Find the following integral using integration by parts:∫ π

0

3t sin(t) dt

Start by writing down the two parts: u = 3t and dv = sin(t)dt

Step 1 Find du and v.

Step 2 Use the integration by parts formula to rewrite the antiderivative problem:

∫ π

0

u dv = uv∣∣∣π0−∫ π

0

v du

Note: The vertical bar with bounds on the right of uv means you have to plug in the bounds intothe formula for uv using the Fundamental Theorem of Calculus.∫ π

0

3t sin(t) dt =

Step 3 Find the new integral using the Fundamental Theorem of Calculus to find the final answer.


WebAssign #6: Find the following antiderivative using integration by parts twice:∫6x2e3x dx

• First, use integration by parts with u = 6x2 and dv = e3x dx.∫6x2e3x dx =

• Then use integration by parts a second time with u = 4x and dv = e3x dx.

• Last, find the final antiderivative (include +C).

WebAssign #7: Use integration by parts twice to find the following integral:∫ 1

0

18x2e3x dx

• Start with the parts u = 18x2 and dv = e3x dx.

• Then use the parts u = 12x and dv = e3x dx on the new integral.


WebAssign #8: Find the following antiderivative using integration by parts.∫ln(x) dx

Use the parts u = ln(x) and dv = dx.

WebAssign #9: Find the following antiderivative using integration by parts.∫x2 ln(x) dx

Answer questions in WebAssign to help determine the u and dv to use.


Find each of the following antiderivatives using integration by parts. Show all work.

• Statement of u and dv. Determine du and v.

• Transformed antiderivative problem using integration by parts formula.

• Computation of the new antiderivative problem for final antiderivative (include +C).

WebAssign #10: Find the following antiderivative using integration by parts.∫(4x+ 3)e−2x dx

WebAssign #11: Find the following antiderivative using integration by parts.∫(5x− 2)e−0.1x dx

WebAssign #12: Find the following antiderivative using integration by parts.∫7x3 ln(x) dx

Trigonometric Integrals Notes and Learning Goals 27

Here are the types of trigonometric integrals you will encounter.

1. Derivatives and resulting antiderivatives of trigonometric functions.

d

dx

(sin(x)

)= cos(x)

∫cos(x) dx = sin(x) + C

d

dx

(cos(x)

)= − sin(x)

∫sin(x) dx = − cos(x) + C

d

dx

(tan(x)

)= sec2(x)

∫sec2(x) dx = tan(x) + C

d

dx

(sec(x)

)= sec(x) tan(x)

∫sec(x) tan(x) dx = sec(x) + C

d

dx

(cot(x)

)= − csc2(x)

∫csc2(x) dx = − cot(x) + C

d

dx

(csc(x)

)= − csc(x) cot(x)

∫csc(x) cot(x) dx = − csc(x) + C

2. Many antiderivatives involving sines and cosines are computable by a substitution of u = sin(x) oru = cos(x).

3. Trigonometric identities are often useful. Everyone must know the following identities:

tan(x) =sin(x)

cos(x)cot(x) =

cos(x)

sin(x)

sec(x) =1

cos(x)csc(x) =

1

sin(x)

sin2(x) + cos2(x) = 1 tan2(x) + 1 = sec2(x)

You are expected to have the above identities memorized.

The following identities are also useful when computing trigonometric antiderivatives. These iden-tities will be provided on the exam formula sheet, so you do not need to memorize them. You willneed to know how to use them.

• Double-angle Identities:

sin(2x) = 2 sin(x) cos(x)

cos(2x) = cos2(x)− sin2(x)

• Half-angle Identities:

sin2(x) =1

2− 1

2cos(2x)

cos2(x) =1

2+

1

2cos(2x)

• Product to Sum Identities:

cos(a) cos(b) =1

2

(cos(a+ b) + cos(a− b)

)sin(a) sin(b) =

1

2

(cos(a− b)− cos(a+ b)

)sin(a) cos(b) =

1

2

(sin(a+ b) + sin(a− b)

)cos(a) sin(b) =

1

2

(sin(a+ b)− sin(a− b)

)4. Trigonometric antiderivatives are computed using a combination of the derivative rules for trigono-

metric functions, trigonometric identities, and integration methods such as u-substitution.

Trigonometric Integrals Worksheet 1 28

WebAssign #1: Find the following antiderivative using the substitution u = sin(x):∫sin2(x) cos(x) dx

WebAssign #2: Find the following antiderivative:∫cos2(x) sin(x) dx


WebAssign #3: Find the following antiderivative:∫cos(x)

sin2(x)dx

WebAssign #4: Find the following antiderivative:∫sin(x)

cos2(x)dx

WebAssign #5: Find the following antiderivative(

Hint: tan(x) =sin(x)

cos(x)

):∫

tan(x) dx


WebAssign #6: Find the following antiderivative, follow the steps in WebAssign:∫sin2(x) dx

WebAssign #7: Find the following antiderivative:∫cos2(x) dx

WebAssign #8: Find the following antiderivative:∫4 cos2(x) dx


WebAssign #9: Find the following antiderivative:∫cos2(3x) dx

WebAssign #10: Find the following antiderivative:∫4 sin2(5x) dx


WebAssign #11: Find the following antiderivative(

Hint: sin3(x) = sin2(x) sin(x))

:∫sin3(x) dx

WebAssign #12: Find the following antiderivative:∫cos3(x) dx

WebAssign #13: Find the following antiderivative∫sin5(x) dx


WebAssign #14: Find the following antiderivative (Hint: Use Product-to-Sum Identity)∫sin(5x) sin(3x) dx

WebAssign #15: Find the following antiderivative:∫cos(5x) cos(3x) dx

Trigonometric Substitution Notes and Learning Goals 34

Trigonometric substitution is a new kind of substitution in which you substitute the x variable with atrigonometric function. We will consider two forms: x = a sin(θ) or x = a tan(θ). These substitutions areto applied to integrals including things of the form

√a2 − x2 or

√a2 + x2, respectively.

1. Transform the antiderivative problem∫x2√a2 − x2 dx using the substitution x = a sin(θ).

Type I: Suppose that x = a sin(θ), then

sin(θ) =x

a=

opposite

hypotenuse

So θ is an angle in a right triangle. By the PythagoreanTheorem the length of the remaining side is

√a2 − x2.

The cosine of the angle is cos(θ) =

√a2 − x2a

.

This gives the following three substitution relations:

x = a sin(θ),√a2 − x2 = a cos(θ), and dx = a cos(θ)dθ

Use these relations to transform the antiderivative problem∫x2√a2 − x2 dx =

∫ (a sin(θ)

)2(a cos(θ)

) (a cos(θ)dθ

)=

∫a4 sin2(θ) cos2(θ)dθ

This new antiderivative problem is a trigonometric integral and can be computed using methodsfrom the previous lesson.

The transformation can also be done algebraically using the identity cos2(θ) = 1− sin2(θ). This gives

√a2 − x2 =

√a2 − a2 sin2(θ) =

√a2(1− sin2(θ)) =

√a2 cos2(θ) = a cos(θ)

Trigonometric Substitution Notes and Learning Goals 35

2. Transform the antiderivative problem

∫x2

a2 + x2dx using the substitution x = a tan(θ).

Type II: Suppose that x = a tan(θ), then

tan(θ) =x

a=

opposite

adjacent

So θ is an angle in a right triangle. By the PythagoreanTheorem the length of the remaining side is

√a2 + x2.

The secant of the angle is sec(θ) =

√a2 + x2

a.

This gives the following three substitution relations:

x = a tan(θ),√a2 + x2 = a sec(θ), and dx = a sec2(θ)dθ

Use these relations to transform the antiderivative problem∫x2(√

a2 + x2)2 dx =

∫ (a tan(θ)

)2(a sec(θ)

)2 (a sec2(θ)dθ)

=

∫a tan2(θ)dθ

This new antiderivative problem is a trigonometric integral and can be computed using methodsfrom the previous lesson.

The transformation can also be done algebraically using the identity tan2(θ) + 1 = sec2(θ). Thisgives

a2 + x2 = a2 + a2 tan2(θ) = a2(1 + tan2(θ)) = a2 sec2(θ)

Trigonometric Substitution Worksheet 1 36

WebAssign #1: Find the following antiderivative using the trigonometric substitution x = sin(θ):∫ √1− x2 dx

a. Find the length of the third side of right triangle at right,then use the right triangle to find cos(θ) in terms of x.

cos(θ) =

b. Take the derivative of the substitution equation, x = sin(θ), to find dx = x′dθ.

Write down the resulting equation for dx.

c. Transform the original antiderivative problem into a new problem written in terms of θ and dθ.

• Use the results in part (a) to substitute√

1− x2 with what it is equal to.

• Use the results in part (b) to substitute dx with what it is equal to.

∫ √1− x2 dx =

d. Find the antiderivative of the transformed antiderivative problem in terms of θ.Hint: The half-angle identity cos2(θ) = 1

2+ 1

2cos(2θ) will be useful.

e. Rewrite the result in terms of the original variable by using the substitution equation x = sin(θ).Hint: The double angle identity sin(2θ) = 2 sin(θ) cos(θ) and the result from (a) will also be helpful.


WebAssign #2: Find the following antiderivative using the trigonometric substitution x = 3 sin(θ):∫x2√

9− x2dx

a. Take the substitution equation x = 3 sin(θ) and solve for

sin(θ) =x

3=

opposite

hypotenuse

Use the result to label right triangle at right, then find

cos(θ) =

b. Take the derivative of the substitution equation, x = 3 sin(θ), to find dx = x′dθ.


c. Transform the original antiderivative problem into a new problem written in terms of θ and dθ.Simplify answer.

• Use the substitution equation to substitute x2 with what it is equal to.

• Use the results in part (a) to substitute√

9− x2 with what it is equal to.


∫x2√

9− x2dx =

d. Find the antiderivative of the transformed antiderivative problem in terms of θ.

e. Rewrite the result in terms of the original variable by using the substitution equation x = 3 sin(θ).


WebAssign #3: Find the following antiderivative using the trigonometric substitution x = 2 sin(θ):∫x√

4− x2 dx

Show all work following the previous questions/worksheets.

WebAssign #4: Find the following antiderivative using the u-substitution u = 4− x2.∫x√

4− x2 dx

Show all work needed for u-substitution.

The above are two different substitution methods to find the same antiderivative. Discuss with your groupthe pros and cons to each method. Which method would you prefer to use?


WebAssign #5: Transform the following antiderivative using two different methods:∫x(1− x2)3/2 dx

Show all work required to transform the antiderivative problem.

Use the trigonometric substitution x = sin(θ):

Use the u-substitution u = 1− x2:

Pick one of the above transformations, and use it to compute the original antiderivative.


WebAssign #7: Find the following antiderivative using the trigonometric substitution x = tan(θ):∫1

x2√x2 + 1

dx

a. Find the length of the third side of right triangle at right,then use the right triangle to find sec(θ) in terms of x.

sec(θ) =

b. Take the derivative of the substitution equation, x = tan(θ), to find dx = x′dθ.


c. Transform the original antiderivative problem into a new problem written in terms of θ and dθ.

• Use the substitution equation to substitute x2 with what it is equal to.

• Use the results in part (a) to substitute√x2 + 1 with what it is equal to.


∫1

x2√x2 + 1

dx =

d. Find the antiderivative of the transformed antiderivative problem in terms of θ.

e. Rewrite the result in terms of the original variable. Hint: Use the associated right triangle.


WebAssign #8: Find the following antiderivative using the trigonometric substitution x = 5 tan(θ):∫ √x2 + 25 dx

a. Take the substitution equation x = 5 tan(θ) and solve for

tan(θ) =x

5=

opposite

adjacent

Use the result to label right triangle at right, then find

sec(θ) =

b. Take the derivative of the substitution equation, x = 5 tan(θ), to find dx = x′dθ.


c. Transform the original antiderivative problem into a new problem written in terms of θ and dθ.Simplify answer. You do not need to compute the transformed antiderivative.

∫ √x2 + 25 dx =


WebAssign #9: Find the following antiderivative using the trigonometric substitution x = 3 tan(θ):∫x√

9 + x2dx

Show all work following the previous questions/worksheets.

WebAssign #10: Find the following antiderivative using the u-substitution u = 9 + x2.∫x√

9 + x2dx

Show all work needed for u-substitution.

The above are two different substitution methods to find the same antiderivative. Discuss with your groupthe pros and cons to each method. Which method would you prefer to use?


WebAssign #11: Find the following antiderivative using the trigonometric substitution x = tan(θ):∫1

1 + x2dx

WebAssign #12: Find the following antiderivative using the trigonometric substitution x = a tan(θ):∫1

a2 + x2dx

Partial Fractions I Notes and Learning Goals 44

1. Know and be able to use the following fact:

“Two polynomials are equal only if their coefficients are equal”

You will use this in your homework when you encounter things like

(A+B)x2 + (B + C)x+ (A+ C) = 4x2 − 2

You should conclude that:

A+B = 4

B + C = 0

A+ C = −2

2. Be able to solve systems of equations. For example, the system above with A, B and C.

3. Given a partial fractions decomposition, determine the coefficients. For example, be able tosolve for A and B if you are given

5x− 1

(2x+ 1)(x− 3)=

A

2x+ 1+

B

x− 3

The process is:

• Rewrite with a common denominator.

• Simplify the numerator.

• Equate coefficients to make a system of equations. (See Item 1 above.)

• Solve for A and B. (See Item 2 above.)

Worked Example: Find A and B, given5x− 1

(2x+ 1)(x− 3)=

A

2x+ 1+

B

x− 3

• Use a common denominator to simplify the right hand side.

A

2x+ 1+

B

x− 3=A(x− 3) +B(2x+ 1)

(2x+ 1)(x− 3)=

(A+ 2B)x+ (−3A+B)

(2x+ 1)(x− 3)

• Compare to the original, without denominators.

5x− 1 = (A+ 2B)x+ (−3A+B)

• Write a system of equations.{A+ 2B = 5 (x-coefficient)−3A+B = −1 (constant coefficient)

• Solve the system to get A = 1 and B = 2.

• Conclude that5x− 1

(2x+ 1)(x− 3)=

1

2x+ 1+

2

x− 3

Partial Fractions I Notes and Learning Goals 45

4. Use all of the above to compute integrals and antiderivatives. For example∫5x− 1

(2x+ 1)(x− 3)dx

Solution:

The previous exam showed how to figure out that

5x− 1

(2x+ 1)(x− 3)=

1

2x+ 1+

2

x− 3

Use this conclusion to rewrite the antiderivative problem∫5x− 1

(2x+ 1)(x− 3)dx =

∫ (1

2x+ 1+

2

x− 3

)dx

Then compute the antiderivative of the rewritten problem. Putting this all together gives

∫5x− 1

(2x+ 1)(x− 3)dx =

∫1

2x+ 1dx+

∫2

x− 3dx

=1

2ln |2x+ 1|+ 2 ln |x− 3|+ C

5. Work Expectations: When showing how to find an antiderivative using partial fractions

• All work used to determine the coefficients:

– Clearly show how to setup and state a system of equations for the unknown coefficients.

– Clearly show how you solved the resulting system of equations.

• A rewritten integral using the partial fractions decomposition.

• Compute the antiderivatives of the rewritten problem.

Partial Fractions I Worksheet 1 46

WebAssign #5: Find the partial fractions decomposition of5x+ 10

(1− x)(x+ 4).

Start with the partial fractions formA

1− x+

B

x+ 4.

• Use algebra to put the partial fractions form over a common denominator and group like terms:

A

1− x+

B

x+ 4=

• Compare the numerator of the result in step 1 to the original fraction, set up a system of equations,and solve the system for A and B.

• Use the solution to the system to write down the partial fractions decomposition.


x(x+ 1)using the above steps.

Assume the fraction can be written in the formA

x+

B

x+ 1.



(x+ 1)(x− 2)(x+ 3).

Start with the partial fractions formA

x+ 1+

B

x− 2+

C

x+ 3.

• Write out the statement that the fraction is equal to its form

4x+ 22

(x+ 1)(x− 2)(x+ 3)=

A

x+ 1+

B

x− 2+

C

x+ 3

Multiply both sides by the denominator, (x+1)(x−2)(x+3), and write down the resulting equation:

• Plug in the nice value x = −1 into both sides of the above equation, write down the resultingequation, then use it to solve for A.

• Use other nice values of x to find the remaining coefficients B and C.

• Use your results to write down the partial fractions decomposition of the original fraction.


WebAssign #8: Find the following antiderivative using partial fractions:

∫1

x(x+ 1)dx

Assume that the integrand can be written in the formA

x+

B

x+ 1.

• Use a method from previous problems to set up a system of equations and solve for A and B.

• Use the partial fractions decomposition to rewrite the antiderivative problem∫1

x(x+ 1)dx =

• Find the antiderivative of the rewritten problem (include +C).

WebAssign #9: Find the following antiderivative using partial fractions:∫10x

(x+ 1)(2x− 3)dx

Follow the above steps. Assume that the integrand can be written in the formA

x+ 1+

B

2x− 3.


Compute the following antiderivative using partial fractions. Show all work as described below.

• All work used to determine the coefficients, including clear statements of the equations that you setup and how you solved them.

• A rewritten antiderivative problem using the partial fractions decomposition.

• Computation of the new antiderivative (include +C).

• WebAssign #10:

∫5

x2(x− 5)dx

Assume the integrand can be written asA

x+B

x2+

C

x− 5.

Warning! Be sure to use the least common denominator.

Partial Fractions II Notes and Learning Goals 50

In the previous lesson, Partial Fractions I, every partial fractions problem began with a given format thatincluded some unknown constants. Today you have to create the form yourself. Here are the rules.

• The numerator is completely irrelevant.6

• The denominator must be completely factored. Make sure that any quadratic factors are irreducible.The most common irreducible form is x2 + k, where k is positive.

• Each linear factor (ax+ b) contributesA

ax+ b

• Each repeated linear factor (ax+ b)2 contributes

A

ax+ b+

B

(ax+ b)2

• Higher order repeats, (ax+ b)3 and so on, extend this pattern.

A

ax+ b+

B

(ax+ b)2+

C

(ax+ b)3

• Each irreducible quadratic factor (ax2 + bx+ c) contributes

Ax+B

ax2 + bx+ c

In many examples b is zero, so the factor is just (ax2 + c)

• Eventually you will encounter a repeated quadratic factor,7 which contributes

Ax+B

ax2 + bx+ c+

Cx+D

(ax2 + bx+ c)2

• If you get a triple quadratic factor your teacher is trying to hurt you.

Example: Write down the partial fractions form of

1

(x+ 1)(x− 2)2

Solution:A

x+ 1+

B

x− 2+

C

(x− 2)2

Note: The problem says “write down the form” so stop here. Do not solve for A, B and C.

6If the degree (highest power) of the numerator is equal to or larger than the degree of the denominator you must first usepolynomial long division to rewrite the problem as a quotient and a remainder term. This will not occur in your homework.

7This will happen in differential equations and some physics or engineering applications. It occurs when an oscillatingobject experiences a resonant frequency.

Partial Fractions II Notes and Learning Goals 51

Additional Items:

• Problems that say “write the form” are intended as very quick short answer problems. The onlywork you could possibly show would be factoring the denominator.

• The number of unknown constants in your answer must always match the degree (highest power ofx) in the denominator.

• For problems that say “find the complete partial fractions decomposition,” first you must create theform. Then proceed as in Partial Fractions I.

• For problems that say “find the antiderivative,” you will need these facts:

∫1

ax+ bdx =

1

aln |ax+ b|+ C

∫1

(ax+ b)2dx becomes

∫(ax+ b)−2 dx, which is elementary.

∫1

x2 + 1dx = tan−1 x+ C

∫1

x2 + a2dx =

1

atan−1

(xa

)+ C

∫x

x2 + a2dx uses substitution with u = x2 + a2.

∫Ax+B

x2 + a2dx splits into two pieces: A

∫x

x2 + a2dx+B

∫1

x2 + a2dx

Partial Fractions II Worksheet 1 52

Find the following antiderivatives using the method of partial fractions. Show all work:

• Clear statement the partial fractions form with unknown constants.

• All work used to determine the coefficients, including clear statements of the equations you set upand how they were solved.

• A correctly written antiderivative problem using the partial fractions decomposition.

• A correct antiderivative. Include +C.

WebAssign #11:

∫x+ 12

x2 − x− 6dx

WebAssign #12:

∫1

4x2 − 3x− 1dx.

Partial Fractions II Worksheet 2 53

Find the following antiderivatives using the method of partial fractions. Show all work:

• Clear statement the partial fractions form with unknown constants.

• All work used to determine the coefficients, including clear statements of the equations you set upand how they were solved.

• A correctly written antiderivative problem using the partial fractions decomposition.

• A correct antiderivative. Include +C.

WebAssign #13:

∫x− 2

x3 + xdx

WebAssign #14:

∫5x− 12

x3 + 4x2dx.

Integration Methods Notes and Learning Goals 54

In order to be ready for Exam 1 you must be able to quickly and accurately recognize which antiderivativeor integration technique is appropriate. There are five situations that you must recognize:

1. Elementary Antiderivatives

2. Integration by Substitution

3. Integration by Parts

4. Partial Fractions

5. Trigonometric Substitution

You must be able to determine which method will be useful to find an antiderivative, and then be ableto execute the method. The previous lessons covered how to execute each method. Today’s lesson willdevelop the skill of selecting which method would be useful. In today’s lesson you will be asked to

• Determine which integration method would be appropriate for a given problem.

• State the starting steps of the chosen method:

– Elementary antiderivatives: Find the antiderivative and include +C.

– Substitution: State the substitution equation, u = formula .

– By Parts: State the u = formula and dv = formula .

– Partial Fractions: State the partial fractions form using unknown coefficients, A,B,C, · · · .– Trigonometric Substitution: State the trigonometric substitution used, x = a sin(θ) orx = a tan(θ).

Note: Many problems have more than one right answer. It is entirely possible that an integral could fallin more than one of the above categories. You get full credit for picking any one of the correct answers.

Here is how to determine if a method will be useful:

• Decide on a method. But DO NOT click anything.

• Get out paper and execute the first few steps of your method.

• If the problem transforms to something simpler (or elementary), you are probably right.

• If it gets worse, you probably need a different method.

The most important recognition skill by far is this:

“You must immediately recognize whether a problem is Elementary or Not.”

Elementary Antiderivatives

• Anything on the Elementary Antiderivatives List is considered elementary, and you can write downthe answer without showing any work.

• Some problems require you doing some algebra (or trigonometric identities) to rewrite it in a formthat is elementary.

• Be able to recognize if a problem is elementary, and if it is compute the answer. If it is not elementary,then you need to use one of the other methods.


Integration by Substitution

• Recognizing integration by substitution is the same as identifying an appropriate choice of u.

• Integration by Substitution II’s notes have some suggestions on how to choose u.

• Once you think of u, write it down and execute the substitution process.

• If you end up with an elementary (or simpler) antiderivative then you were right.

• If not, start over with a different technique or a different u.

Integration by Parts

• Recognizing integration by parts is mostly a matter of experience. Work a lot of examples.

• Parts is often indicated by a product, but not all products need parts.

Example 1:

∫x cosx dx requires parts.

Example 2:

∫x cos(x2) dx requires substitution with u = x2.

Example 3:

∫x2√

1− x2 dx requires trig substitution with x = sin θ.

Example 4:

∫x√

1− x2 dx is easiest with u = 1− x2, but x = sin θ will also work.

• Once you decide that a problem needs integration by parts, write down u and dv.

• Execute the process. If the problem gets simpler (or no worse), you are on the right track.

• Sometimes integration by parts has to be repeated.

Partial Fractions

• These are pretty easy to recognize. Partial fractions is appropriate when these conditions are met.

– The integrand is an algebraic fraction.

– Numerator and denominator are both polynomials.8

– The denominator is factorable into linear factors or irreducible quadratics.

– There is more than one factor.

• Once you decide that partial fractions is appropriate, immediately write down the form of the partialfractions decomposition.

8The degree of the numerator must be less than the degree of the denominator. This technicality has been kept out ofyour homework. You don’t have to worry about it.


Trig Substitution

• This is triggered by the presence of algebraic structures that are too complicated for other techniques.The most common are

1. (a2 − x2), where a is a constant. It often, but not always, occurs inside a square root.

Use x = a sin θ.

2. (a2 + x2), where a is a constant. It often, but not always, occurs in a denominator.

Use x = a tan θ.

• Trig substitution is longer and more difficult than other techniques. It is often worth spending alittle time looking for a quicker substitution. See Example 4 above.

• Once you decide that trig substitution is appropriate, execute the substitution and transform theproblem using appropriate right triangle or trig identities.

If x = a sin θ. Then√a2 − x2 = a cos θ (use a right triangle or sin2 θ + cos2 θ = 1).

If x = a tan θ. Then√a2 + x2 = a sec θ (use a right triangle or tan2 θ + 1 = sec2 θ).

Summary

• Antiderivatives in general can be difficult to find. In order to answer problems involving antideriva-tives, one must be able to recognize which method (if any) would be appropriate.

• There is no formulaic way to recognize which method should be used. The only way to know if amethod works, is to try it. If the method produces a simpler (elementary) antiderivative, then themethod was useful.

• It is vital that you can recognize which antiderivatives are elementary or not. If an antiderivative iselementary, be able to write down the answer. If it is not, you must then try various methods untilyou find one that works.

• Some problems can be done using multiple methods. When choosing a method, choose the one thatis the easiest to use. As such, trigonometric substitution, is best saved for problems in which noother method can work.

Limits Notes and Learning Goals 57

Part I: Vocabulary and Notation

• The notation for limit at infinity is limx→∞

f(x).

You read this: limit of f(x) as x goes to1 infinity.

• There are three basic types of behavior for a function, f(x), as x approaches infinity.

Know how to recognize each type and know how to write the notation for it.

1. Horizontal Asymptote.Notation is:

limx→∞

f(x) = L

where L is a number.We say f has a horizontal asymptote at L.

Examples of asymptotes:

2. Goes to Infinity.Notation is:

limx→∞

f(x) =∞

Or

limx→∞

f(x) = −∞

Examples going to infinity:

3. Everything Else.We say the limit does not exist.Notation is

limx→∞

f(x) DNE

One possible DNE Example:

Part II: Speed to Infinity

• If two functions both go to infinity, it is possible that one does so faster than the other.

Know how to tell which is faster, or if they are the same speed. Also know the notation:

– If f is faster than g the notation is f � g (or g � f).

– If f is slower than g the notation is f � g (or g � f).

– If they are the same speed the notation is f ∼ g.

1Or “approaches”


• Be able to quickly decide the relative speeds of any of the following simple functions: logarithms,roots, powers, and exponentials. Here are the essential facts.

– Roots are just fractional powers. Lower powers are slower than higher powers.

x1/2 � x3 � x5

– Any logarithm is slower than any power, including fractional powers.

log10 x�√x

– Any power is slower than any exponential that has base b > 1.

x100 � 2x

– Lower base exponentials are slower than higher base exponentials.

2x � ex � 10x

– All logs are the same speed.log10 x ∼ log2 x

• Be able to rank any collection of simple functions by speed. You should not have to do any algebraor calculus. Just write down the answer.

Example: Rank by speed. x2, 2x, ln x,√x.

Solution: lnx�√x� x2 � 2x

• There is also an algebraic technique. To compare f and g, assuming both diverge to infinity, computethe limit of f/g.

– If limx→∞

(f

g

)= 0 then f � g.

– If limx→∞

(f

g

)=∞ then f � g.

– If limx→∞

(f

g

)= any non-zero number, then f ∼ g.

• Be able to compute limx→∞

(f

g

). L’Hopital’s Rule will work.

• You should also know how to find the limit using dominant terms of f and g. If f and g havemore than one term you can ignore all but the fastest term of each.

Example: limx→∞

4x2 − 5x+ 1

x− x2 + 2= lim

x→∞

4x2

−x2= −4 (Because the x’s cancel.)

Example: limx→∞

x5 − 5x

2x − 3x= lim

x→∞

−5x

−3x=∞ (Because 5x is faster than 3x.)


Part III: Tail Thickness

• The most important examples are functions that have horizontal asymptotes at zero, as in the figurebelow.

The far right of the graph is called the tail.2

• Be able to tell if two tails have different thickness, and if so, which is thicker.

• Tails are most commonly caused by fractions,1

f(x), where f(x) is any function that goes to infinity.

For these functions the tail thickness rule is:

If f is faster than g, then1

fhas a thinner tail than

1

g.

As with fractions, bigger denominators mean smaller fractions: 4 is bigger than 2, so 14< 1

2.

• Even the notation is the same. If f � g, then the notation for tail thickness is

1

f� 1

g

• Be able to rank any collection of simple functions by tail thickness. You should not have to do anyalgebra or calculus. Just write down the answer, based on the speed of the denominators.

Example: Which has a thicker tail,1

2xor

1

x3?

Solution:1

2x� 1

x3(Because 2x is faster than x3.)

• Be able to replace a complicated function with a simple function that has the same tail thickness.You typically use these two facts:3

– Constant multiples do not affect speed to infinity or tail thickness.

– You can rely on dominant terms in a quotient.

Example: Find the simplest function whose tail is the same thickness as the tail of

4x3 + 2x+ 1

3x5 + 4x3 + 7x∼ 4x3

3x5∼ x3

x5∼ 1

x2

2Sometimes it is referred to as an asymptotic tail.3The mathematically exact method is limit comparison, as with speed to infinity. These two facts cover most limit

situations. However, more complicated examples may need L’Hopital’s rule or even more advanced limit techniques.

Limits Worksheet 1 60

WebAssign #8: All of the following functions go to infinity as x→∞, limx→∞

f(x) =∞:

ex, 3x, 2x

Order these functions by the speed at which they go to infinity using the blanks below.

� �


f(x) =∞:

x2√x, x3, 3

√x


� � �


f(x) =∞:

x5, x1/5, ex, 5x, x2, x1/2, ln(x), 2x


1

�2

�3

�4

�5

�6

�7

�8


When computing more complicated limits of fractions as x → ∞ we can simplify the computation byworking with the dominant terms as follows:

• Step 1: For large values of x a fraction is approximately the ratio of its dominant terms (the onesthat go to infinity the fastest), so rewrite the limit problem as a limit of only the dominant terms.

• Step 2: Simplify the result.

• Step 3: Find the limit of the simplified problem.

Compute the following limits using the above steps by filling in the boxes below:

Note: The mathematical notation below says that the limit of the two fractions are equal (not that thefractions are equal). As such the limit operator remains until the limit is finally computed in step 3.

WebAssign #12:

limx→∞

2x2 − 5x3 + 6

2x3 + x+ 8= lim

x→∞= lim

x→∞=

WebAssign #13:

limx→∞

5x3/2 − 4x5/2 + 12

3x2 − 6x+ 5= lim

x→∞= lim

x→∞=

WebAssign #14:

limx→∞

2x2 − 6x+ 5√4− 6x2 + 9x4

= limx→∞

= limx→∞

=

WebAssign #15:

limx→∞

3− 6 · (2x)5 + x+ 5 · (3x)

= limx→∞

= limx→∞

=


Compute the following limits. Follow the steps on the previous worksheet, using correct mathematicalnotation. Get feedback on your notation from your instructor, learning assistant, or a tutor.

WebAssign #16:

limx→∞

√25x4 + 15x2 + 6

7x2 + 9x+ 10=

WebAssign #17:

limx→∞

3x3 − 5x5 + 1

5x5 − 7 · (2x)=

WebAssign #18:

limx→∞

3x5 + 9x4 − 31x

2x4 − 31x2 + 12=

WebAssign #19:

limx→∞

√9x4 − 3x2 + 5x+ 1

5x2 − 6x+ 1=

WebAssign #20:

limx→∞

3x2 + 5x

4x2 + 6−x + 3x=

Improper Integrals I Notes and Learning Goals 63

An improper integral is an integral for which the area under the curve could possibly be infinite.Although there are many ways for this to happen, we will concentrate on situations like the one below,where the shaded area extends forever to the right.

1. Know how to compute the exact value of

∫ ∞a

f(x) dx. The process is:

• Convert to a proper integral by replacing ∞ with an unknown constant.

Pick your favorite letter. These notes and WebAssign will use R.

• Compute the proper integral. The answer will be a formula in terms of R.

• Find the limit of your formula as R→∞. Use techniques from Limits at Infinity.

Example: Compute

∫ ∞0

e−x/3 dx

Solution:

• Convert to a proper integral.

∫ ∞0

e−x/3 dx −→∫ R

0

e−x/3 dx

• Compute the new integral.∫ R

0

e−x/3 dx =[− 3e−x/3

]R0

(elementary antiderivative)

= −3e−R/3 + 3 (plug in R and 0)

• Write the limit problem and compute the limit.

limR→∞

(−3e−R/3 + 3

)= 0 + 3 (Limits at Infinity)

= 3 (final answer)

2. Know the new vocabulary for improper integrals:

• If the area is finite, the improper integral converges.It’s also appropriate to say the integral is convergent.

• If the area is infinite, the improper integral diverges.It’s also appropriate to say the integral is divergent.

3. Show all work! For full credit on quizzes and exams you must show all work to compute animproper integral as a limit of a proper integral using proper notation.

• Rewrite the improper integral as a proper integral.

• Show all work to compute the proper integral.

• Clearly state and compute the limit of the above result.

Improper Integrals I Worksheet 1 64

WebAssign #1: Consider the improper integral:

∫ ∞2

3

x2dx

Show the following steps to compute this improper integral:

1. Use the Fundamental Theorem of Calculus to compute the related proper integral.

∫ R

2

3

x2dx =

2. Take the limit as R→∞ of the proper integral.

3. Lastly, state if this improper integral is convergent (finite) or divergent(not finite)

WebAssign #2: Consider the proper integral:

∫ ∞4

3√xdx

Show the following steps to compute the improper integral above.

• Use the F.T.C. to compute a related proper integral with upper bound R.

• Take the limit as R→∞ of the proper integral.

• State if this improper integral is convergent (finite) or divergent (not finite)?


WebAssign #3: Consider the improper integral:

∫ ∞0

20e−2xdx

Show all work to compute this improper integral using the same instructions as previous worksheet.

WebAssign #4: Find the area under the curve y = 5/√x on

the infinite interval 1 ≤ x < ∞ as shown at right. This area isrepresented by the improper integral:∫ ∞

1

5√xdx

Show all work to compute this improper integral using the sameinstructions as the previous worksheet.


WebAssign #5: Find the area under the curve y = 7e−0.2x onthe infinite interval 0 ≤ x < ∞ as shown at right. This area isrepresented by the improper integral:∫ ∞

0

7e−0.2xdx

Show all work to compute this area.

WebAssign #6: Find the area under the curve y = x2 on theinfinite interval 0 ≤ x < ∞ as shown at right. This area isrepresented by the improper integral:∫ ∞

0

x2dx

First, make a prediction: Is this improper integral convergent(finite area) or divergent (infinite area)? Then show all work tocompute the area.

Reflection: Was your prediction correct? Could you predict the results for functions with similar graphs?


WebAssign #7: Find the area under the curve y = 4x3/2

on theinfinite interval 1 ≤ x < ∞ as shown at right. This area isrepresented by the improper integral:∫ ∞

1

4

x3/2dx

First, make a predicition: Is this improper integral convergentor divergent. Then show all work to compute the area.

WebAssign #8: Find the area under the curve y = 8x

on theinfinite interval 2 ≤ x < ∞ as shown at right. This area isrepresented by the improper integral:∫ ∞

2

8

xdx


Reflection: Were your predictions correct? Could you predict the results for functions with similar graphs?


WebAssign #9: Find the area under the curve y = 3− 12(x+2)2

onthe infinite interval 0 ≤ x < ∞ as shown at right. This area isrepresented by the improper integral:∫ ∞

0

(3− 12

(x+ 2)2

)dx


Reflection: Was your prediction correct? Could you predict the results for functions with similar graphs?

Improper Integrals II Notes and Learning Goals 69

The previous lesson on improper integrals introduced how to compute an improper integral. The lessonalso introduced the vocabulary convergent (finite area) or divergent (infinite area).Today’s lesson will cover how to test if an improper integral is convergent or divergent by looking at theintegrand. The three tests covered are:

1. Checking for a tail4 is a quick test to see if it is at all possible for

∫ ∞a

f(x) dx to converge.

A function has a tail if limx→∞

f(x) = 0. Check for a tail as follows:

• Compute limx→∞

f(x).

• If you don’t get 0, the function does not have a tail and there is no chance of convergence.

• If you do get 0, the function has a tail and the integral might converge. Or it might not.

NOTE: This is usually obvious if you draw the graph of f(x) and look at the area.

2. Be able to decide if an improper integral converges or diverges by comparing it to one you alreadyknow5. There are four cases. In all cases f is the function that you already know about; g is the oneyou hope to learn about.

• g(x)� f(x) and

∫ ∞a

f(x) dx diverges.

∫ ∞a

g(x) dx must also diverge.

g has a thicker tail and f creates an infinite area, so g’s tail area is infinite.


∫ ∞a

f(x) dx converges.

∫ ∞a

g(x) dx must also converge.

g has a thinner tail and f creates a finite area, so g’s tail area is finite.


∫ ∞a

f(x) dx diverges. Comparison is useless.

g has a thinner tail and f creates an infinite area, so g’s tail area could be finite or infinite.


∫ ∞a

f(x) dx converges. Comparison is useless.

g has a thicker tail and f creates a finite area, so g’s tail area could be finite or infinite.

3. Be able to judge quickly, for simple functions, if

∫ ∞a

f(x) dx converges. Here’s how:

• Know that convergence or divergence depends on tail thickness. Thin tails have less area. Thicktails have more – sometimes an infinite amount.

• Know the tail thickness hierarchy, based on Limits at Infinity.

· · · � 1

ex� 1

2x� · · · � 1

x3� 1

x2� 1

x� 1

x1/2� 1

x1/3� · · · � 1

ln(x)� · · ·

• Be able to correctly position any other simple function in this hierarchy.

Example: Where would you put1

x1.5?

Answer:1

x2� 1

x1.5� 1

x4This is known as the Divergence Test.5This is known as the Comparision Test.

Improper Integrals II Notes and Learning Goals 70

• Use the hierarchy along with at least one integral that you know.

Example: Determine if the following improper integral converges or diverges:∫ ∞1

1

2xdx

In the last lesson we showed that the following converges:∫ ∞1

1

x2dx

And we know that1

2x� 1

x2

Since the bigger area converges, the small area also converges.

Another Example: Determine if the following improper integral converges or diverges:∫ ∞1

1

ln(x)dx

In the last lesson we showed that the following diverges:∫ ∞1

1

xdx

And we know that1

x� 1

ln(x)

Since the smaller area diverges, the bigger area also diverges.

• You will explore this in more detail in the WebAssign exercises. But by the time you take Exam1 you should be able to immediately and correctly decide convergent or divergent for an integralinvolving any simple function.

4. Be able to determine if a more complicated improper integral converges or diverges. Do this bycomparing it to a simple function with the same tail thickness.6. Apply the method of dominantterms from Limits at Infinity.

Example:

∫ ∞2

5x2 + 2x+ 5

2x4 + x2 + 5dx. Convergent or Divergent?

Solution:

5x2 + 2x+ 5

2x4 + x2 + 5∼ 5x2

2x4(locate dominant terms)

∼ x2

x4(ignore constant multiples)

=1

x2(simplify)∫ ∞

1

1

x2dx is known convergent, so

∫ ∞2

5x2 + 2x+ 5

2x4 + x2 + 5dx is CONVERGENT.

6If two integrands have the same tail thickness then both integrals converge or both integrals diverge. This is called theLimit Comparison Theorem

Improper Integrals II Worksheet 1 71


f(x) =∞:

√x, x3/2, x2/3, ex, x2, x, ln(x), x5


1

�2

�3

�4

�5

�6

�7

�8

WebAssign #2: All of the following functions have tails, limx→∞

f(x) = 0:

e−x, x−3/2, x−7,1

x,

1

ln(x), x−1/3,

1

x2,

1√x

Order these functions by their tail thickness using the blanks below.

1

�2

�3

�4

�5

�6

�7

�8

Discussion point: What are the differences and similarities between ordering functions based on their speedto infinity vs tail thickness?


WebAssign #3: Compute the following limits

• limx→∞

20e−3x =

• limx→∞

1− 3x2

x2 + 1=

• limx→∞

15e2x =

• limx→∞

x−3/2 =

• limx→∞

x3/2 =

• limx→∞

x+ 3√x3 + 2

=

• limx→∞

1

x−2/3=

• limx→∞

x

x2 + 1=

• limx→∞

x2 + 1

x=

A function has a tail if limx→∞

f(x) = 0. Indicate which the above functions have tails.


WebAssign #14: We were not able to determine if the following improper integral was convergent ordivergent by comparing it to problems we have done so far.∫ ∞

1

1

x1.1dx

Determine if this improper integral is convergent or divergent by computing it:

• Compute a related proper integral with upper bound R.

• Take the limit of the proper integral as the upper bound goes to infinity, R→∞.

Be sure to show all work and get feedback from your instructor or LA on your work.

Is this improper integral convergent or divergent?



f(x) = 0:

1

x3,

1

x,

1

x1/3,

1

x5,

1

x1.1,

1

x2,

1

x0.1,

1

x1/2,

1

x1/4


1

�2

�3

�4

�5

�6

�7

�8

�9

WebAssign #16: Use the above ordering, along with results from previous problems and comparison, todetermine if the associated improper integrals converge or diverge. Enter answers in WebAssign.


f(x) = 0:

1

2x,

1

x1/5,

1

ln(x),

1

x1.5,

1√x,

1

ex


1

�2

�3

�4

�5

�6

WebAssign #18: Use the above ordering, along with results from previous problems and comparison, todetermine if the associated improper integrals converge or diverge. Enter answers in WebAssign.

Discussion point: How can ordering integrands by their tail thickness help determine if an associatedimproper integral converges or diverges?

Sequences Notes and Learning Goals 75

Almost everything in this lesson has a nearly identical counterpart in Lesson: Limits at Infinity.

Sequences

• A sequence is a discrete function that only has whole number inputs: 0, 1, 2, 3, etc1.

• The best way to understand this is to notice that the graph of a sequence is a pattern of individualdots, not a continuous curve. See examples below.

• The notation for sequences is slightly different than what you usually see for functions.

Function Example: f(x) = x2. Sequence Example: an = n2.

• You can think of a sequence as an infinite list of numbers. For example, if an = n2, the list is

0, 1, 4, 9, 16, 25, . . .

• Vocabulary: We say that an is the n-th term of the sequence.

Limits of Sequences Know the notation, vocabulary, and possible behaviors for the limit of a sequence.This is almost Limits at Infinity. The only difference is that graphs of sequences are discrete.

1. Approaches a Number.

limn→∞

an = L

Examples:

2. Becomes Infinite.

limn→∞

an = ±∞

Examples:

3. Everything Else.

limn→∞

an DNE

The limit does not exist.

Example:

1Sometimes a sequence will start at a value other than 0.

Sequences Notes and Learning Goals 76

Part II: Speed to Infinity

• Everything in Limits at Infinity, Part II is true for sequences.

• There is one new type of sequence: factorial. The notation and formula for n factorial is

n! = n · (n− 1) · (n− 2) · · · 2 · 1

Examples:

6! = 6 · 5 · 4 · 3 · 2 · 1 = 720

1! = 1

0! = 1 (Why 1? Same reason that anything0 = 1.)

• The speed hierarchy is the same as for functions, but with factorials added. Factorials get to infinityeven faster than exponentials.

log � roots � powers � exponentials � factorial

Example:log2 n � n1/3 � n5 � 4n � n!

Tail Thickness

• Everything in Limits at Infinty, Part III is true for sequences.

• Factorial denominators make the thinnest tails.

Example: Rank the following by tail thickness:1

n,

1

n!,

1

2n.

Solution:1

n!� 1

2n� 1

n

Comparing Speed and Thickness

• Be able to use your knowledge of speed and tail thickness to compare two sequences.

• Limits of ratios are exactly as they were in Limits at Infinity:

– limn→∞

(anbn

)= 0 if and only if an � bn (bn is faster, or thicker.)

– limn→∞

(anbn

)=∞ if and only if an � bn. (an is faster, or thicker.)

– limn→∞

(anbn

)= a non-zero number. In this case both have the same speed or thickness.

Sequences Worksheet 1 77

WebAssign #1: Consider the sequence an =n(n+ 1)

2Fill out the following table then graph sequence on grid at right.Sequence graphs must

• Label the horizontal axis n with proper scale.

• Label the vertical axis an with proper scale.

• Have a dot at each point, (n, an) (there are no lines/curvesconnecting the dots).

n 1 2 3 4an

WebAssign #2: Consider the sequence an =n

n+ 1Fill out the following table then graph sequence on grid at right.

n 0 1 2 3an


WebAssign #5: Determine which sequence, an = n3 or bn = n2, goes to infinity faster as n→∞ as follows:

• Compute the limit of their ratio. Write down the computation using correct notation (First writedown the limit, second simplify the limit, third compute the simplified limit).

limn→∞

anbn

= limn→∞

= limn→∞

=

• Use the result to state the conclusion using correct notation with the symbols �, �, or ∼.

WebAssign #6: Determine which sequence, an = 5n3 or bn = 7n3, goes to infinity faster by

• Find the limit of their ratio following the steps above:


WebAssign #7: Determine which sequence, an = n! or bn = 2n, goes to infinity faster by



WebAssign #8: Determine which sequence, an = 2n or bn = n82n, goes to infinity faster by




WebAssign #9: Determine which sequence, an = 2n or bn = n8 + 2n, goes to infinity faster by

• Find the limit of their ratio following the steps on previous worksheet:


WebAssign #10: Determine which sequence, an = n ln(n) or bn = n, goes to infinity faster by



WebAssign #11: Determine which sequence, an = n− ln(n) or bn = n, goes to infinity faster by



WebAssign #13: All of the following sequences go to infinity as n→∞, limn→∞

an =∞:

n!, n, n ln(n), n2, 2n, 3n, ln(n), n3, n1/3, n1/2

Order these sequences by the speed at which they go to infinity using the blanks below. Hint: Use theresults from previous problems. If unsure you can compare two sequences by taking the limit of their ratio.

1

�2

�3

�4

�5

�6

�7

�8

�9

�10


WebAssign #14: Determine which sequence, an = 1/n1/3 or bn = 1/n1/2, has a thicker tail as follows:

• Compute the limit of their ratio. Write down the computation using correct notation (First writedown the limit, second simplify the limit, third compute the simplified limit).

limn→∞

anbn

=


WebAssign #15: Determine which sequence, an = 1/n! or bn = 1/en, has a thicker tail by



WebAssign #16: Determine which sequence, an = 1/ log2(n) or bn = 1/ log3(n), has a thicker tail by

• Find the limit of their ratio following the steps above. Hint logb(n) =ln(n)

ln(b).


WebAssign #17: All of the following sequences have tails, limn→∞

an = 0:

2−n,1

n2,

1

n,

1

n!,

(1

3

)n,

1

ln(n),

1

n1/2,

1

n ln(n)

Order these sequences by their tail thickness using the blanks below.

1

�2

�3

�4

�5

�6

�7

�8

Sigma Sums Notes and Learning Goals 81

Sigma Notation

1. Learn sigma notation for a finite sum. For example:

5∑n=2

1

n2

• The symbol∑

is the capital Greek letter sigma.

•∑

stands for “sum.” It means “add stuff up.”

• The “stuff” is generated by the formula1

n2.

• Start generating stuff with n = 2. Stop when you get to n = 5.

2. Know how to interpret and expand sigma notation.

Example: Expand the sum5∑

n=2

1

n2

Solution: Start at 2. Stop at 5. Add stuff up.

1

22+

1

32+

1

42+

1

52

3. Be able to do this if the formula involves letters as well as numbers.

Example: Expand the sum4∑

n=1

nxn

Solution: Start at 1. Stop at 4. Include x’s.

x+ 2x2 + 3x3 + 4x4

4. Be able to reverse this process. That is, given a sum in expanded form, be able to write it usingsigma notation.

Areas and Sums

1. Suppose you have a finite sum written in sigma notation. For example:

4∑n=1

1

n2/3

Know these facts:

• The formula inside the sigma notation is a Sequence, called the sequence of terms.

an =1

n2/3


• The sequence has a graph. (The dots in the figures below.)

• The sum equals the area under the graph of the sequence (across an appropriate domain).

4∑n=1

1

n2/3= Area under the graph of

1

n2/3

2. Be able to recognize correct graphical representations of a finite sum as an area.

Example: For4∑

n=1

1

n2/3, both of the following pictures are correct representations.

3. Know that there can be more than one correct representation. In the two examples above, one hasrectangles on the right side of each dot while the other has them on the left. Both are correct.2

4. Given a finite sum in sigma notation, be able to create your own correct graphical representations.

5. Given a graphical representation, be able to write the sum, either in expanded or in sigma notation.

Sums and Integrals The rectangles-under-a-graph representation of a finite sum should remind you ofthe area-under-a-graph representation of an integral.

1. Know these two important facts:

• Fact: A sequence is the discrete analog of a continuous function.

The graph of a sequence is discrete dots.

The graph of a continuous function is a curve.

• Fact: A finite sum is the discrete analog of an integral.

A sum is a finite set of rectangles, one per dot, each exactly 1 unit wide.

An integral is an infinite set of rectangles, all with 0 width, forming an area under the curve.

2. Integral and sigma notation are intentionally very similar. Know all the notation and vocabulary inthe table on the next page.

2Other rectangle positions are possible. As long as there are four rectangles, each with width 1, and just high enough totouch the correct four dots, the picture is a correct representation.


3. Given any continuous notation or vocabulary, be able to name and write the discrete analog.

4. Given any discrete notation or vocabulary , be able to name and write the continuous analog.

Continuous Discrete

Vocabulary function sequence

Formula f(x) an

Graph curve dots

“Add up” symbol

∫ ∑Start point x = a n = j

End point x = b n = k

Rectangle height f(x) an

Rectangle width dx 1

Total

∫ b

a

f(x) dxk∑n=j

an

Area

Sigma Sums Worksheet 1 84

WebAssign #1: Consider the following finite sum written in sigma notation:5∑

n=1

2n

2n+ 1:

• Identify the formula for the sequence of terms, an, for this sum.

• Use the formula for the sequence of terms and the bounds from n = 1 to n = 5 to expand this sum.This means write this as a sum of terms for n = 1, 2, 3, 4, 5. The sum must be written using exactnumbers in the correct order.

5∑n=1

2n

2n+ 1=

WebAssign #2: Consider the following finite sum:

103 + 104 + 105 + 106

Write this sum using sigma notation. Your notation must include a sigma sum. The lower bound of thesigma sum should be written as n = 3 and also include a correct upper bound.

WebAssign #3: Consider the following finite sum:

2

3+

4

32+

6

33+

8

34+

10

35+

12

36+

14

37

Write this sum using sigma notation. Your notation must include a sigma sum. The lower bound of thesigma sum should be written as n = 1 and also include a correct upper bound.


WebAssign #4: Consider the sigma sum4∑

n=1

1

2n−1.

Graph this sigma sum on grid at right:

• Label the horizontal axis n, with a scale of 1.

• Label the vertical axis an, with a scale of 0.25.

• Graph the sequence of terms, an =1

2n−1(recall the graph

of an is dots), from n = 1 to n = 4.

• Draw a rectangle to the right of each dot of width 1, usingthe dot as the height above the axis.


n=2

2n

n!.

Graph this sigma sum on grid at right:

• Label the horizontal axis n, with a scale of 1.

• Label the vertical axis an, with a scale of 0.5.

• Graph the sequence of terms, an =2n

n!(recall the graph of

an is dots), from n = 2 to n = 5.

• Draw a rectangle to the left of each dot of width 1, usingthe dot as the height above the axis.


n=1

√n.

Graph this sigma sum on grid:

• Identify the sequence of terms:

an =

• Graph must be properly labeled.

• Draw and shade rectangles that correctly represent thefinite sum as an area.


WebAssign #7: Consider the sequence an =24

n+ 1

1. Graph an from n = 0 to n = 8. Use the grid below. Label your axes properly.

2. In your above graph, draw and shade a set of rectangles whose area matches the sum6∑

n=3

an.

NOTE! Get feedback on your graph from your instructor or LA.


Below you will find six different attempts at a graphical representation of5∑

n=1

1

nas an area.

Decide if each representation is correct or incorrect. If incorrect, explain why in one sentence.

1. CORRECT INCORRECT (Why?)






NOTE! Get feedback on your written explanations from your instructor or LA.

Series Notes and Learning Goals 88

1. Know what a series is.

• Recall: a sequence is the discrete analog of a function.

• Recall: a sigma sum is the discrete analog of a definite integral.

• A series is the discrete analog of an improper integral.

A series is the result of extending a finite sum out to infinity.

2. Know the notation and vocabulary for series.

• A series is also sometimes called an infinite series or an infinite sum.

• There are two notational styles, expanded form and sigma notation.

Sigma Notation Example:∞∑n=1

1

n1/2

Use ∞ as the upper bound.

Expanded Example: 1 +1√2

+1√3

+1√4

+ · · ·

Use “· · · ” to suggest continuation to infinity.

• The formula in the sigma notation generates a sequence called the sequence of terms.

3. A series can be represented graphically, like a finite sum, but the rectangles continue to infinity.

4. We say a series is convergent if the infinite summation adds up to a finite total.3 Equivalently, ifthe area of all the rectangles is a finite amount. Otherwise it is divergent.

3Technically, it means that longer and longer finite sums approximate the total to within any desired tolerance.


5. There is a very strong relationship between infinite series and improper integrals. Suppose that f(x)and an have the same formula. For example:

f(x) =1

x1/2and an =

1

n1/2

• Critical Fact: ∫ ∞c

f(x) dx converges if and only if∞∑n=j

an converges.

• That is, either both areas are finite, or both areas are infinite.

• Note that this statement is only a statement about the tails of the integral and the series. Wedon’t care what the starting point is, or whether or not they start together.

• The relationship is due to the close connection between the graphical representations of seriesand integrals, as shown here:

• These pictures have a lot of information. Roughly, the smooth areas are integrals and therectangles are the series. What this all means is explored in your homework.

6. Know how convergence of a series depends on its tail thickness.

• A series can’t possibly converge unless it has a tail. That is∑

an is divergent unless limn→∞

an = 0.

• Just like improper integrals, a series converges if its tail is “thin enough”. Series with fat tailsdiverge, series with skinny tails converge, and the dividing line4 is about 1/n.

• Be able to quickly sort and rank simple series by tail thickness, as you did in previous lessons.

• Given any simple series, be able position it correctly within the hierarchy below, and also knowif it is convergent or divergent.

4There is no exact boundary. For any divergent series, there is a series with a thinner tail that is also divergent. Also forany convergent series, there is a series with a thicker tail that is also convergent.


∑ 1

lnn�∑ 1

n1/3�∑ 1

n1/2�∑ 1

n︸︷︷︸Divergent

�∑ 1

n2�∑ 1

n3�∑ 1

2n�∑ 1

3n�∑ 1

n!︸︷︷︸Convergent

7. Be able to reduce a complicated series to a simple series with the same tail thickness. This may beasked for directly, or it may be necessary to answer a question about convergence.

Example: Does this series converge or diverge?∞∑1

5n2 + 6n√4n5 + 12n2

Solution:

5n2 + 6n√4n5 + 12n2

∼ 5n2

2n5/2(locate dominant terms)

∼ n2

n5/2(ignore constant multiples)

=1

n1/2(simplify)

∞∑1

5n2 + 6n√4n5 + 12n2

is divergent. (final answer)

8. Series that sit near the boundary of convergence/divergence can be very subtle.

Example: Does this series converge or diverge?∞∑2

1

n lnn

Non-Solution: It is easy to position this series in the standard hierarchy of tail thickness:∑ 1

n�∑ 1

n lnn�∑ 1

n2

But this does not help. It only tells us that

DIVERGENT >∑ 1

n lnn> CONVERGENT

This just means our series is somewhere between infinite and finite. Most things are.

Solution:

• Replace with an improper integral:

∫ ∞2

1

x lnxdx

• Convert to a proper integral.

∫ ∞2

1

x lnxdx −→

∫ R

2

1

x lnxdx

• Compute the proper integral. Maybe u = lnx works?

• Take the limit as R goes to infinity.

• Convergent or divergent? You will find out in your homework.

Series Worksheet 1 91

WebAssign #5: Consider the harmonic series:∞∑n=1

1

n

This series converges if and only if the related improper integral,

∫ ∞1

1

xdx, converges.

• Compute this improper integral. Show all work using correct notation (compute a proper integralwith upper bound R, then find the limit as R→∞ of the proper integral).

• Do both the improper integral and series converge or diverge?

WebAssign #6: Consider the series:∞∑n=1

2

n3/2

This series converges if and only if the related improper integral,

∫ ∞1

2

x3/2dx, converges.

• Compute this improper integral. Show all work using correct notation (compute a proper integralwith upper bound R, then find the limit as R→∞ of the proper integral).

• Do both the improper integral and series converge or diverge?


WebAssign #10: Consider the following series:

∞∑n=1

1

n5,

∞∑n=1

n−2/3,

∞∑n=1

n3/2,

∞∑n=1

1

n0.99,

∞∑n=1

n−1.01,

∞∑n=1

1√n3

Order these series using the series hierarchy (by tail thickness) using the blanks below.Note: The answers are series, so you must use the sigma sum symbol in your ordering.

Hint: Rewrite each series as a p-series in the form∞∑n=1

1

np.

� � � � �

Clearly indicate which of the above series are convergent and which are divergent.

WebAssign #11: A p-series is any series of the form:

∞∑n=1

1

np

• State the values of p for which these series are divergent:

• State the values of p for which these series are convergent:


WebAssign #12: This worksheet explores how to determine if the series∞∑n=2

1

n lnnconverges or diverges.

1. First let’s try to compare this series to a p-series. Start by ranking the series below based on tailthickness. Use � or � notation. Get feedback to see if you are using it correctly.

∞∑n=2

an =∞∑n=2

1

n4

∞∑n=2

bn =∞∑n=2

1

n1/2

∞∑n=2

cn =∞∑n=2

1

n lnn

2. Is∑

an convergent or divergent? Circle your answer. This should be quick.

3. What do Problems 1 and 2 tell you about convergence or divergence of∑

cn? Why?

4. Is∑

bn convergent or divergent? Circle your answer. This should be quick.

5. What do Problems 1 and 4 tell you about convergence or divergence of∑

cn? Why?

Worksheet continued on next page...


6. Since it is not possible to compare this series to any p-series (why?), lets compare it to an improperintegral. Compute the improper integral below. Follow the hint and steps on the last page of theNotes and Learning Goals. ∫ ∞

2

1

x lnxdx

Show all work. Instructions for work are posted here. Get feedback on your work. (Something likethis could appear on Exam 1.)

7. Conclusion: Is∞∑n=2

1

n lnnconvergent or divergent?


WebAssign #15: Consider the following series:

∞∑n=0

1

2n,

∞∑n=0

(2

3

)n,

∞∑n=0

(6

5

)n,

∞∑n=0

(6

5

)−n,

∞∑n=0

(2

3

)−n,

∞∑n=0

1n

Order these series using the series hierarchy (by tail thickness) using the blanks below.Note: The answers are series, so you must use the sigma sum symbol in your ordering.

Hint: Rewrite each series as a geometric series in the form∞∑n=0

rn.

� � � � �

Clearly indicate which of the above series are convergent and which are divergent.

WebAssign #16: A geometric series is any series of the form:

∞∑n=0

rn

Answer the following assuming that r is a positive real number.

• State the values of r for which these series are divergent:

• State the values of r for which series are convergent:

Convergence Notes and Learning Goals 96

1. A series is convergent if it has a thin enough tail5. One method to determine if a series diverges orconverges is to compare the series to a known series.

• A p-series is any series of the form∑

1/np. These series diverge when p ≤ 1 and converge6

when p > 1. Ordering these series by their tail thickness gives

· · · �∑ 1

n100�∑ 1

n5�∑ 1

n1.001︸︷︷︸Convergent

� · · · �∑ 1

n�∑ 1

n1/2� · · ·︸︷︷︸

Divergent

• A geometric series is any series of the form∑

rn. If |r| ≥ 1, these series do not have a tail and

must diverge. If |r| < 1 these series are thinner than p-series and must converge.

· · · �∑(

1

5

)n�∑(

1

2

)n�∑ 1

n2︸︷︷︸Convergent

� · · · �∑ 1

n�∑

1n �∑

2n � · · ·︸︷︷︸Divergent

2. To determine if a series is convergent or divergent your primary technique should be to try to placeyour series in the appropriate location in the tail ordering:

· · · �∑ 1

n!�∑ 1

en�∑ 1


� · · · �∑ 1

n�∑ 1

lnn� · · · �

∑(an with no tail)︸︷︷︸

Divergent

• If a series has a thinner tail than∑

1/np for some p > 1, then it must converge.

• If a series has a thicker tail than∑

1/n, then it must diverge.

3. The series∑

1/n is not an absolute boundary. There are series in between that can be convergent,

and others that are divergent.

· · · �∑ 1


� · · · �︸︷︷︸Unknown

∑ 1

n� · · ·︸︷︷︸

Divergent

Two examples7 of series in the unknown region are:

•∑ 1

n ln(n)can be shown to be divergent.

•∑ 1

n(

ln(n))2 can be shown to be convergent.

5A series has a tail if its terms approach zero, limn→∞ an = 0, and the resulting infinite series,∑an, converges if the

terms approach zero fast enough.6 Comparing a p-series to an improper integral can determine when these series converge, and when they diverge.7Both these series have thinner tails than

∑1/n, but thicker tails than

∑1/np for all p > 1.

Convergence Notes and Learning Goals 97

4. Using similar methods from limits at infinity and sequences, determine if a series converges or divergesby comparing the series to one with about the same tail thickness. For example:

∑ 5n2 − 2n+ 1

n2 + 3n4∼∑ 5n2

3n4∼∑ 1

n2

By comparing dominant terms, the series is similar to∑

1/n2, which is convergent. Thus the

original series is also convergent.

5. A variable series is a series which depends on some independent variable, such as x. Some values ofx will make a series converge, while other values8 of x will make the series diverge. The values of xcan be written using interval notation, called the interval of convergence.

Example: Determine what values of x > 0 the following series converges:∑(x

2

)nThis geometric series converges when x/2 < 1 or x < 2.

Thus the interval of convergence is (0, 2).

8x could be any real number, but for this lesson we’ll restrict to positive values: x > 0.

Exam 1 Review Notes and Learning Goals 98

Antiderivatives and Integrals

• An antiderivative of a function is any function whose derivative is the original function. In mathnotation, an antiderivative of f(x) is any function F (x) such that F ′(x) = f(x).

• An antiderivative problem,∫f(x)dx, asks to find all possible antiderivatives. Answers to antideriva-

tive problems will contain an unknown constant, +C.

• An integral problem,∫ baf(x)dx, asks to compute an integral. Integrals can be computed using the

Fundamental Theorem of Calculus:∫ b

a

f(x)dx = F (x)∣∣∣ba

= F (b)− F (a)

where F (x) is any antiderivative of f(x).

• Be able to find antiderivatives and compute integrals, without calculator aid, using the methods:

Elementary antiderivatives, Integration by u-substitution, Integration by parts,Trigonometric integrals, Trigonometric substitution, Partial fractions.

• When given an antiderivative or integral problem, be able to determine an appropriate method tofind the antiderivative, then use the method.

• In all cases, be able to clearly show the required work to obtain the answer.

Limits

• Be able to find limits as x→∞ of elementary functions.

• Be able to order functions by their speed or tail thickness as x→∞.

• Be able to use dominant terms to find limits of more complicated functions.

Improper Integrals

• One type of improper integrals are integrals with an upper bound of ∞.

• An improper integral is the limit of a proper integral.∫ ∞a

f(x)dx = limR→∞

∫ R

a

f(x)dx

– First compute a proper integral,

∫ R

a

f(x)dx.

– Then take the limit as R→∞ of the result.

• An improper integral converges if the limit is finite. Otherwise it diverges.

Sequences

• A sequence, an, is an infinite list of values: a0, a1, a2, a3, a4, · · · .


• The limit of a sequence, limn→∞

an, is the value (if it exists) the sequence approaches as n→∞.

• Limits for sequences are computed in a very similar way to limits of functions.

• Sequences can be ordered by their speed to infinity or tail thickness.

• Dominant terms can be used to compute more complicated limits.

Series

• A finite sum of a sequence written in sigma notation is

n=m∑n=k

an = ak + ak+1 + ak+2 + · · ·+ am

• Be able to write finite sums using sigma notation. Also be able to expand a sum into sigma notation.

• A series is a sum with an infinite number of terms, which is a limit of finite sums:

∞∑n=k

= limN→∞

N∑n=k

an

• The series converges if the limit is finite. Otherwise it diverges.

Convergence

• A series (or improper integral) can only converge if it has a tail (limit of terms is zero). Thus anyseries without a tail must diverge:

If limn→∞

an 6= 0, then∑

an must diverge.

• Not all series with tails converge. A series with a tail only converges if the tail is thin enough:

– If a series converges, than any series with a thinner tail also converges:

If bn � an and∑

an converges, then∑

bn also converrges.

– If a series diverrges, then any series with a thicker tail also diverges:

If an � bn and∑

an diverges, then∑

bn also diverges.

– A p-series (or improper integral) is of the form

∞∑n=1

1

npor

∫ ∞1

1

xpdx

∗ If p ≤ 1 the series (or improper integral) diverges.

∗ If p > 1 the series (or improper integral) converges.

• Comparing series to p series serves as a starting point to determine if a series has a thin enough tail1

to converge or not.

1 There are series with thinner tails than∑

1n that still diverge. In general there is no single series whose tail thickness

is on the boundary between convergent and diverget series.

Antiderivative Review Worksheet 1 100

WebAssign #3: Integration by Substitution work requirements:

• Statement of u. Computation of du.


• Correct antiderivative of the transformed integral.

Write down the integral for problem #3 then show all work. Get feedback on your work.

WebAssign #4: Integration by Parts work requirements:

• Statement of u and dv. Computation of du and v.

• Result from integration by parts formula.

• Antiderivative of the new integral in the by parts formula.



WebAssign #6: Integration by Partial Fractions work requirements:

• Clearly stated partial fractions form using unknown coefficients.

• Work to setup and solve a system of equations for the unknown coefficients.

• A rewritten integral using the partial fractions form.

• Antiderivative of the new integral.


WebAssign #7: Trigonometric Integral work requirements:

• Clearly show how any trigonometric identities are used to rewrite the problem.

• All work for any integration method used (such as substitution).

• Antiderivative from the method used.



WebAssign #8: Write down the integral in the problem. Determine an appropriate method to use to findthe antiderivative. Show the appropriate work. Get feedback on your work.

WebAssign #9: Trigonometric Substitution work requirements:

• Statement of your substitution formula, x = a sin(θ) or x = a tan(θ). Computation of dx.

• Transformed integral written entirely in terms of θ and dθ. A right triangle maybe useful here.

• Correct antiderivative of the transformed integral showing all work for a trigonometric integral.


Taylor Polynomials I Notes and Learning Goals 103

• Know the vocabulary, notation, and standard form for Taylor Polynomials.

Constant T0(x) = a0

Linear T1(x) = a0 + a1(x− c)Quadratic T2(x) = a0 + a1(x− c) + a2(x− c)2

Cubic T3(x) = a0 + a1(x− c) + a2(x− c)2 + a3(x− c)3

4th Degree T4(x) = a0 + a1(x− c) + a2(x− c)2 + a3(x− c)3 + a4(x− c)4...

...

nth Degree Tn(x) = a0 + a1(x− c) + a2(x− c)2 + · · ·+ an(x− c)n

• Other vocabulary:

– In the formulas above, c is called the center point.

– We say a Taylor polynomial is centered at x = c.

– The degree of a Taylor polynomial is the highest power of x that appears in it.

• Taylor Polynomials are used to approximate functions. The approximation is best near the centerpoint. For example, here is a graph of f(x) =

√x, approximated by a quadratic Taylor polynomial

centered at x = 1. Near x = 1, both graphs are very close together.

• The close match of T2(x) and√x, at least near x = 1, is due to three facts:

– The values of both functions match at x = 1. That is, T2(1) = f(1).

– The slopes of both functions match at x = 1. That is, T ′2(1) = f ′(1).

– Their concavity matches at x = 1. That is, T ′′2 (1) = f ′′(1).

• Higher degree Taylor polynomials, and better approximations, are obtained by extending this ruleto higher derivatives.

Taylor Polynomials I Notes and Learning Goals 104

• Know how to use this rule to compute the coefficients of a Taylor polynomial.

Example: Find the quadratic Taylor polynomial for f(x) =√x, centered at x = 1. (The one shown

above.)

Solution:

1. Write the standard form of T2(x) using unknown coefficients:

T2(x) = a0 + a1(x− 1) + a2(x− 1)2

2. Match values at x = 1.

T2(1) = a0 + a1(1− 1) + a2(1− 1)2 (plug in x = 1)

= a0 (simplify)

f(1) =√

1 = 1 (plug in x = 1)

a0 = 1 (set equal)

3. Match slopes at x = 1.

T ′2(x) = a1 + 2a2(x− 1) (symbolic derivative)

T ′2(1) = a1 + 2a2(1− 1) (plug in 1)

= a1 (simplify)

f ′(x) =1

2x−1/2 (symbolic derivative)

f ′(1) =1

2(plug in 1)

a1 =1

2(set equal)

4. Match second derivatives at x = 1.

T ′′2 (x) = 2a2 (second derivative)

T ′′2 (1) = 2a2 (plug in 1)

f ′′(x) = −1

4x−3/2 (second derivative)

f ′′(1) = −1

4(plug in 1)

2a2 = −1

4(set equal)

a2 = −1

8(solve for a2)

5. Write the final answer in standard form.

f(x) ≈ 1 +1

2(x− 1)− 1

8(x− 1)2

• Higher degree examples will be explored in your WebAssign homework. Eventually you might startcomputing coefficients using this formula:

an =f (n)(c)

n!

Taylor Polynomials I Worksheet 1 105

WebAssign #2: The 2nd degree Taylor polynomial centered at x = c written in standard form is:

T2(x) = a0 + a1(x− c) + a2(x− c)2

1. Write the quadratic (2nd degree) Taylor polynomial centered at c = 4 in standard form and take thefirst and second derivative of it.

T2(x) =

T ′2(x) =

T ′′2 (x) =

2. Evaluate the Taylor polynomial and its derivatives at the center point, c = 4.

T2(4) =

T ′2(4) =

T ′′2 (4) =

Note: Pay attention to the notation’s meaning. For example T ′2(x) means the formula for the firstderivative of the 2nd degree Taylor polynomial, while T ′2(4) means the value of the first derviative whenx = 4.


WebAssign #3: Suppose you want to approximate the functionf(x) =

√x with a quadratic Taylor polynomial centered at x =

4, as shown in graph at right. The standard form of this Taylorpolynomial is T2(x) = a0 + a1(x− 4) + a2(x− 4)2.

The function and the Taylor polynomial agree as follows:

• T2(4) = f(4) [Same value at x = 4.]

• T ′2(4) = f ′(4) [Same slope at x = 4.]

• T ′′2 (4) = f ′′(4) [Same concavity at x = 4.]

Use these conditions to find values of the coefficients a0, a1, and a2 as follows:

1. Compute f(4) and set it equal to T2(4). Write down the resulting equation and solve it for a0.

2. Compute f ′(4) and set it equal to T ′2(4). Write down the resulting equation and solve it for a1.

3. Compute f ′′(4) and set it equal to T ′′2 (4). Write down the resulting equation and solve it for a2.

4. Use the values for a0, a1, and a2 to write down the resulting Taylor polynomial in standard form.

5. Use this Taylor polynomial to approximate√

5 ≈ T2(5). (Accurate to 5 decimal digits.)


WebAssign #4: Suppose you want to approximate the functionf(x) =

√x with a cubic Taylor polynomial centered at x = 4,

as shown in graph at right. The standard form of this Taylorpolynomial is T3(x) = a0 + a1(x− 4) + a2(x− 4)2 + a3(x− 4)3.

The function and the Taylor polynomial agree as follows:

• T3(4) = f(4) [Same value at x = 4.]

• T ′3(4) = f ′(4) [Same slope at x = 4.]

• T ′′3 (4) = f ′′(4) [Same concavity at x = 4.]

• T (3)3 (4) = f (3)(4) [Same third derivative at x = 4.]

Use these conditions to find the values of the coefficients a0, a1, a2, and a3 as follows:

1. Following the steps from the previous worksheet, compute f (n)(4) and set it equal to T(n)3 (4), for

n = 0, n = 1, n = 2, and n = 3. In each case write down the resulting equation and solve it for an.

2. Use the results from above to write down T3(x) in standard form.

3. Use this Taylor polynomial to approximate√

5 ≈ T3(5). (Accurate to 5 decimal digits.)

4. Which Taylor polynomial, T2(x) or T3(x), gives a better approximation of√

5?


WebAssign #9: Suppose you want to approximate the functionf(x) = ex with Taylor polynomials centered at x = 0, as shownin graph at right.

Write down the standard form for a cubic Taylor polynomialcentered at x = 0 using unknown coefficients a0, a1, a2, and a3.

T3(x) =

Use the conditions f (n)(0) = T(n)3 (0) to find the values of the coefficients a0, a1, a2, and a3.

1. Fill out the following table to compute the unknown coefficients.

n T (n)(0) f (n)(x) f (n)(0) an

0

1

2

3

2. Use the above table to write down the formulas for the first, second, and third degree Taylor poly-nomials of this function in standard form.

T1(x) =

T2(x) =

T3(x) =

Discussion Point: How are these different degree Taylor polynomials similar? different? How does thedegree of the Taylor polynomial affect the approximation?


WebAssign #10: Find the quadratic Taylor polynomial approx-imation to f(x) = sin(x) centered at x = π/2.

T2(x) =

WebAssign #11: Find the quadratic Taylor polynomial approx-imation to f(x) = sin(x) centered at x = 5π/6.

T2(x) =

WebAssign #12: Find the quadratic Taylor polynomial approx-imation to f(x) = sin(x) centered at x = π.

T2(x) =

Discussion Point: How does the center point affect the Taylor polynomial approximation?


WebAssign #13: Find the cubic Taylor polynomial approxima-tion to f(x) = ln(x) centered at x = 1. Start by writing downthe Taylor polynomial in standard form. Show all work.

T3(x) =

WebAssign #14: Find the quartic (4th degree) Taylor polyno-mial approximation to f(x) = 1/x centered at x = 2. Writedown the Taylor polynomial in standard form. Show all work.

T4(x) =

Taylor Polynomials II Notes and Learning Goals 111

In this lesson you must determine whether a Taylor polynomial provides a good or bad approximation toa given function.

• Good and bad are subjective judgment calls, not mathematically defined terms.

• All questions will be answered by looking at graphs. Although good and bad are not preciselydefined, the guidelines for this assignment are:

– Good means the graph of Tn(x) appears to match the graph of f(x).

– Bad means the graphs do not appear to match.

• WARNING! All WebAssign questions will have severely limited submissions.

DO NOT GUESS. DO NOT WASTE your limited submissions.

Work in teams. It is probably best for one person on a team to have a computer to show thegraphs, and another person to record answers on paper.

• The good/bad question can be asked in three different ways.

1. It will be asked about a single location that is not the center point.

Example: Suppose f(x) =√

25− x2 and T1(x) isthe linear Taylor polynomial centered at x = 2, asshown at right.

Question: At x = 4, is T1(x) a good approximationof f(x) or a bad approximation?

Answer: Bad. Clearly the graphs are well separatedat x = 4.

2. It will be asked about an interval. The Taylor polynomial center point will be either an endpointof the interval or exactly in the middle.

Example: The figure shows f(x) =√

25− x2 andT4(x) centered at x = 2.

Question: On the interval [0, 4], is T4(x) a goodapproximation of f(x) or a bad approximation?

Answer: Good. On this interval the graphs areindistinguishable.

Question: How about on the interval [2, 5]? Goodor bad?

Answer: Bad. The graphs clearly separate beforethey get to x = 5.

Taylor Polynomials II Notes and Learning Goals 112

3. It will be asked by specifying a distance from the center point.

Example: The figure shows f(x) =√

25− x2 andT5(x) centered at x = 2.

Question: At a distance of 2 from the center point,is T5(x) a good approximation of f(x) or a bad ap-proximation?

Answer: Good. You have to check on x = 0 andx = 4. Both look good.

Question: How about at a distance of 3?

Answer: Bad. If either x = −1 or x = 5 is bad,then the answer is bad. Although x = −1 is a closecall, x = 5 is clearly bad.

Question: How far from the center can you go and still have T5(x) be a good approximation?

Answer: Probably the best you can say is

“Somewhere between 2 and 3.”

This is too subtle and open-ended for a WebAssign question. However, it is likely to occuron worksheets, quizzes, and exams, so you should be thinking about it. WebAssign will havemultiple-choice versions of this question.

• By the end of this assignment you should be able to answer these questions:

1. For a target approximation location (or interval, or distance from center), how big does n haveto be to get a good approximation?

2. For a target approximation location (or interval, or distance from center), is there always a bigenough n?

3. For a given n, how wide an interval (or what locations, or what distances) can I use and stillhave a good approximation?

Taylor Series Notes and Learning Goals 113

1. Know what a Power Series is.

• A power series is a series that has coefficients and powers of x, instead of just numbers.

Expanded Example:

1 +x

2+x2

4+x3

8+x4

16+ · · ·

Sigma Notation Example:∞∑n=0

xn

2n

• Think of a power series as an infinitely long polynomial.

• A power series can also be shifted, or centered at x = c.

General Formula:

∞∑n=0

an(x− c)n = a0 + a1(x− c) + a2(x− c)2 + a3(x− c)3 + a4(x− c)4 + · · ·

2. Know what a Taylor series is.

• A Taylor series is an infinitely long Taylor polynomial.

Example: The Taylor series for1

xcentered at x = 2 is

1

x=

1

2− 1

4(x− 2) +

1

8(x− 2)2 − 1

16(x− 2)3 +

1

32(x− 2)4 + · · ·

Using sigma notation:1

x=∞∑n=0

(−1)n

2n+1(x− 2)n

3. Given f(x) and a center point, x = c, be able to compute the first few terms of a Taylor series, onecoefficient at a time. Here are two methods:

• Equate derivatives, as in Taylor I.

– Compute the symbolic derivative of f(x). Plug in x = c.

– Compute the symbolic derivative of the general Taylor polynomial. Plug in x = c.

– Set equal and solve for the coefficient.

• Compute one coefficient at a time, using this formula:

an =f (n)(c)

n!

Taylor Series Notes and Learning Goals 114

4. Be able to use the following four formulas. Usage is explored in your homework.

Note: These formulas will appear on the formula sheet for Exam 2. Be ready to use them.

ex =∞∑n=0

xn

n!= 1 + x+

x2

2+x3

6+x4

24+ · · ·

sin(x) =∞∑n=0

(−1)nx2n+1

(2n+ 1)!= x− x3

6+

x5

120− x7

7!+ · · ·

cos(x) =∞∑n=0

(−1)nx2n

(2n)!= 1− x2

2+x4

24− x6

6!+ · · ·

1

1− x=

∞∑n=0

xn = 1 + x+ x2 + x3 + x4 + · · ·

Example: Find the Taylor series for e−x2/2, centered at zero.

First the Taylor series for eu centered at zero is (from above)

eu =∞∑n=0

un

n!= 1 + u+

u2

2+u3

6+u4

24+ · · ·

Comparing eu to e−x2/2 we see that we can substitute u = −x2/2 to get the new series

e−x2/2 =

∞∑n=0

(− x2/2

)nn!

=∞∑n=0

(−1)nx2n

2n · n!= 1− x2

2+x4

8− x6

48+

x8

384− · · ·

Taylor Series Worksheet 1 115

WebAssign #1: Compute the 6th degree Taylor polynomial for ex centered at x = 0. Follow this process:

1. Write the standard form for T6(x) using the coefficients a0, a1, · · · , a6.

2. Fill out the following table to find the values of an.

Suggestion: Where appropriate, write answers using factorial notation.

Example: 5 · 4 · 3 · 2 · 1 = 5!

n n-th deriv of ex n-th deriv of ex at 0 an

0 ex 1 a0 = 1

1 ex

2

3

4

5

6

T6(x) =

3. If you wrote answers for an using factorial notation, you should see a clear pattern. Guess theseanswers based on that pattern. Use factorial notation.

a7 = a25 =

4. Write T6(x) using sigma notation. You will need to use factorial notation.


WebAssign #2: Compute the 8th degree Taylor polynomial for f(x) = cos(x) centered at x = 0.

1. Write the standard form for T8(x) using the coefficients a0, a1, · · · , a8.

2. Fill out the following table to find the values of an.

n f (n)(x) f (n)(0) an

0

1

2

3

4

5

6

7

8

T8(x) =

3. If you used factorial notation in the table above, you should see a clear pattern. Guess the 10th

derivative of cos(x), evaluated at x = 0.

f (10)(0) =

4. Using factorial notation, find the 10th degree coefficient and the 20th degree coefficient.

a10 = a20 =


WebAssign #5: Compute the 15th degree Taylor polynomial for f(x) = sin(x) centered at x = 0.

1. Write the standard form for T15(x) using sigma sum notation.

2. Find the values of a0, a1, · · · , a15 following the previous worksheets. Start by creating a table and fillout enough rows until you see a pattern, then use the pattern to guess the remaining coefficients.

3. From your work above, what are the 5th, 10th, and 15th degree coefficients?

a5 = a10 = a15 =


WebAssign #6: Compute the first few terms of the Taylor series for f(x) = 1/x centered at x = 1.

1. Compute the 4th degree Taylor polynomial for f(x) = 1/x centered at x = 1. Show all work. Startby writing the standard form of T4(x). Then show how you computed the coefficients.

T4(x) =

2. Look for a pattern in your above work. Use this pattern to find the 7th and 20th degree coefficients.

a7 = a20 =


WebAssign #10: Consider the Taylor series for ex centered at x = 0: ex =∞∑n=0

xn

n!

1. Use this known series to write down the third degree Taylor polynomial for eu centered at u = 0.

eu ≈

2. Use the substitution u = −x2 in the above to find the first few terms of the series for e−x2.

e−x2 ≈

3. Simplify the result and write the answer in standard form.

e−x2 ≈

4. What is the degree of the resulting Taylor polynomial? Is this the same as the degree of the Taylorpolynomial you started with?

WebAssign #11: Find the fourth degree Taylor polynomial for e4x centered at x = 0 as follows.

1. Use a known series to write down the first few terms of the series for eu.

eu ≈

2. Identify a “u =” substitution to use in the known series, then substitute it into the above Taylorpolynomial and simplify to find the fourth degree Taylor polynomial for e4x.

e4x ≈


WebAssign #12: The geometric series (centered at x = 0) is1

1− x=∞∑n=0

xn

Use the geometric series to find the 6th degree Taylor polynomial of1

1 + x2centered at x = 0.

1. Use this known series to write down the first few terms for the series1

1− u:

1

1− u≈

2. Identify a “u =” substitution to use in the known series, then substitute it into the above Taylor

polynomial and simplify to find the sixth degree Taylor polynomial for1

1 + x2.

1

1 + x2≈

WebAssign #13: Find the 5th degree Taylor polynomial, centered at x = 0, for −3 sin(2x). Do this byusing the known series of sin(u).

−3 sin(2x) ≈

WebAssign #14: Find the 4th degree Taylor polynomial, centered at x = 0, for1

2 + x=

1

2· 1

1− (−x/2).

1

2 + x≈


WebAssign #15: Find the 5th degree Taylor polynomial, centered at x = 0, forx− sin(x)

x2as follows:

1. Write down the 7th degree Taylor polynomial approximation of sin(x), centered at x = 0.

sin(x) ≈

2. Substitute the 7th degree Taylor polynomial approximation of sin(x) into the given expression, andsimplify the result, until it is written as a 5th degree Taylor polynomial in standard form.

x− sin(x)

x2≈

WebAssign #16: Find the 5th degree Taylor polynomial, centered at x = 0, forcos(x)− 1

x.

cos(x)− 1

x≈

Binomial Series Notes and Learning Goals 122

1. A binomial series is the Taylor series, centered at 0, for any function of the form (1 + x)k.

2. In order to work with binomial series, you have to be able to compute binomial coefficients. Thenotation and formula are: (

k

n

)=

n factors︷︸︸︷k(k − 1)(k − 2) · · · (k − n+ 1)

n!

In this formula n must be an integer, but k can be any number.

Example 1:

(9

5

)=

5 factors︷︸︸︷(9)(8)(7)(6)(5)

5!

= 126

Example 2:

(−3

7

)=

7 factors︷︸︸︷(−3)(−4)(−5)(−6)(−7)(−8)(−9)

7!

= −36

Example 3:

(12

4

)=

4 factors︷︸︸︷(1

2

)(−1

2

)(−3

2

)(−5

2

)4!

= − 5

128

Example 4:

(k

0

)= 1. Just like zero factorial.

Binomial Series Notes and Learning Goals 123

3. The full binomial series for (1 + x)k is

(1 + x)k =∞∑n=0

(k

n

)xn =

(k

0

)+

(k

1

)x+

(k

2

)x2 +

(k

3

)x3 + · · ·

4. Binomial series are often used to create simple, low order approximations for expressions involvingsquare roots.

Example: Find the quadratic approximation of√

1 + x.

Solution: Rewrite as √1 + x = (1 + x)1/2

Write out the form of the approximation

(1 + x)1/2 =

(1/2

0

)+

(1/2

1

)x+

(1/2

2

)x2

Compute binomial coefficients(1/2

0

)= 1

(1/2

1

)=

1

2

(1/2

2

)= −1

8

Answer: √1 + x ≈ 1 +

1

2x− 1

8x2

5. Some expressions must be manipulated to look like (1 + something)k before expansion.

Example: Find the quadratic approximation of√

25 + x.

Solution:

√25 + x = (25 + x)1/2 (rewrite)

= (25)1/2(

1 +x

25

)1/2(factor out 251/2)

= 5(

1 +x

25

)1/2(simplify 251/2)

≈ 5

[1 +

1

2

( x25

)− 1

8

( x25

)2](binomial expansion)

≈ 5 +1

10x− 1

1000x2 (simplify)

Binomial Series Worksheet 1 124

The binomial series centered at x = 0 is (1 + x)k =∞∑n=0

(k

n

)xn.

Where

(k

n

)are called the binomial coefficients.

WebAssign #4: Find the 3rd degree Taylor polynomial for (1+x)2/3 by using the binomial series as follows:

1. The third degree Taylor polynomial for this series, using the binomial series is:

T3(x) =

(2/3

0

)+

(2/3

1

)x+

(2/3

2

)x2 +

(2/3

3

)x3

2. Compute the binomial coefficients. Fully simplify them as exact fractions (no decimals)(2/3

0

)=

(2/3

1

)=

(2/3

2

)=

(2/3

3

)=

3. Use the results to write down the 3rd degree Taylor polynomial approximation for (1 + x)2/3.

(1 + x)2/3 ≈

WebAssign #5: Find the 3rd degree Taylor polynomial for (1+x)−4 by using the binomial series as follows:

1. Use the binomial series to write down the third degree Taylor polynomial using binomial coefficients:

T3(x) =

2. Compute the binomial coefficients. Fully simplify them as exact fractions (no decimals)

3. Use the results to write down the 3rd degree Taylor polynomial approximation for (1 + x)−4.

(1 + x)−4 ≈


WebAssign #6: Find the 6th degree Taylor polynomial approximation for (1 + x)−1 by using the binomialseries. Show all work following the steps on the previous worksheet.Hint: Look for a pattern that can be used to compute higher degree coefficients.

(1 + x)−1 ≈

WebAssign #7: Find the 3rd degree Taylor polynomial for (1 + u)1/2 by using the binomial series.

(1 + u)1/2 ≈

Use the above Taylor polynomial and an appropriate ‘u =’ substitution to find a Taylor polynomialapproximation of

√1 + 4x2.

√1 + 4x2 ≈

What is the degree of the Taylor polynomial for√

1 + 4x2? Is it the same/different than the Taylorpolynomial for (1 + u)1/2. Why or why not?


WebAssign #8: Find the 3rd degree Taylor polynomial for (1 + x/3)3/2 centered at x = 0. Show all work.Hint: Use an appropriate binomial series (1 + u)k.

(1 +

x

3

)3/2≈

WebAssign #9: Find a 6th order Taylor polynomial approximation of (1− x2)3/2.

(1− x2

)3/2 ≈Use the above Taylor polynomial approximation to find an 8th order approximation of x2(1− x2)3/2.

x2(1− x2

)3/2 ≈


WebAssign #10: Use the binomial series to find the the 3rd degree Taylor polynomial approximation for(1 + x)−1/2 centered at x = 0. Show all work.

(1 + x)−1/2 ≈

Use the above Taylor polynomial to find the 4th degree Taylor polynomial approximation for

2x√1 + x

≈

WebAssign #11: Find the quadratic Taylor polynomial approximation for√

16 + x centered at x = 0.Hint: Factor out a 16 and use an appropriate binomial series (1 + u)k.

√16 + x ≈


WebAssign #12: Find a quadratic degree Taylor polynomial approximation, centered at x = 0, for

(27 + 6x)1/3 ≈

WebAssign #13: Find a fourth degree Taylor polynomial approximation, centered at x = 0, for

x2√

9 + x ≈

Numerics Notes and Learning Goals 129

In today’s lesson you will use knowledge of Taylor polynomials to solve a numerical computation problem.

Problem: Compute the value of e, accurate to 4 decimal places, without touching the ex

button on your calculator.

You will need knowledge and skills from previous lessons:

• Know how to compute Taylor polynomials.

• Know that higher degree Taylor polynomials are more accurate.

• Know that accuracy is better at points close to the center point.

The solution to the problem is not hard. There are only two steps:

1. Replace ex with a Taylor polynomial, Tn(x).

2. Plug x = 1. That is, compute Tn(1).

Note: You can compute Tn(1) using only +, −, × and ÷ on your calculator.There are some details, though:

• You have to pick a center point. x = 0 is a good first choice.

• You need a suitably high degree. How high is the subject of this lesson.

You will need new knowledge and skills.

1. Know that the term stepsize refers to the distance between the center point and the evaluationpoint.

2. Given f , center point, and fixed stepsize, determine the minimum degree needed for a desired accu-racy.

3. Given f , center point, and a choice of higher degree or lower stepsize, determine which choices willor will not deliver the desired accuracy.

Numerics Worksheet 1 130

WebAssign #3: In this problem we want to approximate e = e1 using Taylor polynomials.

• Write down the 2nd degree Taylor polynomial for ex centered at x = 0.

• Use the second degree Taylor polynomial to estimate e1. Round answer to 4 decimal digits.

• Write down the 5th degree Taylor polynomial for ex centered at x = 0.

• Use the fifth degree Taylor polynomial to estimate e1. Round answer to 4 decimal digits.

WebAssign #4: The following table looks at different degree Taylor polynomial approximations of e = e1,all rounded to four decimal digits.

degree 0 1 2 3 4 5 6 7 8 9 10

e ≈ 1.0000 2.0000 2.6667 2.7083 2.7181 2.7183 2.7183 2.7183 2.7183

• Use your results from the above problem to fill in the missing blanks.

• Discussion Point: Do higher degree Taylor polynomials give better or worse approximations?

• To four decimal places, use the above table to determine the best possible approximation of e.

• What is the minimum degree that achieves this approximation?

• At some point in this table, the approximations no longer change. What is the minimum degree atwhich an approximation, to four decimal places, is the same as the approximation to its left?


WebAssign #5: In this problem we want to approximate√e = e0.5 using Taylor polynomials.

• The Taylor series for ex centered at x = 0 is∞∑n=0

xn

n!.

Use the sigma sum notation to expand the Taylor series by writing the first few terms, then endingit with an ellipsis, + · · · .

Discussion Point: How can the sigma sum notation for the Taylor series be used to find Taylorpolynomials centered at the same point?

• Fill out the following table using Taylor polynomials of degrees 0 through 8, centered at x = 0, toestimate e0.5. Round all estimates to 4 decimal digits. Once an approximation, to 4 decimal digits,is the same as the approximation to its left, write ‘NO CHANGE’ or ‘NC’ for the remaining cells.

degree 0 1 2 3 4 5 6 7 8

e0.5 ≈

• To four decimal places, what is the best possible approximation of√e?

• What is the minimum degree that achieves this approximation?


WebAssign #6: The Taylor series for cos(x), centered at x = 0, is cos(x) =∞∑n=0

(−1)nx2n

(2n)!

Use the sigma sum notation to write out the first few terms of the Taylor series.

Fill out the following table using Taylor polynomials of different degrees to approximate cos(1) to fourdecimal digits and find the minimum degree that achieves the approximation. Do not use the cos(x)button on your calculator for this. Why are there only even degrees listed?

degree 0 2 4 6 8 10 12 14 16

cos(1) ≈

WebAssign #7: The Taylor series for ln(x), centered at x = 1, is ln(x) =∞∑n=0

(−1)n

n+ 1(x− 1)n+1

Write out the first few terms of this Taylor series.

Fill out the following table using Taylor polynomials of different degrees to approximate ln(1.2) to fourdecimal digits and find the minimum degree that achieves the approximation. Do not use the ln(x) buttonon your calculator for this.

degree 0 1 2 3 4 5 6 7 8

ln(1.2) ≈


WebAssign #8 and #9: In this worksheet you will explore how stepsize and Taylor polynomial degreeaffect the quality of an approximation. All questions use Taylor polynomials for ex centered on x = 0.

The stepsize is the distance between the center point and the evaluation point.

All questions in WebAssign #8 refer to the table below. Complete the table below before answering thequestions in WebAssign. The final result should be a cleanly completed table.

degree 2 3 4 5 6 7 8

e1.0 ≈ 2.5000 2.6667 2.7083 2.7167 2.7181 2.7183 No Change

e0.5 ≈

e0.2 ≈

e0.1 ≈

e0.05 ≈

1. Discussion point: The numerical entries in the row for e1.0 are from WebAssign Problem 4. Com-pare them to your answers. Why does the degree 8 cell say “No Change” (NC)?

2. Look up your answers to WebAssign Problem 5 (on previous worksheet). Use your answers to fill inthe row for e0.5.

• Write all numerical answers rounded to 4 decimal places.

• Write “No Change” in the appropriate cells.

• When you are done with this row, answer parts (a)–(c) of WebAssign Problem 7.

Warning! Limited submissions.

• If your answers to the WebAssign questions exposed errors in your table, fix your table. Printa clean copy if necessary. You will want a clean table for upcoming problems.

3. Use Taylor polynomials of the appropriate degree to fill out the e0.2 row.

• Be sure to use “NO CHANGE” (NC) appropriately. When you are done with this row, answerparts (d)–(f) of WebAssign Problem 8. Warning! Limited submissions.

• Clean up your table if necessary.

4. Complete the rest of the table, then answer the rest of the problems in WebAssign #8.

5. Keep your clean and complete table nearby. You will need it for WebAssign #9.

Convergence II Notes and Learning Goals 134

A Taylor series is a series that converges for some values of x and diverges for others. The primary examplesused in this lesson are related to a geometric series:

A geometric series is any series of the form

c

1− r=∞∑n=0

crn

This series converges if −1 < r < 1, namely, when |r| < 1.

In this lesson you will investigate when a Taylor series converges and diverges by comparing it to a geometricseries.

• The Taylor series for1

1− xcentered at x = 0 is a geometric series

1

1− x=∞∑n=0

xn

• The interval of convergence is the interval of values for which a Taylor series converges. Forexample, the interval of convergence for the above Taylor series (a geometric series) is

(−1, 1) = {x : −1 < x < 1}

• When a Taylor series converges to a function for particular values of x, then the related Taylorpolynomials give better and better approximations with higher degree.1

• An alternating series is one which switches sign with each term. Example:

∞∑n=0

(−1)n

2n= 1− 1

2+

1

4− 1

8+ · · ·

• An alternating series converges2 if a corresponding series with positive numbers converges.

Example:∞∑n=0

(−1)n

2nconverges because

∞∑n=0

1

2n=∞∑n=0

(1

2

)nconverges.

1 It is possible that a Taylor series converges without converging to a correct function value. This case is not included inthis lesson. Every converging Taylor series in this lesson converges to the correct function value.

2 It is possible that an alternating series converges when the corresponding positive series diverges. These cases are notcovered in this lesson. If interested, you should look up conditional and absolute convergence.

Convergence II Notes and Learning Goals 135

• Be able to determine when a series converges or diverges by comparing it to a geometric series.

Example: Consider the following Taylor series centered at x = 2.

1

x=∞∑n=0

(−1)n

2n+1(x− 2)n

If x = 1 this series becomes

∞∑n=0

(−1)n

2n+1(1− 2)n =

∞∑n=0

(−1)n(−1)n

2n+1=∞∑n=0

1

2n+1=∞∑n=0

1

2 · 2n=∞∑n=0

1

2

(1

2

)n

This is a geometric series with r =1

2, so it converges.

If x = 5 this series becomes

∞∑n=0

(−1)n

2n+1(5− 2)n =

∞∑n=0

(−1)n(3)n

2n+1=∞∑n=0

(−3)n

2n+1=∞∑n=0

(−3)n

2 · 2n=∞∑n=0

1

2

(−3

2

)n

This is a geometric series with r =−3

2, so it diverges.

• Be able to find an interval of convergence by comparing a Taylor series to a geometric series.

Example: The Taylor series for3

3 + xcentered at x = 0 is

3

3 + x=∞∑n=0

(−1)n

3nxn

Rewrite this series in geometric form

∞∑n=0

(−1)n

3nxn =

∞∑n=0

((−1)x

3

)n=∞∑n=0

(−x3

)nSo this is a geometric series with r = −x/3.

Use that a geometric series converges when −1 < r < 1 to find the interval of convergence

−1 < r < 1

−1 <−x3

< 1

−3 < −x < 3

3 > x > −3

−3 < x < 3

So the interval of convergence of this series is (−3, 3).

Convergence II Worksheet 1 136

WebAssign #9: The Taylor series for1

1− xcentered at x = 0 (called a geometric series) is:

1

1− x=∞∑n=0

xn

The interval of convergence is the values of x for which the above series converges. Use the previousproblems in WebAssign to write down the interval of convergence in two ways.

• Write down the interval of convergence using inequalities.

• Write down the interval of convergence using interval notation.

WebAssign #10: Suppose the Taylor series to some function is f(x) =∞∑n=0

(x− 3

5

)nDetermine the interval of convergence for this series as follows:

• Write down an inequality that the geometric series∞∑n=0

rn converges using r.

• Identify an appropriate ‘r =’ substitution to write the given series as a geometric series.

• Substitute the formula for r into the above inequality. Then solve the inequality for x and write theinterval of convergence using interval notation.



xcentered at x = 3 is:

3

x=∞∑n=0

(−1)n

3n(x− 3)n

Determine the interval of convergence for this series by comparing it to a geometric series∞∑n=0

rn as follows:

• Identify an appropriate ‘r =’ substitution needed to write the series as a geometric series.

• Use the ‘r =’ substitution to write down an inequality for the values of x for which the series convergesthen solve the inequality. Last, write interval of convergence using interval notation.


5− xcentered at x = 0 is:

1

5− x=∞∑n=0

xn

5n+1

Determine the interval of convergence for this series by comparing it to a geometric series∞∑n=0

crn as follows:

• Identify what c and r are needed to write the series as a geometric series.

• Use the ‘r =’ substitution to write down an inequality for the values of x for which the series convergesthen solve the inequality. Last, write interval of convergence using interval notation.



xcentered at x = 2 is:

1

x=∞∑n=0

(−1)n

2n+1(x− 2)n

• If x = 0.5, write a simplified form of this series in geometric form∞∑n=0

crn.

Identify what c and r are, and if the series is convergent or divergent.

• If x = 4.5, write a simplified form of this series in geometric form∞∑n=0

crn.

Identify what c and r are, and if the series is convergent or divergent.

• Rewrite this series in geometric form. Identify r and use it to find the interval of convergence.

Area I Notes and Learning Goals 139

This lesson is about the computation of area, such as the one shown below. However, the learning goalsare NOT about computation of area. Learning goals are:

1. Know that area computation is achieved by slicing, and then integrating.

2. Know how to correctly label the height and width of a slice in a diagram.

3. Know how to describe the height, width, and area of a slice with appropriate algebraic notation.

4. Know how to write an integral that would give the total area.

In nearly every homework problem you are expected, and in fact required, to do all of these things:

• Label the slice.

• Give algebraic descriptions of height, width, and area of the slice.

• Write an integral for the total area.

There are two common methods of slicing and integrating to find area.

• Method I: Slice into thin vertical rectangles. Then integrate along the x-axis.

1. Imagine slicing the region into many thin vertical rectangles of width dx.

2. Write a formula for the area of your typicalrectangle.

height: h(x) = top− bottom

width: dx

area: dA = h(x)dx

3. Determine the bounds of integration

a = left bound, b = right bound

4. Use an integral to add up all the rectangle ar-eas.

A =

∫ x=b

x=a

dA =

∫ b

a

h(x)dx

Area I Notes and Learning Goals 140

• Method II: Slice into thin horizontal rectangles. Then integrate along the y-axis.

1. Imagine slicing the region into many thin horizontal rectangles of height dy.

2. Write a formula for the area of your typicalrectangle.

height: dy

width: l(y) = right− left

area: dA = l(y)dy

3. Determine the bounds of integration

a = lower bound, b = upper bound

4. Use an integral to add up all the rectangle ar-eas.

A =

∫ y=b

y=a

dA =

∫ b

a

l(y)dy

• If the area is split into multiple regions where thetop/bottom functions switch places you have tocalculate the area of each region separately andadd those areas together for the total area.

A = A1 + A2

Area I Worksheet 1 141

WebAssign #5: Consider the region bounded between the curves y = x+ 2 and y = x2 − 4.

Find the area of this region by integrating along the x-axis as follows:

1. The area is split into thin vertical rectangles of width dx and height h. A typical slice is locatedat the variable position x as shown in the figure.

2. Identify and label the width and height of the typical slice in the above figure.

Use arrows to clearly show what lengths represent the height and width of the slice.

3. Find the area of a typical rectangle. Write your answer as “dA = formula ”.

Your formula will involve both x and dx.

4. Find the left and right bounds for the location of the typical slices.

5. Set up the definite integral that sums up all the thin rectangles to find the area of this region.

A =

∫


WebAssign #9: Consider the region bounded between the curves y =√x+ 2, y = x, and the x-axis.

Find the area of this region by integrating along the y-axis as follows:

1. The area is split into thin horizontal rectangles of height dy and width w. A typical slice is locatedat the variable position y as shown in the figure.


Use arrows to clearly show what lengths represent the height and width of the slice.


Your formula will involve both y and dy.

4. Find the top and bottom bounds for the location of the typical slices.



WebAssign #10: Consider the region bounded between the curves y = 2− 2x2, y = 3x, and the y-axis.

Find the area of this region by integrating along the x-axis as follows:

1. A typical slice of width dx is shown in the figure.





Consider the region bounded between the curves y = ln(x), x = e, and the x-axis.

WebAssign #12: Find the area of this region byintegrating along the y-axis.

• Identify and label the width and height of a typical slice.

• Find the area of a typical slice.

• Write an integral for the total area of the region.

WebAssign #14: Find the area of this region byintegrating along the x-axis.

• Identify and label the width and height of a typical slice.

• Find the area of a typical slice.

• Write an integral for the total area of the region.

You have written two different integrals that compute the same area. Pick one of those integrals and useit to compute the area.

Area II Notes and Learning Goals 145

Recall. There is a five step process for computing the area of a region.

1. Make a graph of the region.

2. Determine if you are going to slice the region along the x- or y-axis. Then draw a typical rectangle.

3. Find the area of the typical rectangle, dA. Write it in terms of your variable of integration.

4. Find the bounds of integration and set up an integral to find the area.

5. Evaluate your integral using antiderivatives or a calculator or computer.

For Area I you only had to do steps 3, 4 and 5. Today you will also do steps 1 and 2. That is, you have to:Create a Slicing Strategy

• Sketch the region. This is a required element of each problem, even though WebAssign can’t gradeit.

• Choose an axis of integration. Note: this will also determine the variable of integration.

• Slices must line up along the axis of integration. “Line up along” can mean different things indifferent contexts. For this lesson it means

Along x-axis: Slices are vertical rectangles with width = dx

Along y-axis: Slices are horizontal rectangles with height = dy

• Decide if your integral has to be broken into more than one region.

• In each region, draw a typical rectangle. This is also a required element of each problem, even thoughWebAssign cannot grade it or give feedback.

After you complete all steps of the slicing strategy, you still have to complete Steps 3–5 to find the area ofthe region.

Area II Worksheet 1 146

WebAssign #1: A region is bounded by the y-axis, the line y = 2 and the line y = 23x as shown.

Find the shaded area by integrating along the x-axis using thin vertical slices.

1. In the shaded area draw a typical slice (that is a thin vertical rectangle that goes from the bottomboundary to the top boundary).

2. Correctly label the height and width of your slice in the above figure. Get feedback.

3. Write a formula for the area of your slice.

dA =

4. Write an integral for the total shaded area. Enter your answer in WebAssign.

A =


WebAssign #2: A region is bounded by the y-axis, the line y = 2 and the line y = 23x as shown.

Find this area by integrating along the y-axis using thin horizontal slices.

1. In the shaded area draw a typical slice (that is a thin horizontal rectangle that goes from the leftboundary to the right boundary).



dA =


A =


WebAssign #3: A region is bounded by the x-axis, the line x = 2 and the curve y = 12

√x+ 2 as shown.

Find this area by integrating along the x-axis using thin vertical slices.

1. In the shaded area draw a typical slice (that is a thin vertical rectangle that goes from the bottomboundary to the top boundary).



dA =


A =


WebAssign #4: A region is bounded by the x-axis, the line x = 2 and the curve y = 12

√x+ 2 as shown.

Find this area by integrating along the y-axis using thin horizontal slices.

1. In the shaded area draw a typical slice (that is a thin horizontal rectangle that goes from the leftboundary to the right boundary).



dA =


A =


WebAssign #5: A region is bounded by the curves x = y3 and y = x2 in the first quadrant as shown.

Find the area of this region by following the steps below.

1. Communicate your slicing strategy. That is,

• Choose an axis of integration (x or y). Write your choice on this worksheet.

• Draw a typical slice.

• Correctly label the height and width of your slice in the above figure.




WebAssign #6: A region is bounded by the curves y = x2 and y = 2x+ 3.


1. Graph both the curves on the axes above and shade in the region in question. Use proper labeling.

2. Use the x-axis as the axis of integration.

3. Draw a typical slice in the region that agrees with the axis of integration.

4. Correctly label the height and width of your slice in the above figure.




WebAssign #7: A region is bounded by the curves y = sin(x) and y = cos(x) on the interval π4≤ x ≤ 5π

4.


1. Graph both the curves on the axes above and shade the region in question. Use proper labeling.


3. Draw a typical slice in the region that agrees with the axis of integration.

4. Correctly label the height and width of your slice in the above figure.

5. Write a formula for the area of your slice. Enter your answer in WebAssign.

6. Write an integral for the total shaded area.


WebAssign #8: A region is bounded by the curves y = sin(x) and y = sin(2x) on the interval 0 ≤ x ≤ πas shown.



2. What is the minimum number of regions you need to split this area into?

3. Draw a typical slice in each region you need.

4. Correctly label the height and width of your slice(s) in the above figure.

5. Write formulas for the areas of the slices in your sketch.

6. Write an integral or a sum of integrals for the total shaded area.


WebAssign #12: A region is bounded by the curves y2 = x and x− y = 2 as shown.




• Determine how many regions you will have to split the problem into.

• Draw typical slices in each region.

• Correctly label the height and width of each slice.

2. Write formulas for the area of each slice.



WebAssign #13: A region is bounded by the curves x = (y − 1)2, 4y = x2 and x+ y = 3 as shown.




• Determine how many regions you will have to split the problem into.

• Draw typical slices in each region.

• Correctly label the height and width of each slice.

2. Write formulas for the area of each slice.


Density Notes and Learning Goals 156

You now have a process that you can use to compute the area of a region. The most important parts ofthe process are

• Choose and communicate a slicing strategy.

• Draw and correctly label a typical slice.

• Write a formula for the area of that slice.

After that you can use an integral to add up the areas of the slices.In today’s homework you will use this process to compute the mass of a two-dimensional object. Forpurposes of this assignment, density will always be mass per unit area.1

1. If density is constant, mass is easy to compute:

mass = density · area

m = ρ · A

2. If density is variable, then you have to slice the object into thin slices. If you do this correctly, youcan safely assume density is constant on each slice. Then you can write a formula for the mass ofa thin slice:

dm = ρ · dA

Note: As in previous homework, dA will be a formula with a variable and a differential in it.Typically, ρ will also be a formula.

Note: If ρ is a formula, then the variable in that formula must also be the variable of integration.This will affect your slicing strategy.

3. Once you have a formula for dm, use an integral to add up the masses of the slices. Total mass is

m =

∫ b

a

dm

1Technically this is called area density or surface density.

Density Worksheet 1 157

WebAssign #4: A rectangular cutting board isbuilt from strips of different types of wood, each4 cm wide and 24 cm long, as shown at right.

• The density of oak is 0.74 g/cm2.

• The density of walnut is 0.57 g/cm2.

• The density of cherry is 0.63 g/cm2.

1. What is the area of a typical strip?

2. What is the mass of an oak strip?

3. What is the mass of a walnut strip?

4. What is the mass of a cherry strip?

5. What is the mass of this cutting board?


WebAssign #5: A rectangular block is formed by the region inthe first quadrant, below the line y = 8 and to the left of the linex = 15. This block has variable density given by the function

ρ(x) = 0.06x(15− x) g/cm2

Find the mass of this block by integrating along the x-axis.

1. What is the area of the typical slice shown?

dA =

2. What is the mass of the typical slice shown?

dm =

3. Write an integral to compute the total mass of the block, then compute the total mass.

m =

WebAssign #6: A rectangular block is formed by the region inthe first quadrant, below the line y = 8 and to the left of the linex = 15. This block has variable density given by the function

ρ(y) = 3.5 sin(π

8y)

g/cm2

Find the mass of this block by integrating along the y-axis.

1. What is the area of the typical slice shown?

dA =

2. What is the mass of the typical slice shown?

dm =

3. Write an integral to compute the total mass of the block, then compute the total mass.

m =


WebAssign #15: A triangular plate is in the shape of theregion bounded by x = 0, y = 4−x, and y = x/3, as shownat right. Both axes are measured in meters and the plate ismade of a material with variable density

ρ(x) = 17.4− 0.3x2 kg/m2

Follow all the steps below to find the mass of this plate.

1. Communicate your slicing strategy by stating the axisof integration (make a full statement, don’t just writedown the variable) and draw a typical slice.

2. Label the dimensions of the typical slice (be sure to use arrows to clearly show the dimensions).

3. Find the area of a typical slice. Write answer as dA = formula .

4. Find the mass of a typical slice. Write answer as dm = formula .

5. Write an integral for the total mass of the plate. Write answer as m = integral formula .

6. Compute the integral to find the mass of the plate.


WebAssign #17: A plate is in the shape of the regionbounded by y = 2−x2/2 and the x-axis, as shown. Both axesare measured in meters and the plate has variable density

ρ(x) = 5.4− 0.2x2 kg/m2


1. Communicate your slicing strategy by stating the axisof integration (make a full statement, don’t just writedown the variable) and draw a typical slice.

2. Label the dimensions of the typical slice (be sure to use arrows to clearly show the dimensions).

3. Find the area of a typical slice. Write answer as dA = formula .

4. Find the mass of a typical slice. Write answer as dm = formula .

5. Write an integral for the total mass of the plate. Write answer as m = integral formula .

6. Compute the integral to find the mass of the plate.


WebAssign #18: A tapered piece of thin steel is shaped as shownat right. Both axes are measured in centimeters and the platehas a variable density

ρ(y) = 1.4− 0.1y g/cm2


1. Communicate your slicing strategy by stating the axis ofintegration (make a full statement, don’t just write downthe variable) and draw a typical slice.

2. Label the dimensions of the typical slice (be sure to usearrows to clearly show the dimensions).

3. Find the area of a typical slice.

4. Find the mass of a typical slice.

5. Write an integral for the total mass of the plate. Then compute the integral to find the mass.


WebAssign #19: A plate is in the shape of the regionbounded by y = 2−x2/2 and the x-axis, as shown. Both axesare measured in meters and the plate has variable density

ρ(y) = 3.6− 0.5y kg/m2

Find the mass of this plate. Show all work. Include all stepslisted on previous worksheets.


WebAssign #20: A triangular plate is in the shape of theregion bounded by x = 0, y = 4−x, and y = x/3, as shownat right. Both axes are measured in meters and the plate ismade of a material with variable density

ρ(y) = 6y(4− y) kg/m2

Find the mass of this plate. Show all work. Include all stepslisted on previous worksheets.

New Slices Notes and Learning Goals 164

Recall. There is a general process for computation using an integral.

1. Pick a slicing strategy. I.e., choose an axis of integration and draw a typical slice.

2. Write a formula for whatever you are measuring on that slice. (This could be area, mass, etc.)

3. Determine bounds of integration.

4. Write an integral for the total amount of whatever you are trying to compute.

5. Compute your integral either by hand using antiderivatives or use your calculator (there will bespecific rules for how you must

Today. Learn two new slicing strategies. Both strategies will be used with density. However, today’sproblems will expand the notion of density.

1. Density

• Previous problems used density in the form of mass per unit of area, so that

total mass = density · area

• Today you will encounter linear mass density, which is mass per unit of length. In thiscontext

total mass = (linear density) · length

• There will be quantities besides mass. For example, electrical charge per unit of length.

total charge = (charge density) · length

• More generally, in this assignment density can mean any quantity measured per unit of area orlength. You can tell what sort of density you have by looking at the “per” units. The generalformulas are

stuff = (stuff per unit area) · area

stuff = (stuff per unit length) · length

• Finally, today’s assignment includes beams with distributed loads. This is force per unit length,so it works just like linear density.

2. One-dimensional slices

• Use when your data is attached to a one-dimensional object, like a thin rod or a beam.

• There is only one choice for an axis of integration — along the object.

• Slices are tiny bits of the object, with tiny length dx.

• A small amount of stuff is usually

d stuff = (linear density) · dx

New Slices Notes and Learning Goals 165

3. Radial slicing

• Use when your object is two-dimensional (or three) and your data depends on distance from acenter point.

• Axis of integration is along a radius extending from the center point.

• Slices are thin rings.2 You will need to know the formula for the area of a thin ring:

dA = 2πr dr

• A small amount of stuff is usually

d stuff = (area density) · 2πr dr

2For now. Later there will be partial rings.

New Slices Worksheet 1 166

WebAssign #2: A 1.5 meter steel rod is manufactured such that it has a variable linear density given by

ρ(x) = 1168 sin(1.12x+ 1.019) g/m

where x is the distance (in meters) from the left end of the rod. Find the mass of this rod as follows:

1. Find the length of a typical slice (in meters).

2. Find the mass of a typical slice (in grams). Write answer as dm = formula .

3. Write an integral to compute the total mass of the rod. Write answer as m = integral formula .

WebAssign #3: A steel rod of length 50 cm is manufactured such that it has a variable density given by

ρ(x) = 36− kx g/cm

where x is the distance (in cm) from the left end of the rod and k is measured in g/cm2.

1. Find the length of a typical slice (in meters).


3. Write an integral to compute the total mass of the rod. Write answer as m = integral formula .

4. Find the value of k so that the mass of the rod is exactly 1500 g. Include units.


WebAssign #4: A steel rod of unknown length, L, has a variable linear density given by

ρ(x) = 54.8− 0.247x g/cm

where x is the distance (in cm) from the left end of the rod.

1. Draw a typical slice and label its length on the provided figure.


3. Write an integral to compute the total mass of the rod. Write answer as m = integral formula .Then find the length of the rod so that its mass is exactly 3500 g. Include units.

WebAssign #5: A thin rod of length 12 cm carries an electrical charge with a variable charge density

λ(x) = 2− 2e−0.3x nC/cm

where λ is in nanocoulombs per centimeter (nC/cm) and x is the distance (in cm) from the left end.

1. Draw a typical slice and label its length on the provided figure.

2. Find the charge on a typical slice (in nC). Write answer as dC = formula .

3. Write an integral to compute the total charge on the rod. Write answer as C = integral formula .Then compute the total charge. Include units.


WebAssign #8: A four meter long beam carries a distributed load (force density) given by

w(x) = 200− 12.5x2 N/m

where x is the distance (in m) from the left end of the rod.

1. Identify and label the length of a typical slice in the provided figure.

2. Find the force on a typical slice (in Newtons). Write answer as dF = formula .

3. Write an integral to compute the total force on this beam. Then find the total force on the beam.

WebAssign #9: An 80 foot tower experiences a wind load (force density) of

λ(y) = 2.2y1/3 lbs/ft

where y is the distance (in ft) from the bottom of the tower.

1. Identify and label the length of a typical slice in the provided figure.

2. Find the force on a typical slice (in pounds).

3. Write an integral to compute the total force on this tower.Then find the total force on the tower.


WebAssign #11: A population density of foxes decreases with the distance r from the center of theirterritory, given by the population density function ρ(r) = 7.3− r/2, in foxes per square kilometer.

1. Identify and label the radius r and thickness dr of a tiny circularring (typical slice) in the provided figure.

2. Complete WebAssign #10. Use result to state the area of thisring in terms of r and dr. Write answer as dA = formula .

3. Find the population of foxes on this tiny ring.Write answer as dP = formula .

4. Write an integral for and compute the total population of foxes in a circle with a 5 km radius.

WebAssign #12: A circular ring has an inner radius of 6 cm and an outer radius of 18 cm as shown. Thedensity of this ring varies with the distance r from the center given by ρ(r) = 162/r2 g/cm2.

1. Draw a typical slice of radius r and thickness dr in the providedfigure. Then identify and label the dimensions of the slice.

2. State the area of a typical slice.

3. Find the mass of a typical slice.

4. Write an integral for and compute the total mass of this ring.

1D Moments Notes and Learning Goals 170

Moment of a Concentrated Force

The figure below shows a beam anchored to a wall at a point B. A force of 100 lbs pushes down on thebeam at a point 4 feet from the wall.

This force creates a physical quantity called torque or moment.3 The numerical value of the moment inthis example is

(4 ft)(100 lbs) = 400 ft·lbs

The Pivot Point is Important

There is no such thing as a moment without a specified pivot point. In the example pictured above thepivot point is B. The correct vocabulary is the moment about B. The notation is MB. The formula is

MB = rF

where F is the force and r is the distance from the force point to the pivot point.

The Distance is Critical!

Every time you compute a moment you should execute the following three step process:

• Find the distance from the force to the pivot.

• Find the force.

• Compute: Moment = (Distance)× (Force).

Sign Conventions

Force, distance and moment are signed quantities,4 meaning they might be positive or negative dependingon context. For this lesson we will adhere to the following sign conventions:

• Treat all forces as positive.

• Treat all distances as positive.

• All counter-clockwise moments are positive.

• All clockwise moments are negative.

3Physics books usually use the word “torque.” Engineering books prefer the word “moment.” Calculus books use theword “moment,” but in a more abstract way: they use mass instead of force, they ignore units, and they sometimes reversethe counter-clockwise-is-positive rule.

4Actually, they are vector quantities, but that’s for a different class.


Distributed Loads

Suppose that a beam carries a distributed load as shown below.

• The distributed load is described by a function, w(x) = 8x− x2 lbs/ft.

• The function only makes sense if there is an x-axis. In this and many similar situations the beamitself is the x-axis.

• O is the origin on the x-axis.

• The figure shows a slice of the beam.

• The slice is located at x.

Example: For the beam and load shown above, write an integral for MB.Solution: Slice the beam up into tiny slices, and then find the (tiny) moment caused by the (tiny) forceon just one slice:

Use the following process:

• The axis of integration is the x-axis.

• The (tiny) length of the slice is dx.

• The (tiny) force on the slice is dF = (8x− x2) dx.

• The distance from the slice to pivot point is (4− x)

• The (tiny) moment caused by the (tiny) force on this slice is

dMB = (distance)(force) = (4− x)(8x− x2) dx


• Finish by writing an integral for the total moment. The bounds of integration run from the left endof the load to the right end of the load.

MB =

∫ 4

0

(4− x)(8x− x2) dx

1D Moments Worksheet 1 173

WebAssign #8: A 12 foot beam carries a distributed load of 20 pounds per foot (lbs/ft).

A typical slice of length dx is located at position x as shown.

1. Identify and label the length of a typical slice in the above figure.

2. Given a force density of ω(x) = 20 lbs/ft, find the force on a typical slice.

Write answer as dF = formula .

3. Identify and label the distance between a typical slice and the pivot at B. What is this length?

4. Find the moment about B caused by a typical slice. Write answer as dMB = formula .

5. Write an integral to compute the total moment about B. Write answer as MB = integral formula .Then compute the moment. Include correct units on final answer.


WebAssign #9: In each of these 3 figures/ws, a 4 meter long beam carries a distributed load of ω(x) = 150xwhere x is the distance, in meters, from the left end of the beam. A typical slice is drawn at location x.

Answer the following question for each figure:

1. In each figure, notice the location of the origin. Is it always in the same location?

2. In each figure, notice the location of the pivot point. Is it always in the same location?

3. In each figure, draw arrows between a typical slice and the pivot, and label it with the formula forthe distance between the two locations.

4. In each figure, does the force on the typical slice drawn create a positive or negative moment?


WebAssign #10: A 4m beam carries a distributed load given by w(x) = 150x N/m, as shown.

Find the moment caused by this distributed load about a pivot located at B as follows:

1. Communicate your slicing strategy:

• State your axis of integration.

• Identify and label the length of a typical slice above.

2. Find the force on a typical slice. Write answer as dF = formula .

3. Identify, find, and label the distance from a slice to the pivot B in the above figure.

4. Find the moment about B caused by a typical slice. Write answer as dMB = formula .

5. Write an integral to compute the total moment about B. Write answer as MB = integral formula .Then compute the moment. Include correct units on final answer.



Find the moment caused by this distributed load about a pivot located at A as follows (use proper notation):



• Draw a typical slice and label its length on above figure.

2. Find the force on a typical slice.

3. Identify, find, and label the distance from a slice to the pivot A in the above figure.

4. Find the moment about A caused by a typical slice.

5. Write an integral to compute the total moment about A. Then compute the moment.



Find the moment caused by this distributed load about a pivot located at A. Show all work.Does the moment cause the beam to tip clockwise, counter-clockwise, or remain balanced?

WebAssign #14: A 12ft beam carries a distributed load given by w(x) = 2x2 lbs/ft, as shown.

Find the moment caused by this distributed load about a pivot located at A. Show all work.Does the moment cause the beam to tip clockwise, counter-clockwise, or remain balanced?


WebAssign #16: A 4m beam carries a distributed load given by w(x) = 100√x N/m, as shown.

Find the moment caused by this distributed load about a pivot located at B. Show all work.


Find the moment caused by this distributed load about a pivot located at O. Show all work.


Moment about an Axis

In 1D Moments you computed moments about a pivot point. Today you will compute moments about apivot axis.

• The notation is similar, except that there is a pivot axis, not a point. In the example above themoment is about the y-axis. The notation and formula are:

My = (2 m)(50 N)

= 100 N m

• The pivot axis is also called the moment axis.

• The sign convention is determined by the blue arrow curled around the pivot axis. In the exampleabove the moment is positive because it agrees with the curled arrow.5

• Distance must be constant!

In the example above the load is distributed. However the load is distributed parallel to the pivotaxis, so the distance from the load to the pivot axis is always 2 meters.

• Moments can occur about any axis. In your homework you will encounter moments about the y-axis,about the x-axis, and about other pivot axes that are offset from the x- or y-axes.

5This is consistent with the right hand rule, for rotation about the y-axis.


2-D Representation.The example on the previous page was drawn in three-dimensional perspective. Your homework, and mostexamples in future applications, will be presented as 2-D drawings.

• The figure above shows a 20× 20 foot platform centered on an x-y axis system.

• The platform is allowed to pivot around the y-axis.

• The vertical gray bar represents a 200 pound load.

• The downward direction of gravity is “into the page,” or into the computer screen.

• The result is a moment about the y axis of My = 600 ft lbs.

• The moment is positive because the force of gravity (into the page) would cause the platform to tipin a direction that agrees with the blue arrow.

The 2-D representation makes it more difficult to visualize the rotation, but it is useful for problemsinvolving distributed loads.The most common distributed load is the weight of the plate itself.


Distributed Weight

Suppose that weight is spread across a two-dimensional region. For example, the re-gion shown at right is bounded by

y = 1− x2

in the first quadrant. It represents a thinplate made of steel that weighs 40 lbs/ft2.

This distributed weight can be sliced ver-tically as shown above.

Each slice generates a moment about they-axis. These can be added up with anintegral, as outlined below.

• The axis of integration is the x-axis.

• A typical slice is drawn with height 1− x2 and width dx.

• The (tiny) area of the slice isdA = (1− x2) dx

• The (tiny) weight of the slice exerts a (tiny) force

dF = 40(1− x2) dx

• The distance from the slice to pivot axis is

distance = x

• The slice causes a (tiny) moment about the y-axis

dMy = distance · force

= x · 40(1− x2) dx

• The total moment about the y-axis is obtained by integrating across the plate.

My =

∫ 1

0

40x(1− x2) dx


WebAssign #4: Consider the region in the first quadrantbounded by y = 2−x as shown at right. The figure representsa thin plate made of steel that has a density of 41 lbs/ft2.Both axes are measured in feet. Follow all steps below tocompute the total moment of this plate about the y-axis.

Slice the region into thin vertical slices as shown.

1. State the axis of integration.

2. Label the height and width of a typical slice.Find the (tiny) area of the slice, dA.

3. Find the (tiny) weight of this slice, dF .(Note: We use F for weight, because weight is the downward force gravity exerts on the plate.)

4. Find the distance between the slice and the y-axis.

5. Weight will cause a moment about the y-axis. Find the (tiny) moment caused by this slice, dMy.

6. Find the bounds of integration then write an integral for the total moment, My, of this plate aboutthe y-axis. Round answer to one decimal place. Include units.


WebAssign #5: Consider the region in the first quadrantbounded by y = x2 + 1, the x-axis, and the vertical linesx = 2 and x = −1. The figure represents a thin plate of oakthat weighs 147 N/m2. Both axes are measured in meters.

Find the moment of this plate about the y-axis.

1. State an axis of integration and draw a typical slice.

2. Label the height and width of the slice.State the area of a typical slice.

3. Find the (tiny) weight of this slice.

4. Identify and label the distance between the slice and the moment axis (y).

5. Find the (tiny) moment about the y-axis caused by a typical slice.

6. Write an integral for the total moment about the y-axis.Find this moment accurate to the nearest whole number. Include units.


WebAssign #6: Consider the region in the first quadrantbounded by y = 1− (x− 2)2 and the x-axis. The figurerepresents a thin plate made of plywood that weighs 1.42lbs/ft2. Both axes are measured in feet.

Find the moment caused by this plate about the y-axis.Show all work with the steps on previous worksheets.


WebAssign #7: Consider the region in the first quadrantbounded by y =

√x, the x-axis, and the vertical line x = 9.

The figure represents a thin plate made of pine that weighs 2.75lbs/ft2. Both axes are measured in feet.

Find the moment of this plate about the line l, located at x = 4.Slice the region into thin vertical slices as shown.


2. Label the height and width of a typical slice and find its (tiny) area.

3. Find the (tiny) weight of a typical slice.

4. Identify and label the distance between a typical slice and the moment axis.

5. Find the (tiny) moment about l caused by this slice.

6. Write an integral to compute the total moment. Find the total moment. Include units.



Find the moment of this plate about the line, l, at x = 1.Show all work as outlined on previous worksheets.


WebAssign #9: Consider the region in the first quadrantbounded by y = 2−x as shown at right. The figure representsa thin plate made of steel that has a density of 41 lbs/ft2.Both axes are measured in feet. Follow all steps below tocompute the total moment of this plate about the x-axis.

Slice the region into thin horizontal slices as shown.


2. Label the height and width of a typical slice.Find the (tiny) area of the slice, dA.

3. Find the (tiny) weight of this slice, dF .

4. Find the distance between the slice and the x-axis.

5. Find the (tiny) moment caused by this slice, dMx.

6. Write an integral for the total moment, Mx, of this plate about the x-axis. Round answer to onedecimal place. Include units.


WebAssign #10: Consider the region bounded by y = 1−x2,the y-axis, and the line y = −3. The figure represents a thinplate made of steel that has a density of 41 lbs/ft2. Both axesare measured in feet. Follow all steps below to compute thetotal moment of this plate about the line l at y = −2.

Slice the region into thin horizontal slices as shown.


2. Label the height and width of a typical slice.Find the (tiny) area of the slice.


4. Identify and label the distance between the slice and the moment axis.

5. Find the (tiny) moment caused by this slice.

6. Write an integral to compute the total moment of this plate about the line l.


WebAssign #11: Consider the region bounded by the curvey = x2 − 4 and the line y = 5. The figure represents a thinplate made of pine that weights 2.75 lbs/ft2. Both axes aremeasured in feet.

Find the moment of this plate about the line, l, at y = 1.Show all work as outlined on previous worksheets.



1. Find the moment of this plate about the line l at un-known location x = b. Show all work. Work and an-swers will contain the unknown constant b.

2. Find b such that the plate is perfectly balanced.


WebAssign #14: Consider the region bounded by the curvey = x2 − 4 and the line y = 5. The figure represents a thinplate made of pine that weights 2.75 lbs/ft2. Both axes aremeasured in feet.

1. Find the moment of this plate about the line l at y = b.Show all work.

2. Find b such that the plate is perfectly balanced.

Volumes I Notes and Learning Goals 192

Volume

• The volume of an object of height h and crosssectional area A as shown is

V = Ah

• Similarly, the volume of a little slice of height dhwould have a little bit of volume

dV = Adh

• Learn to split a volume into little slices and find the total volume.

1. Determine how you are going to slice the ob-ject and define a variable of integration (y inthis figure).

2. Find the volume of a typical slice: dV =πr2dy

(a) Draw and label a typical slice with ap-propriate arrows.

(b) Draw the x-y cross section and the cor-responding typical slice with consistentlabeling.

(c) Write the volume of a typical slice interms of the variable of integration.

3. Use the integral to add the volume of the slices to get total volume

V =

∫ y=b

y=a

dV =

∫ b

a

πr(y)2dy

Volumes I Notes and Learning Goals 193

Volumes of Rotation

• Rotating a curve y = f(x) about the x-axis creates a 3D object.

1. The axis of integration is the x-axis.

2. Slice the object into little discs as shown.The volume of a slice is

dV = πr2dx = π[f(x)]2dx

3. Then total volume is

V =

∫ x=b

x=a

dV = π

∫ b

a

[f(x)]2dx

• Rotating a region about the x-axis creates a hollowed out 3D object.

1. The axis of integration is the x-axis.

2. Slice the object into little washers as shown.The volume of each slice is

dV = π(r22 − r21)dx

3. Then total volume is

V =

∫ x=b

x=a

dV = π

∫ b

a

(r22 − r21)dx

• You can also rotate curves or regions about the y-axis or any other line (axis) of rotation.

Volumes I Worksheet 1 194

WebAssign #6: A three-dimensional 10 inch long tapered dowel is created by rotating the line y = 2−0.1xabout the x axis as shown. Find the volume of this dowel as follows

1. Determine an axis to slice the object along into tiny circular discs

Axis of integration:

2. Determine the volume of a typical slice

(a) Draw a typical slice then identify and label its radius r and thickness with appropriate arrows.Find the volume of the slice in terms of these variables.

dV =



(b) Draw a graph of the x-y cross section. In the graph, draw a typical slice that matches your slicein part (a) and label its dimensions with appropriate arrows. Labels must be consistent withthe variables used on the figure of a typical slice.

(c) Find the radius in terms of the variable of integration:

Write the volume of a typical slice in terms of the variable of integration.

dV =

3. Find the bounds of integration and write the integral that gives the total volume.


WebAssign #7: A three-dimensional object is created by rotating the curve y = x2/3 about the x-axis fromx = 0 to x = 8 as shown. Find the volume of this object as follows

1. Determine an axis to slice the object along into tiny circular discs

Axis of integration:

2. Determine the volume of a typical slice

(a) Draw a typical slice then identify and label its radius r and its thickness with appropriatearrows. Find the volume of the slice in terms of these variables.

dV =



(b) Draw a graph of the x-y cross section. In the graph, draw a typical slice that matches your slicein part (a) and label its dimensions with appropriate arrows. Labels must be consistent withthe variables used on the figure of a typical slice.

(c) Write the volume of a typical slice in terms of the variable of integration.

dV =

3. Find the bounds of integration and write the integral that gives the total volume.


WebAssign #8: A three-dimensional funnel is created by rotating the curve y = 15 ln(x) about the y-axisfrom y = 0 to y = 25 as shown.

1. Axis of integration:

2. Find the volume of a typical slice. Use theblank space below to

(a) Draw a typical slice and label its dimen-sions with appropriate arrows.

(b) Draw the x-y cross section, the corre-sponding slice and label its dimensions.Labeling must be consistent.

(c) Find the volume of the typical slice interms of the variable of integration.

Your drawings should look similar to what wasgiven on Worksheet 1.

=


WebAssign #9: A three-dimensional vase is created by rotating the curve x = 1 + 0.5 sin(0.7y) about they-axis from y = 0 to y = 9 as shown.






Your drawings should look similar to what wasgiven on Worksheet 1.

=


WebAssign #12: A three-dimensional object is created by rotating the area between y = 2.5 and y =√x

from x = 0 to x = 4 about the x-axis as shown.






=


WebAssign #13: A three-dimensional object is created by rotating the area between the curves y = x3/2

and y = x4 about the y-axis as shown.






=

Volumes II Notes and Learning Goals 202

• These lessons continue with three-dimensional volumes. Volume I introduced finding volumes by thefollowing process:

1. Communicate a slicing strategy. Required elements of this are

– State your axis of integration.

– Draw a typical slice as a stand-alone 3-D picture.

– Label this picture.

– Draw a 2-D picture of the entire object, in which your slice is clearly located.

– Label the 2-D picture consistently with your labeling of the 3-D picture.

2. Compute the volume of your slice. Write your answer in terms of your integration variable.

3. Write an integral for the total volume.

4. Compute your integral.

• Hand drawn pictures with proper labeling are a required part of written solutions. Thisincludes

– The 3-D slice must be properly labeled with appropriate arrows. The tiny thickness must belabeled dx or dy and the other dimensions can use any variable of your choice.

– The 2-D cross section must be properly labeled (axis and scale) and include your typical slice.The dimensions of the slice must be properly labeled using the same variables as the 3-D slice.

– The labeling must be such that it is clear how you then computed the formula for the volumeof a typical slice.

• This assignment is mostly done on worksheets which require you to draw and label pictures asdescribed. This element is the most important skill for this unit. Make sure you get feedback onyour pictures.

Volumes II Worksheet 1 203

WebAssign #1: A three-dimensional object is cre-ated by rotating the curve y = 4

xabout the line

y = 4 from x = 1 to x = 4 as shown to the right.






Label the dimensions of the slice with appropriate arrows. Then identify the dimensions on the x-y crosssection and find the radius and volume of the slice.

r = dV =


WebAssign #2: A three-dimensional object is cre-

ated by rotating the curve y = 32

sin(π4x)

about the

line y = 3 from x = 0 to x = 4 as shown to the right.






=


WebAssign #3: A three-dimensional object is cre-ated by rotating the curve y = 4x2 + 4 about theline x = −3 from x = 0 to x = 2 as shown.






=


WebAssign #4: A three-dimensional object is cre-ated by rotating the curve y = 4x2 + 4 about theline x = −3 from x = −2 to x = 0 as shown.

1. Discussion Point: What is similar and whatis different between this problem and the pre-vious problem.






=


WebAssign #5: A three-dimensional object is created by rotating the area between y = 18

(ex + e−x

)and

the x-axis about the line y = 1.5 from x = −2 to x = 2 as shown.






=


WebAssign #6: A three-dimensional object is cre-ated by rotating the area between y = 2x2, y =8x − 16, y = 8 and the x-axis about the line x = 4as shown. Here x and y are measured in inches.






=

Volumes III Worksheet 1 209

WebAssign #1: A four sided pyramid (upside down) is shown at right. The pyramid is 10 cm tall.Cross-sections are squares. The top end is 5 cm on a side.

1. Use the y-axis for the axis of integration.

2. Find the volume of a typical slice.

(a) Draw a typical slice and label its dimensions with appropriate arrows as shown.

Write the volume in terms of the dimensions used

=



(b) Put a coordinate system on this problem. There are two main choices, put the origin at the tipor put the origin at the base of the pyramid.

For your choice of coordinate system, locate and label the dimensions of the typical slice usingappropriate arrows as shown.

(c) Find the volume of the typical slice in terms of the variable of integration.

=

3. Find the bounds of integration and write an integral for the total volume of this pyramid.


WebAssign #2: A three sided pyramid (upside down) is shown at right. The height (measured from thetip to the midpoint along the base of a side as shown) is 5 inches. Cross-sections are equilateral trianglesperpendicular to the vertical axis shown in green. The top end is 3 inches on a side.

1. Use the y-axis for the axis of integration.


(a) Draw a typical slice and label its dimensions with appropriate arrows as shown.

Write the volume in terms of the dimensions used

=



(b) Put a coordinate system on this problem. There are two main choices, put the origin at thetip or put the origin at the base of the pyramid. The y axis must move along the vertical axisshown.

Draw an x-y cross section using your choice of coordinate system, then locate and label thedimensions of the typical slice using appropriate arrows.


=

3. Find the bounds of integration and write an integral for the total volume of this pyramid.


WebAssign #3: The shape pictured below has a triangular base. The base is between the lines y = 2−0.5xand y = 0.5x− 2 where x and y are in feet. Above the base, cross-sections are semicircles.

1. Axis of Integration:


(a) Draw a typical slice and label its dimensions with appropriate arrows.

Write the volume in terms of the variables used

=



(b) Draw a graph of the x-y cross section, then draw and label a typical slice using appropriatearrows.


=

3. Find the bounds of integration and write an integral for the total volume of this object.


WebAssign #4: The shape pictured at right has a base bounded by the curves y = −2+0.25x, y = 2−0.25x,x = 0 and x = 4 where x and y are measured in meters. Above the base, cross-sections are semicircles.






=


WebAssign #5: The shape pictured at right has a base bounded by the curves y = −4 + 0.25x2, y =4−0.25x2, and x = 0 where x and y are measured in meters. Above the base, cross-sections are semicircles.






=

Centroids Notes and Learning Goals 217

1. Learn the efficient method for locating a balance line in two-dimensions. The process is:

• Find dA for an appropriate slice.

• Compute Area:

∫dA

• Compute Moment about y-axis:

∫x dA

• Balance line is located at x =Moment

Area

Example: Find the location of the balance line shown below:

Solution:

• Area of slice: dA =1

2x dx

• Area:1

2(8)(4) = 16

• Moment:

∫ 8

0

x · 1

2x dx =

1

6x3∣∣∣∣80

=256

3

• Location: x =256/3

16=

16

3


2. Know that this method works for horizontal balance lines also. A horizontal line is located at somey-coordinate. The formula is

y =

∫y dA∫dA

Both integrals are computed using horizontal slices.

3. The two balance lines cross at a point called the centroid, as shown below:

The coordinates of the centroid, (x, y), are given by these formulas, where dA is the area of anappropriate slice.

x-coordinate = x =

∫x dA∫dA

y-coordinate = y =

∫y dA∫dA


4. There is a common application of centroids to distributed loads, like the example below, where

w = 100x1/2 lbs/ft

Fact: A distributed load is equivalent to a concentrated force that acts through thecentroid of the load. The force is the integral of the load.

Example: Replace the distributed load above with a concentrated force.

Solution:

• Total Force:

∫ 10

0

100x1/2 dx =200

3x3/2

∣∣∣∣100

≈ 2108 lbs.

• Moment:

∫ 10

0

x · 100x1/2 dx = 40x5/2∣∣100≈ 12469 ft-lbs.

• Centroid: x =12469

2108≈ 6 ft.

The distributed load can be replaced with this:

Centroids Worksheet 1 220

WebAssign #1: Find the location of the vertical balance line forthe shape shown at right. Show all work.

• State the axis of integration.

• Draw and label the dimensions of a typical slice.

• Find the area of a typical slice, dA.

• Compute both∫dA and

∫xdA.

• Use the result from above to find the centroid, x =

∫xdA∫dA

.

WebAssign #2: Find the location of the vertical balance linefor the shape shown. Show all work following above steps.


WebAssign #3: Find the location of the horizontal balance linefor the shape shown at right. Show all work.

• State the axis of integration.

• Draw and label the dimensions of a typical slice.

• Find the area of a typical slice, dA.

• Compute both∫dA and

∫ydA.

• Use the result from above to find the centroid, y =

∫ydA∫dA

.

WebAssign #4: Find the location of the horizontal balanceline for the shape shown. Show all work following above steps.


WebAssign #5: Find the location of the vertical balanceline for the shape shown at right. Show all work.

WebAssign #6: Find the location of the horizontal balanceline for the shape shown. Show all work.

Using the results from the two problems above, write down the centroid point, (x, y), of this shape.


WebAssign #7: A 12 foot long beam carries a distributed load of w(x) = 50x lbs/ft.

• Find the total force, F , pushing on the beam.

• This distributed load is equivalent to a concentrated force whose magnitude is the total force, pushingon the centroid of the force distribution as shown:

Find the centroid point, x, on the beam the force acts on.

• Find the counter-clockwise moment at the wall caused by this load by using the concentrated force:

– Find the distance from the concentrated force to the wall at the right end of the beam.

– Find the counter-clockwise moment at the wall caused by this force.


WebAssign #10: Find the centroid, (x, y), of the object pictured atright. Show all work.

Hint: One of the two coordinates can be found by symmetry (nointegrals needed). If you find a centroid by symmetry, provide aclear statement indicating the coordinate and how you found it.

WebAssign #11: Find the centroid, (x, y), of the object pictured atright. Show all work.


Taylor Polynomials and Series

• Recall that there are three standard methods for computing Taylor polynomials.

1. Compute coefficients one at a time using derivatives of the function. After you write eachderivative of f(x) you can either:

– write each symbolic derivative of standard form Tn(x), equate the center point values f (n)(c)and Tn(c), and solve for each an,

– or use the formula an =f (n)(c)

n!Both the standard form and the coefficient formula are on the Exam 2 formula sheet.

2. Substitute into a known series for one of the common functions. There are four very commonexamples. All are provided on the Exam 2 formula sheet.

3. Use binomial expansion, perhaps with a substitution. Binomial expansion and binomial coeffi-cient formulas are on the Exam 2 formula sheet.

• Write all Taylor polynomials in standard form, with fully simplified coefficients, and with zero coef-ficient terms omitted.

• Recall that a Taylor series is a Taylor polynomial extended to infinity. Conversely, a Taylor polyno-mial uses a finite number of terms taken from the corresponding Taylor series.

Integral Applications

• A slicing strategy must be clearly communicated. This means stating an integration axis or variable,sketching needed slice(s), and labeling slice dimensions clearly and accurately.

• Know the types of slices that might be used. These include small lengths, small areas, small volumes,and small radial slices.

• Know how to determine other physical quantities for slices, such as mass, weight, and force, basedon a given density. Know how to determine moment with force and a pivot distance.

• Be able to write and compute integrals that provide total amounts, based on underlying slice infor-mation. Calculator use is acceptable, as long as such use is fully described.

• Know and use proper notation for all of the above. Examples include d[something] for slice charac-teristics (such as dA and dF ), V for total volume, and Mx for moment about the x-axis.

Notation Review Worksheet 1 226

WebAssign #1: A beam carries a distributed load given by w(x) = 80x N/m, as shown.

Write an integral to compute the moment caused by this distributed load about the pivot at A. Show allwork. Use proper notation. Get feedback on your work from your instructor, LA, or tutor.



• Draw a typical slice and label its length on above figure.

2. Find the force on a typical slice.

3. Identify, find, and label the distance from a slice to the pivot A in the above figure.

4. Find the moment about A caused by a typical slice.

5. Write an integral to compute the total moment about A.

Notation Review Worksheet 2 227

WebAssign #2: A thin plate is in the shape of the regionshown. The bounding curves are y = 1−x2, y = −3, and they-axis. The plate is made of wood that weighs 6 lbs/ft2.

Write an integral for the moment about the line l, located aty = −1. Show all work following the steps below. Be sure touse correct notation. Get feedback on your work.


2. Draw and label a typical slice.

3. Find the (tiny) area of the slice.


4. Identify and label the distance between the slice and the moment axis.

5. Find the (tiny) moment caused by this slice.

6. Write an integral to compute the total moment of this plate about the line l.

Slicing Review Worksheet 1 228

The shape pictured below is formed by rotating y = 5x and y = 1.25(x− 3)2 around the axis x = 3. Alsopictured are a typical slice and a 2-D cross section.

Write a short English phrase to answer each of the following questions:

1. What does r1 measure? (Sample phrases are available in WebAssign Problem 1.)

2. What does r2 measure?

3. What does π(r22 − r21) dy measure?

Get feedback on your writing. Then answer WebAssign questions 1–6.


The shape pictured at right is formed by rotating x = y2,from 0 ≤ y ≤ 2, around the axis y = 2.

Communicate a slicing strategy that could be used tofind the volume of the shape. At a minimum you must:


• Draw and label typical 3D slice.

• Draw a 2-D picture of the x-y cross-section of theshape, in which your slice is clearly located.

• Label your 2-D picture consistently with the label-ing of your 3-D picture.

• State the volume of the slice, dV , in terms of thevariable of integration.

Get feedback on your pictures and labels. Then answer WebAssign questions 7–12.


WebAssign #15: Two students answer the problem “Write an integral for the area of the region below.”

1. Alice solves the problem by writing ∫ 4

0

(4y − y2

)dy

Figure out what slicing strategy she must have used.

• What was her axis of integration?

• Draw her slice in the figure above. Label it correctly.

• Find the area of her slice.



2. Bob solves the problem by writing ∫ 4

0

(2√

4− x)dx

Figure out what slicing strategy he must have used.

• What was his axis of integration?

• Draw and label his slice in the figure below.

• Find the area of his slice.

3. Discussion Questions:

• Which slicing strategy do you prefer? Why?

• Suppose that Alice and Bob were trying to find the centroid. Whose slicing strategy will workbetter? Why?


A metal bushing is formed in the shape shown below. The outer curve is y = x2 + 1. The inner bore hasconstant radius 0.5.

1. Draw and label both figures to communicate a slicing strategy for a volume integral along the x-axis.You can use the empty axes for your 2D cross-section.

2. Write an integral for the volume. Your work must include:

• A formula for the inner radius of your slice.

• A formula for the outer radius of your slice.

• A formula for the volume of your slice.

• An integral for the total volume.

3. Answer WebAssign questions 16–19.

Vectors I Notes and Learning Goals 233

Vocabulary and Notation: This lesson has a lot of new vocabulary. Here is an initial reference list ofthe terms that you have to learn. Definitions, notation, and examples follow below.

• Vector

• Scalar

• Initial Point

• Terminal Point

• Component and/or Components

• Component Form

• Position Vector

• Magnitude

• Unit Vector

1. Definition: A vector is a quantity that has both magnitude and direction.

It is useful to think of a vector as an arrow. The length of the arrow represents the magnitude of thequantity, and the arrow gives the direction.

Physical Examples:

• An airplane is flying north-east at 500 mph. Its velocity is a vector:

• A force of 300 lbs acts downward at the point shown below.


2. Definition: A scalar is a quantity that has only magnitude.

It’s useful to think of scalars as just ordinary numbers.

Examples:

• 75 degrees Celsius

• −2

• C, where C is any unknown constant.

• x, if x is any variable that you have encountered previously.

3. Know the difference between a scalar, a vector,and a point. Scalars are not hard to identify,since they are ordinary numbers. But vectors andpoints are subtly and closely related.

One way to think about this is, “A point is a lo-cation. A vector is an arrow that points to thatlocation.”

Example: In the figure at right:

• P is the point (−1, 2).

• The arrow pointing to P is a vector.

4. The fact that a vector points to a point gives rise to more vocabulary, as well as a common notationfor vectors.

Definition: The terminal point of the vector is the point at the end of the arrow. This is alsocalled the head of the vector.

Definition: The initial point of the vector is the point at the beginning of the arrow. This is alsocalled the tail of the vector.

Example: In the example above, if O is the origin, then

• O is the initial point.

• P is the terminal point.

• The vector is denoted−→OP .


Example: There is no requirement that the ini-tial point be the origin. In the figure at right,

• The initial point is Q = (−1, 0).

• The terminal point is R = (1, 1).

• The vector is denoted−→QR.

5. Definition: Suppose that a vector has initialpoint at the origin and terminal point P = (a, b).That is, the vector points from the origin to (a, b).

We say that a and b are the components of thevector. The component form is a notationalway of writing the vector that distinguishes itfrom the point that it points to.

The notation uses angle brackets:

−→OP = 〈a, b〉

In the example at right,

−→OP = 〈−1, 2〉

Note: Component form also applies to vectorsthat are not based at the origin.

Think of component form as

〈how far over, how far up〉

For the example at right:

−→QR = 〈2, 1〉

6. Definition: When a vector is based at the origin, say−→OP , then the vector is called the position

vector for the point P .

7. Often vectors are written as single letters. In order to distinguish this from a scalar, the letter willbe bold face, or it will have an arrow over it, or both.

Examples:−→v = 〈2, 1〉 or −→w =

−→OP


8. Definition: The magnitude of a vector is the length of the arrow. The notation is the same asabsolute value.

Example: In the figure above the magnitude of−→QR is

|−→QR| =

√12 + 22 =

√5

General Formula:|〈a, b〉| =

√a2 + b2

9. Definition: A unit vector is a vector that has magnitude = 1.

Vectors I Worksheet 1 237

WebAssign #4: Write down the following vectors in component form using correct notation.

1. Graph the points P = (1, 1) and Q = (3, 4) to the right.

Draw an arrow that starts at P and ends at Q.

Write the vector−→PQ in component form. Use correct labeling

and notation. Write answer as−→PQ = vector formula .

2. Graph the points P = (1, 1) and Q = (3, 4) to the right.

Draw an arrow that starts at Q and ends at P .

Write the vector−→QP in component form. Use correct labeling

and notation. Write answer as−→QP = vector formula .

3. Write the vector with initial point R = (1.5, 1) and terminal point S = (0.25, 2) in component form.

4. Write the vector −→v with initial point (a, b) and terminalpoint (c, d) in component form.


Graph the point P = (4, 2) to the right. Label the point P .

Draw an arrow that starts at origin O and ends at P .Label this vector −→r .

Write the position vector −→r =−→OP in component form.

Write answer as −→r = vector formula .

Discussion Points:

• State the differences between the point P and the vector (arrow) −→r in the graph.

• State the differences between the notation used to write down the point, P = (4, 2), and the positionvector, −→r = 〈4, 2〉.

• Why is it important to distinguish the difference between a point and a position vector that goesfrom the origin to that point?

WebAssign #6: Find the position vector for each of the following points. Use correct notation.

1. P = (4, 2):

2. Q = (0, 6):

3. R = (−11.07, 126):

4. S = (x, y):


WebAssign #9: Complete the following worksheet on vector magnitude.

The graph of the vector −→v = 〈2, 3〉 is at right.

Create a right triangle with −→v as the hypotenuse as follows:

• Draw a line that gives the horizontal length of this vector,and label the distance.

• Draw a line that gives the vertical length of this vector,and label the distance.

• Identify the right angle in the triangle that was created.

Find the magnitude (length) of this vector by using the right triangle you drew above.

Write answer as |−→v | = scalar . Use an exact answer, with√

symbol as appropriate.

The graph of the vector −→u = 〈−2,−3〉 is at right.

• Create a right triangle as above.

• Use the right triangle to find the magnitude of the vector.Write and label answer using correct notation.

Discussion Points:

• Does a negative component affect the direction a vector points? Explain.

• Does a negative component affect the magnitude of a vector? Explain.



Find the magnitude of the vector −→v = 〈a, b〉.Write and label answer using correct notation.

Consider the points P = (1, 6) and Q = (4,−5).

• Graph the points P and Q and the vector−→PQ at right.

• Find the magnitude of the vector−→PQ.

• Write and label answer using correct notation.

Consider the vectors −→w1 = 〈5, 0〉 and −→w2 = 〈0, 3〉.

• Graph and label these two vectors at right.

• Find the magnitude of each vector.


WebAssign #10: Answer the following questions.

Consider the two vectors −→v = 〈3, 3〉 and −→u = 〈1, 1〉.

• Graph the vector −→v with initial point (1, 1).

• Graph the vector −→u with initial point (1, 2).

• Find the magnitude of both vectors.

• Write a statement relating the direction of the two vectors.

• Write a statement relating the magnitude of the two vectors.

A unit vector is a vector whose magnitude is 1 (unit).

• Is −→v a unit vector? Why or why not?

• Is −→u a unit vector? Why or why not?

• Find a unit vector that points in the direction of −→v . Write answer as v = vector .

Find a unit vector that points in the same direction as the vector −→w = 〈−1, 2〉.Write and label answer using correct notation.


WebAssign #11: In the figure below, sketch the following vectors:

1. −→v 1 has initial point P = (1, 2) and the same magnitude and direction as −→r .

2. −→v 2 has initial point P = (1, 2) and the same magnitude, but the opposite direction as −→r .

3. −→u 1 is a unit vector with initial point Q = (2, 3) and with the same direction as −→r .

4. −→u 2 is a unit vector based at the origin and making an angle of 60◦ measured counter-clockwise fromthe positive x-axis.

Vectors II Notes and Learning Goals 243

1. Two vectors can be added together. The vector sum of −→v and −→w is denoted −→v +−→w .

2. There two standard geometric interpreta-tions of vector addition.

Suppose that −→v and −→w are as shown atright.

Tip-to-Tail Addition

• Draw −→v based at the origin.

• Draw −→w based at the tip of −→v .

• The vector from the origin to the newtip of −→w is

−→v +−→w

Vectors II Notes and Learning Goals 244

Parallelogram Rule.

• Draw −→v and −→w based at the origin.

• Draw a parallelogram as shown atright.

• The vector from the origin to the farcorner of the parallelogram is

−→v +−→w

3. In component form, vectors add algebraically. In the example above:

−→v = 〈2, 3〉−→w = 〈3,−1〉

−→v +−→w = 〈5, 2〉

The general formula is〈a, b〉+ 〈c, d〉 = 〈a+ c, b+ d〉

4. A vector can be multiplied by a scalar. The scalar product of c and −→v is denoted c−→v .

• Geometrically, scalar multiplication is a streching or shrinking of −→v , perhaps reversing directionas well. This is explored in homework and worksheets.

• Algebraically, scalar multiplication is

c〈a, b〉 = 〈ca, cb〉

Vectors II Worksheet 1 245

WebAssign #1: Suppose that −→v = 〈3,−2〉 and −→w = 〈2, 4〉.

1. Sketch the vector −→v , based at the point (−1, 1). Label it −→v .

2. Locate the terminal point (tip) of the vector −→v you drew.

3. Sketch the vector −→w , based at the terminal point (tip) of the vector −→v . Label it −→w .

4. Draw a new vector that starts at the initial point of the vector −→v and goes to the terminal point ofthe vector −→w . Label this vector −→v +−→w .

5. Write down the new vector you found in component form. Write answer as −→v +−→w = vector .

You have just drawn a tip-to-tail diagram, which shows how to add two vectors geometrically.

Discussion Points:

• Explain how to use a tip-to-tail diagram to add two vectors geometrically: −→v +−→w .

• Is there an algebraic formula you can use to find −→v +−→w from their components?



Suppose that −→v = 〈3,−2〉 and −→w = 〈2, 4〉.

1. Sketch the vector −→w , based at the point (−1, 1). Label it −→w .

2. Locate the terminal point of the vector −→w you drew.

3. Sketch the vector −→v , based at the terminal point of the vector −→w . Label it −→v .

4. Draw a new vector that starts at the initial point of the vector −→w and goes to the terminal point ofthe vector −→v . Label this vector −→w +−→v .

5. Using the figure write down the new vector you found. Write answer as −→w +−→v = vector .

Discussion Points:

• How is the tip-to-tail diagram above different from the previous page? How is it similar?

• Does the order you draw the tip-to-tail diagram affect the tip-to-tail graph?

• Does the order affect the resulting vector sum?


WebAssign #2: Suppose that −→u = 〈1, 1〉 and −→v = 〈4, 2〉.

1. Sketch the vector −→u , based at the origin. Label it.

2. Find a new vector 3−→u by multiplying both components of −→u by 3. Write answer as 3−→u = vector .

3. Sketch the vector 3−→u , based at the origin. Label it.

4. Sketch the vector −→v , based at the origin. Label it.

5. Find a new vector 12−→v by multiplying both components of −→v by 1

2. Use correct notation.

6. Sketch the vector 12−→v , based at the origin. Label it.

This operation of multiplying a scalar by a vector is called scalar multiplication.

Discussion Points:

• Describe how to compute scalar multiplication algebraically.

• Describe how scalar multiplication alters a vector geometrically.


WebAssign #3: Let −→u = 〈2,−3〉, −→v = 〈−5, 1〉, and −→w =⟨−1

2, 32

⟩.

Compute the following. Show work using correct notation.

• −→u +−→v

• −→v +−→u

• 5−→u

• 2−→u + 3−→v

• 2−→u + 4−→w

• −→u −−→v + 2−→w

• |−→v +−→w |


WebAssign #4: Suppose that −→v = 〈2, 1〉 and −→w = 〈1, 4〉.

1. Sketch and label the vectors −→v and −→w , based at the origin.

2. −→v and −→w form two sides of a parallelogram. Sketch and label vectors for the remaining two sides.

3. Sketch and label the vector −→v +−→w in the parallelogram.

4. Sketch and label the vector −−→v , based at the origin.

5. Use the parallelogram method to compute −−→v +−→w .

Discussion Point:

• What does the above parallelogram tell us about −→v +−→w and −→w +−→v ?


WebAssign #7: Suppose that −→v =

⟨1,

1

2

⟩and −→w = 〈−2, 3〉, as shown.

• Sketch the vectors 3−→v and −4−→v .

• Sketch the vectors −−→w and1

2−→w .

The following sentence describes how −4−→v isobtained from −→v .

“Stretch −→v to four times itslength, then reverse its direction.”

Write descriptions of the following situations.Descriptions must state both the magnitudeand direction of the resulting vector. Use theabove as a model (if you want).

• How 3−→v is obtained from −→v .

• How −−→w is obtained from −→w .

• How1

2−→w is obtained from −→w .


WebAssign #15: Suppose that −→v = 〈1,−1〉 and −→w = 〈1, 2〉.Consider the position vector

−→r (t) = −→w + t−→v

Different values of t give different position vectors.

1. Fill out the table at right for seven different values of t.

2. Plot the seven position vectors that you just computed.

3. Sketch the shape formed by connecting their terminal points.

Discussion Point: What is the shape you sketched?

Note: This worksheet will be used in the ‘more’ assignment.

t −→r (t) = −→w + t−→v

−3

−2

−1

0

1

2

3

Vector Lines Notes and Learning Goals 252

• Recall: From algebra, here are two ways to represent a line.

Slope-intercept form: y = mx+ b

Point-slope form: y = m(x− x0) + y0

Recall, also, that the required ingredients for either of these forms are

1. Slope, m, and

2. One point on the line, either (x0, y0) or (0, b).

• Vector Form: A line can be represented by the vector equation

−→r (t) = −→r0 + t−→v

In this form the required ingredients are

1. A direction vector, −→v . This can be any vector that is parallel to the line.

2. One point on the line, but expressed as a position vector −→r0.

• Note: In all forms, there are two required pieces of information.

1. Something that gives a direction for the line.

2. Something that gives one point on the line.

• Geometric Example:Suppose that the direction vector is −→v = 〈1,−1〉 and theposition vector is −→r0 = 〈1, 2〉.

The line represented by

−→r (t) = −→r0 + t−→v

is graphed below. Note that the direction vector, −→v , isparallel to the line.

You can think of the line as the result of tip-to-tail addi-tion of −→r0 and scalar multiples of −→v .

Vector Lines Notes and Learning Goals 253

• Algebraic Expression:

Specific examples of vector equations are usually written in component form. For example,

−→r (t) = 〈3,−2〉+ t〈−5, 7〉

Often it is useful to rewrite this as two component functions:

−→r (t) = 〈3,−2〉+ t〈−5, 7〉= 〈3,−2〉+ 〈−5t, 7t〉 scalar multiplication

= 〈3− 5t,−2 + 7t〉 vector sum

The component functions are

x = 3− 5t

y = −2 + 7t

• Things you have to know how to do:

1. Given a line, determine its direction vector.

Be able to do this if the line is give to you in any of the following forms:

– Vector form

– Component equations.

– Slope-intercept form.

– Point-slope form.

– Graph.

Note: Any vector parallel to the line will work, so there are an infinite number of correctanswers.

2. Given direction and a point, represent the line in vector form.

Note: Any vector parallel to the line can serve as direction, and any point on the line can bethe position vector, so there are a doubly infinite number of correct answers.

3. Given any of the above, graph the correct line.

Note: There is ONLY ONE correct answer.

4. Use any appropriate algebraic form of a line to answer questions about the line.

Vector Lines Worksheet 1 254

WebAssign #1: Suppose that −→r 0 = 〈0, 3〉 and −→v = 〈2,−1〉.Consider the position vector

−→r (t) = −→r0 + t−→v

Different values of t give different position vectors.

1. Fill out the table at right for four different values of t.

2. Plot the four position vectors that you just computed.

3. Sketch the shape formed by connecting their terminal points.

Discussion Point: What is the shape you sketched?

t −→r 0 + t−→v

−1

0

1

2


WebAssign #2: The vector form of a line is −→r (t) = −→r0 + t−→v .

• What vector represents the direction vector for this line (a vector parallel to the line)?

• What vector represents a point on the line (an initial position vector)?

WebAssign #3: A line is given in vector form: −→r (t) = 〈3, 2〉+ t 〈−4, 1〉

• Write down the direction vector for this line.

• Write down a position vector representingan initial point on this line.

• Graph this line as follows:

– Draw the initial position vector for thisline. Label it −→r 0.

– At the tip of−→r 0 draw the direction vec-tor for this line. Label it −→v .

– Extend the line (in both directions)along the vector −→v .

WebAssign #4: A line is parallel to the vector −→v = 〈1, 2〉 and passes through the point P = (−1, 3).

Write down a vector form of this line in the form −→r (t) = vector formula .


WebAssign #5: Graph the line represented by −→r (t) = 〈−1, 1〉+ t〈2, 1〉

After your graph is complete:

1. Find two vectors other than 〈2, 1〉 that would also be correct direction vectors.

−→v1 = −→v2 =

2. Find two vectors other than 〈−1, 1〉 that represent points on the line.

(−→r0)1 = (−→r0)2 =

3. Write two different vector forms of this line.

−→r1(t) =

−→r2(t) =

Discussion Point: What is the difference between the different vector forms of this line? What happensif you change −→r0? What about −→v ?


WebAssign #7: The parametric equations for a line are{x = 2ty = t+ 2

1. Use the parametric equations to fill out table at right.

2. Plot the points, (x, y), from the table. Label each point

with an appropriate t = number on the graph.

3. Sketch the line connecting all the points.

4. Write this line in vector form. Use correct notation,−→r (t) = vector .

t x y

−1

0

1

2

Discussion Points:

• What do the t = number labels add to the graph?

• How are the parametric equations of a line and the vector form of a line related?


WebAssign #8: A line has vector form −→r (t) = 〈6t− 1, 9t+ 2〉.

• Identify the two components of the vector and write the line as a system of parametric equations.

x =

y =

WebAssign #9: A line has vector form −→r (t) = t 〈2, 0〉+ 〈3,−5〉.

• Use vector algebra (scalar multiplication and vector addition) to write this line as a single vector.

• Identify the two components of the vector and write the line as a system of parametric equations.Write answer using correct parametric equation notation.

WebAssign #10: The parametric equations for a line are{x = 2t+ 7y = t− 4

• Write this line in vector form. Then use vector algebra to rewrite this line in the form−→r (t) = −→r0+t−→v .

• Use the rewritten form to identify the direction vector and initial point on this line used.


WebAssign #17: A line is of the form−→r (t) = −→r0 + t−→v

Where −→r0 and −→v are provided in graph below.

• Graph this line on the graph below.

• Write a correct vector equation for this line in component form.

• Write the parametric equations for this line. Use correct notation.

• What are the x and y intercepts of this line? Write answers as points.


WebAssign #21: Consider the two lines in vector form:

−→r1(t) = 〈t, 3t− 1〉 and −→r2(s) = 〈2− 2s, 3s+ 1〉

• Graph both these lines on the axes at right.

• Identify the intersection point on the graph at right.

• Set up a system of equations to find this intersection point.Note the vector lines use different parameters (t vs s).

• Solve the system of equations for t and s.

• What is the intersection point?Write answer using exact fractions.

Discussion Points:

• Why are two different parameters t and s needed to find the intersection point?

• If these vector forms described two different particles traveling down their respective lines, explainthe difference between the paths intersecting and the two particles colliding into each other.

Vector Valued Functions I Notes and Learning Goals 261

1. Know the definition. A vector valued function is a function whose outputs are vectors.

Examples:

• −→r (t) = 〈t2 − t, t+ 1〉. Input t = 1. The output will be the vector 〈0, 2〉.• Every vector formula for a line vector form (from previous lesson).

2. Be able to graph a vector valued function. Here is the process:

• Create a table of values: one column for values of t, and one column for the correspondingvectors.

• For each entry in the table, graph the terminal point of the vector.

• Connect the dots.

• Include arrows to indicate the direction of increasing t.

• Label a few dots with their t-values. (This last part is optional in many settings, but it providesuseful additional information.)

Example: −→r (t) = 〈t2 − t, t+ 1〉.

t −→r−1 〈2, 0〉0 〈0, 1〉1 〈0, 2〉2 〈2, 3〉3 〈6, 4〉

3. Know the vocabulary terms parametrized and parametric.

• The graph of a vector valued function, −→r (t), is a path or curve.

• We say that the curve is parametrized by −→r (t).

• The component functions of −→r (t) are called parametric equations for the curve.

4. Know that many different functions can parametrize the same curve. For examples, refer to anyproblem from Vector Lines that had more than one correct answer.

5. Know that vector valued (or parametric) functions contain more information than just the curve thatresults as a graph.

• The curve can be thought of as the track of a moving object.

• Vector valued functions determine the direction of motion along the curve.

• They also determine when the object reaches any particular point.

• And they determine how fast the object moves.

6. The outputs of a vector valued function is are position vectors, but the position vectors are rarelyincluded in the graph.

Vector Valued Functions I Worksheet 1 262

WebAssign #2: Here is a vector form of a line:

−→r (t) = 〈1− 3t, 2 + 2t〉

Graph this line as follows:

• Fill out table at right with the corresponding positionvector for each value of t.

• Plot the terminal point of each position vector. Labeleach point with an appropriate t = value .

• Draw arrows along the line showing the direction ofincreasing t.

• Draw a single position vector for t = 1. Label it −→r (1).

t −→r

−1

0

1

Discussion Points:

• What is the difference between a position vector and its terminal point graphed on the line?

• What do the t = value labels add to the graph?

• What do the arrows on the line add to the graph?


WebAssign #3: Consider the vector valued function

−→r (t) = 〈t, t2〉

Graph this function as follows:

• Compute some position vectors for this curve by filling outtable at right.

• Plot the terminal points of each position vector.

– Label and scale both axes.

– Label each point with an appropriate t = value .

• Sketch the graph that passes through these points.

• Indicate the direction of increasing t.


t −→r

−2

−1

0

1

2



−→r (t) = 〈t3, t〉



• Sketch the graph that passes through these points.Use proper labeling.



t −→r

−2

−1

0

1

2



−→r (t) = 〈t2 + 1, t3 − 1〉



• Sketch the graph that passes through these points.Use proper labeling.



t −→r

−2

−1

0

1

2


WebAssign #8 - #11: Consider the following vector valued functions

−→r (t) = 〈4 cos(t), 4 sin(t)〉, −→s (t) = 〈4 cos(2t), 4 sin(2t)〉, −→w(t) =⟨

4 cos(π

2− t), 4 sin

(π2− t)⟩

• Fill in the table of values for all three functions

t −→r −→s −→w

0

π

6

π

4

π

2

π

• Then graph all three functions on axes below. Use proper labeling.

Hint: If the table is not complete enough to figure out the shape of the graphs, add more points tocomplete the graph.

Discussion Point: Describe the similarities and differences between the three graphs.


WebAssign #12 - #17: The remaining problems in WebAssign ask you to graph functions. Graph thefunctions on the following grids, then choose the correct graph in WebAssign to check yourself.

Vector Valued Functions II Notes and Learning Goals 269

1. Know how to compute the derivative of a vector valued function. If

−→r (t) = 〈x(t), y(t)〉

Then the derivative of −→r is−→r ′(t) = 〈x′(t), y′(t)〉

2. Know these standard notations for the derivative of a vector valued functions:

−→r ′(t), d

dt

(−→r (t))

ord−→rdt

3. Know the geometric interpretation of −→r ′(t) at a specific point on the graph of −→r (t).

At the instant t0,−→r ′(t0) is a vector that is tangent to the graph of −→r (t)

at the location −→r (t0).

Example: Suppose that −→r (t) = 〈t2 − 1, t〉. At the instant t = 1, the position vector is −→r = 〈0, 1〉and the derivative is −→r ′ = 〈2, 1〉The graph, position vector, and tangent vector are shown below, with the tangent vector placed atthe tip of the position vector.

Vector Valued Functions II Notes and Learning Goals 270

4. Know the physical interpretation of −→r ′(t).

If −→r (t) represents the position of a moving object, then −→r ′(t) is thevelocity of the object.

• This means that velocity is a vector quantity, meaning it has both magnitude and direction.

• In this context, it is common to use the notation −→v (t) in place of −→r ′(t), since −→v is standardnotation for velocity.

• The direction of −→v is the direction of travel for the object.

• The magnitude, |−→v |, is the speed of the object. That is, |−→v | is the number that would appearon, say, a speedometer or airspeed indicator.

• The derivative of −→v is the acceleration of the object. It is often denoted −→a (t). Graphically,−→a (t0) originates at the location −→r (t0), just like velocity.

5. Given −→r ′(t), be able to determine −→r (t), up to unknown constants. That is:

• Be able to antidifferentiate each component function of −→r ′

• Know that the result may include an unknown constant vector, −→r 0.

• Know that this is actually two unknown constants, 〈x0, y0〉.

Example: If −→r ′(t) = 〈4t3,−2e−2t〉, find −→r (t).

Solution: −→r (t) = 〈t4, e−2t〉+ 〈x0, y0〉

6. Given additional information, such as one point on −→r , be able find the exact formula for −→r .

Example: If −→r ′(t) = 〈4t3,−2e−2t〉 and −→r (0) = 〈5, 3〉, find −→r (t).

Solution:

−→r (t) = 〈t4, e−2t〉+ 〈x0, y0〉 (as above)−→r (0) = 〈04, e−2·0〉+ 〈x0, y0〉 (plug in 0)

= 〈x0, 1 + y0〉 (simplify)

〈5, 3〉 = 〈x0, 1 + y0〉 (given)

This gives two equations to solve for x0 and y0.

x0 = 5

1 + y0 = 3

You get x0 = 5 and y0 = 2, so the final answer is

−→r (t) = 〈t4 + 5, e−2t + 2〉

Vector Valued Functions II Worksheet 1 271

WebAssign #1: The graph below is the curve described by the vector valued function: −→r (t) = 〈2t, t2− 2〉

A vector valued function describes the position of an object that moves along this curve.

Add the following to the graph to show how this object moves along the curve.

• Show the location and motion of the object at time t = 0:

– Compute the position vector −→r (0). Then sketch the resulting vector and label it −→r (0).

– The velocity vector is found by taking the derivative: −→r ′(t) = 〈x′(t), y′(t)〉 = 〈2, 2t〉– Compute −→r ′(0). Then sketch the velocity vector based at the tip of −→r (0). Label it −→r ′(0).

• Show the location and motion of the object at time t = 1 by repeating the above steps.

Discussion Points:

• Why do we graph the velocity vector at the tip of the position vector?

• How is the direction of the velocity vector related to the curve the object moves along?


WebAssign #2: Find the following derivatives (velocity vector). Write answers using correct notation.

• −→r (t) = 〈cos(2t), sin(2t)〉

• −→r (t) = 〈t2 − 1, t3 − 2t+ 1〉

• −→r (t) =⟨t, 6et/3

⟩

• −→r (t) = 〈ln(t), t−1〉


WebAssign #3: The graph below is the curve described by the vector valued function:

−→r (t) = 〈4t− t3, 2t2 − 5〉


• Show the location and motion of the object at time t = 0:


– The velocity vector is found by taking the derivative: −→v (t) = −→r ′(t)– Compute −→v (0). Then sketch the velocity vector based at the tip of −→r (0). Label it −→v (0).

• Show the location and motion of the object at time t = 1 by repeating the above steps.


WebAssign #4: The graph at right is described by

−→r (t) = 〈cos(t), sin(t)〉

Compute both −→r ′(0) and −→r ′(π/2) then graph them at right.Hint: Recall the correct location to graph a velocity vector.


−→r (t) = 〈cos(2t), sin(2t)〉

Compute both −→r ′(0) and −→r ′(π/2) then graph them at right.


−→r (t) = 〈cos(t/2), sin(t/2)〉

Compute both −→r ′(0) and −→r ′(π/2) then graph them at right.

Discussion Point: What are the differences (if any) between the above three vector functions that alldescribe the same curve?


WebAssign #9: The graph below is the curve described by the vector valued function:

−→r (t) = 〈t2 − 2, 3t− t2 + 3〉


• Show the position, velocity, and acceleration of the object at time t = 1:


– The velocity vector is found by taking the derivative: −→v (t) = −→r ′(t)– Compute −→v (1). Then sketch the velocity vector based at the tip of −→r (1). Label it −→v (1).

– The acceleration vector is found by taking the second derivative: −→a (t) = −→v ′(t) = −→r ′′(t)– Compute −→a (1). Then sketch the acceleration vector based at the tip of −→r (1). Label it −→a (1).

• Show the position, velocity, and acceleration of the object at time t = 2 by repeating the above steps.

Discussion Points:

• Why do we graph both the velocity and acceleration vectors at the tip of the position vector?

• How is the direction of the acceleration vector related to the curve the object moves along?


WebAssign #10: An object moves along a path represented by the vector valued function

−→r (t) =⟨100t, 10 + 50t− 16t2

⟩• The velocity of this object is the derivative of the position: −→v (t) = −→r ′(t).

Find the velocity of this object as a function of time, t.

• The speed of this object is the magnitude of the velocity: |−→v (t)|.Find the speed of this object as a function of time, t.

• The acceleration of this object is the derivative of the velocity: −→a (t) = −→v ′(t) = −→r ′′(t)Find the acceleration of this object as a function of time, t.

WebAssign #11: An object moves along a path represented by the vector valued function

−→r (t) =

⟨8 + 8 cos

(t

2

),−8 sin

(t

2

)⟩• Find the velocity of this object as a function of time, t.

• Find the speed of this object as a function of time, t. Fully simplify the result.

• Find the acceleration of this object as a function of time, t.

Discussion Point: What is the difference between velocity and speed?


The velocity of an object is given by −→v (t) = −→r ′(t) = 〈− sin(t), cos(t)〉

WebAssign #15: Write down all possible position vectors (family of antiderivatives) of this object.

Hint: You need to use a different constant for each component (C and D) or a single vector constant (−→C).

WebAssign #16: Find −→r (t) given the initial condition −→r (0) = 〈1, 0〉.Hint: Use the previous question and the initial condition to solve for the unknown constants.

WebAssign #17: Find −→r (t) given the initial condition −→r (0) = 〈2, 1〉.

WebAssign #18: Find −→r (t) given the initial condition −→r (0) = 〈0, 0〉.

WebAssign #19: Find −→r (t) given the initial condition −→r (0) = 〈−1, 1〉.

Discussion Point: All of the above objects have the same velocity, but different initial condition.How does the initial condition affect the path the object travels along?

Vector Applications Notes and Learning Goals 278

This lesson will apply old knowledge from Trigonometry and recent skills from Vector Valued Functionsto some physical applications.

1. Know that there are two standard descriptionsof a vector. For the vector pictured at right:

• The component form is

−→v = 〈2.2, 3.1〉

• The descriptive form, using magnitudeand direction, is

3.8 meters at an angle of 54.6◦

The component form is better for math. Thedescriptive form is better for writing.

Note: The descriptive form usually involves angles. Degrees are the standard angle measure whenwriting or describing.

Warning! However, mathematical computations, especially in Calculus, often require that you useradians. Be careful with your calculator settings.

2. Given magnitude |−→v |, and direction θ, be able find the component form. The component formulasare:

vx = |−→v | cos θ

vy = |−→v | sin θ

Warning! The formulas only work if θ is measured counter-clockwise from the positive x-axis.Often, particularly for vectors outside the first quadrant, you have to use common sense to get theright components.

Vector Applications Notes and Learning Goals 279

3. Given component form, be able to convert to descriptive form. The formulas for magnitude anddirection are:

|−→v | =√v2x + v2y

θ = arctan

(vyvx

)

Warning! The formula for θ is only true in the first or fourth quadrant AND only if θ is measuredcounter-clockwise from the positive x-axis. Often, particularly for vectors in other quadrants, youhave to use common sense to get the right angle.

4. Recall notation, vocabulary, and graphing conventions for position, velocity and acceleration fromVector Functions II. In particular:

• Velocity: −→v = −→r ′

• Speed: |−→v |• Acceleration: −→a = −→v ′ = −→r ′′

5. Given the derivative of a vector valued function, and one point on the function, be able to find thefunction. In particular,

• Given −→v (t) and one position vector, −→r0, find the formula for −→r (t).

• Given −→a (t) and one velocity vector, −→v0, find the formula for −→v (t).

Vector Applications Worksheet 1 280

WebAssign #1: A vector, −→w , has a magnitude of 10meters and a direction given by the angle θ = 50◦.

Find the components of this vector, wx and wy.Round answers to two decimal places and include units.

• Begin by drawing two sides of a right triangle.

• Then use trigonometry to find two side lengths.

WebAssign #2: Suppose −→w = 〈4, 1〉 as shown. Find themagnitude and direction (angle θ) of −→w .

• Begin by drawing a right triangle.

• Round magnitude to two decimal places with units.

• Compute θ in degrees with one decimal place.

Discussion Points:

• Explain how to use the magnitude and direction (angle θ) to write a vector in component form.

• Explain how to find the magnitude and direction (angle θ) of a vector from component form.


WebAssign #3: An object is shot into the air with aninitial speed of 20 m/s and an angle of θ = 40◦.

Write the velocity vector of this object, at the instantit is shot into the air, in component form. Use correctnotation. Round to two decimal places and include units.

Hint: Recall that speed is the magnitude of velocity.

WebAssign #4: An object is shot into the air with theinitial velocity vector:

−→v0 = 〈−45, 35〉 m/s

At the instant t = 0 find the speed and direction (angleθ) of this object. Be accurate to one decimal place.

Note: θ is measured from the positive x-axis.

Discussion Point: Assuming the above objects are being pushed downward by gravity, how is thetrajectory related to the initial velocity?


WebAssign #5: An object is shot into the air.Its position is given by the position vector

−→r (t) = 〈12t, 10t− 4.9t2〉 meters (m)

with t in seconds. The graph shows the trajectory of thisobject and the tangent vector to the curve at t = 0.

Describe the velocity vector at time t = 0 as follows.Round all answers to one decimal place and include units.

• Find the velocity of this object at time t = 0.

• What is the x-component of the velocity, vx, when t = 0?

• What is the y-component of the velocity, vy, when t = 0?

• What is the speed, | −→v |, when t = 0?

• What is the direction when t = 0? That is, find the angle of inclination, θ, in degrees.


WebAssign #6: A fixed-wing drone is traveling on a trajectorygiven by −→r (t) = 〈200t2, 1000t3 − 30t〉 km, with t in hours.

Answer the following questions at the instant t = 0.2 hours.

• Find both the x and y components of the velocity, vx and vy.

• Find the speed of this drone when t = 0.2.

• Find the heading when t = 0.2. The heading is the angle, θ,clockwise from due north (y-axis) as shown.

WebAssign #7: The trajectory of an object is given by

−→r (t) = 〈5 cos(3t)− 4, 5 sin(3t) + 6〉 m

with t in seconds.

At the instant t = 2.5 seconds, what is the speed and direction(angle θ as shown), of this object. Note: The position vectorfunction uses radians while θ is in degrees.


WebAssign #8: An object shot into the air follows thepath given by −→r (t) = 〈120t, 224 + 80t − 16t2〉 ft, with tin seconds.

• At what angle is the object shot into the air? Inother words, what is the direction at time zero?Answer in degrees, accurate to one decimal place.

• At what time t does this object hit the ground? Hint: Write down a statement (equation) aboutthe height of the object when it hits the ground and solve for t.

• At what angle does the object hit the ground? Note: Answer should be negative, accurate to onedecimal place and in degrees.


WebAssign #11: The velocity of a moving object is given by −→v (t) = −→r ′(t) = 〈2, t〉.

Find the vector valued function, −→r (t), that gives the position of the object at time t given that the positionof this object the instant when t = 2 is −→r (2) = 〈4,−1〉.

• Use −→v (t) to write the position vector function of this object using unknown constants.

• Use the initial condition, −→r (2) = 〈4,−1〉, to solve for the unknown constants.

• Write the position vector function for this object as −→r (t) = vector .

WebAssign #12: An object travels in a straight line asshown. Find the vector valued function, −→r (t), that givesthe position of this object at time t using these facts:

• −→v (t) = 〈5,−4〉 m/s.

• −→r (3) = 〈−3, 2〉 m.

Discussion Point: Why are both the velocity and an initial position needed to find an object’s trajectory?


WebAssign #13: An object is attached to a 5 meter long cord andspun in a circle around the point (-4,6) as shown at right. Theobject’s trajectory is given by

−→r (t) = 〈5 cos(3t)− 4, 5 sin(3t) + 6〉 m

with t in seconds.

At t = 2.5 seconds the cord is cut. From that instant onward, theobject moves in a straight line, −→s (t), in the direction of its tangentvector, with constant velocity. Find −→s (t) as follows:

• Find the position vector at t = 2.5 s, −→s (2.5) = −→r (2.5).

• Find the velocity vector at t = 2.5 s, −→s ′(2.5) = −→r ′(2.5).

• Use the constant velocity and initial position above to find −→s (t).Hint: Use the process from the previous worksheet.

• Where is the object located 2 seconds after the cord is cut?


WebAssign #15: An object is fired into the air at anangle of 35◦ and a speed of 65 ft/s. This object has aconstant acceleration of −→a (t) = −→r ′′(t) = 〈0,−32.2〉 ft/s2

due to gravity. Answer the following questions. Keep twodecimal places of accuracy for all numerical coefficients.

• Find the initial position vector of this object, −→r (0).

• Find the initial velocity vector of this object, −→v (0).

• Use the acceleration and initial velocity to find the velocity function for this object, −→v (t).

• Use the velocity and initial position to find the position function for this object, −→r (t).


WebAssign #16: A BMX bike jumps off a 10meter tall ramp, with an angle of inclination of35◦, and a speed of 12 m/s as shown.

Let t be time in seconds, and place a coordinatesystem (with x and y in meters) at the bottomof the 10 m ramp as shown.

At time t = 0 the bike leaves the ramp and follows the trajectory given by the dashed line. While in theair, the bike experiences a constant acceleration (due to gravity) of −→a (t) = 〈0,−9.8〉 m/s2.

Use the above to find the position vector function, −→r (t), for this bike while it is in the air.

Vector Design Notes and Learning Goals 289

This lesson will use vector equations of motion to solve some design problems. The physical situation willbe very similar to situations explored in Vector Apps. However, the problem solving will be significantlydifferent.In this assignment you will need the following skills:

• Use vector equations that involve unknown physical constants.

• Discover algebraic equations that you could use to solve for unknown physical constants.

• Solve those equations.

Vector Design Worksheet 1 290

WebAssign #1: An object is launched into the air fromthe origin, −→r (0) = 〈0, 0〉, with an initial velocity

−→v 0 = −→r ′(0) = 〈a, b〉 m/s

where a and b are unknown constants.

Let t be time in seconds.

After the object is launched, it has an acceleration (due to gravity) of −→a (t) = 〈0,−9.8〉 m/s2.

• Use the acceleration and the initial velocity to find the velocity function of this object as a functionof t. Answer will include the unknown constants a and b.

• Use the velocity and the initial position to find the position function of this object as a function oft. Answer will include the unknown constants a and b.


WebAssign #2: An object shot into the air followsa path given by

−→r (t) =⟨at, bt− 4.9t2

⟩m

with t in seconds and a and b unknown constants.

The object is launched with a speed of 500 m/s at an angle of 30◦.Find the x-coordinate when the object hits the ground as follows:

• Find the initial velocity of this object in two ways:

– Use the formula for −→r (t) to find the initial velocity in terms of a and b.

– Use the launch information to write down the initial velocity of this object.

• Use the above to determine a and b and write down the position vector function for this object.

• Use −→r (t) to find the x-coordinate when the object hits the ground downrange.


The next two problems use the following: An object shot into the air follows a path given by

−→r (t) =⟨at, bt− 4.9t2

⟩m

with t in seconds and a and b unknown constants.

WebAssign #3: An object is launched into the airwith an unknown speed at an angle of 30◦. Find thespeed needed to launch the object into the air suchthat it hits the ground 15,000 meters downrange.

WebAssign #4: An object is launched into the airwith a speed of 500 m/s at an unknown angle. Findthe angle needed to launch the object into the air suchthat it hits the ground 14,000 meters downrange.

Discussion Point: How do unknown constants in formulas help answer more complicated questions?


WebAssign #5: An object is attached to a 5 meter long cord andspun in a circle around the origin. The object’s trajectory is

−→r (t) =

⟨5 cos(t), 5 sin(t)

⟩m

with t in seconds.

The instant the object reaches the location (a,b), the cord is cut.From that instant onward the object moves in a straight line, −→s (t),in the direction of its tangent vector.

• Find the tangent vector at the instant the cord is cut. Writeanswer using a and b without trigonometric functions.

• Write a formula for −→s (t) using the above velocity and initial condition −→s (0) = 〈a, b〉.Your answer will involve a, b, and t.

WebAssign #6: Consider the object and motion described above. Find the release point, (a, b), that willresult in the object traveling through the point (0,10)?


WebAssign #7: A BMX bike jumps off a 10meter tall ramp, with an angle of inclination of35◦, and a speed of 12 m/s as shown.

Let t be time in seconds, and place a coordinatesystem (with x and y in meters) at the bottomof the 10 m ramp as shown.

At time t = 0 the bike leaves the ramp and follows the trajectory given by the dashed line. While in theair, the bike experiences a constant acceleration (due to gravity) of −→a (t) = 〈0,−9.8〉 m/s2.

In the last lesson you found the vector formula for the trajectory of this bike. Use that result to design alanding ramp for the bike that is 6 meters high. Find the distance L and the angle θ such that the rampis tangent to the bikes trajectory the instant it lands.


WebAssign #8: An object is shot into the air with aninitial speed of 500 meters per second and an initialheight of 1000 meters above the ground as shown.

This object has an acceleration (due to gravity) of

−→a (t) = 〈0,−9.8〉 m/s2

Find the launch angle θ such that the object hits the ground 14,000 meters downrange.Hint: Review the first three worksheets of this lesson.

Arc Length Notes and Learning Goals 296

Suppose that a particle travels along a path with vector valued position function −→r (t).This lesson is concerned with the distance traveled along that path, also known as arc length.Here are things you should know.

1. For constant speed in a straight line,

distance = (speed)× (time)

2. If either speed is variable OR the path is curved, then you must use a tiny amount of time, dt, andyou must compute a tiny amount of distance, ds. The formula is

ds = (speed) dt

In vector notation this isds = |−→v | dt

Note: You will not be asked to communicate slicingstrategy for this. However, if you were asked....

• The axis of integration is time, or t.

• A tiny slice of time is dt. But since there is no timeaxis visible you can’t label this.

• A tiny slice of distance is pictured and labeled atright.

3. Total distance traveled, or total arc length, is computed using an integral.

distance =

∫ b

a

|−→v | dt

Since the axis of integration is time, the bounds are time values:

a = Starting time

b = Ending time

Arc Length Notes and Learning Goals 297

Example: An object is launched from a height of 192 feet. Its position as a function of time is

−→r (t) = 〈80t, 192 + 64t− 16t2〉

with t in seconds. The object’s trajectory is shown below. It lands on the ground at the instant t = 6seconds. Compute the total distance the object travels from launch to landing.

Solution:

1. Compute tiny distance.

−→r (t) = 〈80t, 192 + 64t− 16t2〉 (given)−→v (t) = 〈80, 64− 32t〉 (derivative of −→r )

|−→v (t)| =√

802 + (64− 32t)2 (length of −→v )

ds =√

802 + (64− 32t)2 dt (speed × time)

2. Write an integral for total distance.

distance =

∫ 6

0

√802 + (64− 32t)2 dt

3. Use a calculator or a computer to finish the computation.1

distance ≈ 602.5 m

1A fanatical Calculus II student might attempt this by hand, starting with the trig substitution t = 2 + 2.5 tan θ.

Arc Length Worksheet 1 298

WebAssign #1: An object moves along an elliptical pathwith position given by

−→r (t) =

⟨3 cos(t), 2 sin(t)

⟩ft

with t in seconds and both axes measured in feet.

Find the arc length on the interval 0 ≤ t ≤ π/2 by slicingits length into tiny segments of length ds. Find ds asfollows:

• Find the speed of the parameterization, |−→r ′(t)|.Answer is a formula involving t.

• Multiply the speed by a tiny time, dt, to get distance: ds = |r′(t)|dt.Write a formula for the tiny bit of distance, ds, shown in the figure.

• Find the bounds of integration, the values of t that describe both ends of the curve.

• Write an integral for the total arc length.

• Compute the integral to find the total distance traveled on the time interval.

Note: Often times (such as this example) arc length integrals cannot be computed by hand. Inthese cases, students can use computers/calculators to approximate the integral.


WebAssign #2: A fixed-wing drone aircraft is travelingon a trajectory given by

−→r (t) =

⟨200t2, 1000t3 − 30t

⟩km

with t in hours. The trajectory is shown, along with theposition and velocity vector at time t = 0.2 hrs.

Show the work to find the total distance the aircraft trav-els on time interval 0 ≤ t ≤ 0.2. Start by slicing the dis-tance into tiny lengths ds. Be accurate to two decimalplaces and include units.

WebAssign #3: An object shot into the air has positiongiven by

−→r (t) =

⟨120t, 224 + 80t− 16t2

⟩ft

with t in seconds.

Find the total distance this object traveled. Be accurateto one decimal place and include correct units.


WebAssign #4: A fixed-wing drone aircraft is travelingon a trajectory given by

−→r (t) =

⟨200t2, 1000t3 − 30t

⟩km

with t in hours.

When the aircraft reaches the coordinates (18 km, 18km), what is the total distance that it has traveled sinceleaving the origin? Show all work.

WebAssign #5: An object shot into the air follows thepath given by

−→r (t) =

⟨at, bt− 4.9t2

⟩m

with t in seconds and a and b are unknown constants.

The launch velocity is 500 m/s at an angle of 30◦. What is the total distance traveled through the air?


WebAssign #6: A BMX bike jumps off a 10 meter tallramp, with an angle of inclination of 35 degrees, and aspeed of 12 m/s as shown.

In previous problems you should have found the equa-tions of motion and designed the landing ramp. Youwill need your answers from those problems to work thisproblem. Show all work and find the total distance thatthe bike traveled while airborne.

WebAssign #7: An object shot into the air follows thepath given by

−→r (t) =

⟨at, 1000 + bt− 4.9t2

⟩m

with t in seconds and a and b are unknown constants.

The launch speed is 500 m/s. The launch point is at a height of 1000 meters, and the object lands 20,000meters downrange. Show all work and find the total distance the object traveled through the air.

Vectors Review Notes and Learning Goals 302

There is a short review of vectors. Vectors will only cover about 30% of the final exam. Be sure to readthe exam docs (rules, study guide, formula sheet, and topic list) for a full list of what to study.

• Be able to tell the difference between vectors, scalars, and points.

• Be able to find the magnitude and direction (angle) of a vector in component form. In addition beable to find the components of a vector given a magnitude and direction (angle).

• Be able to add vectors and multiply them by a scalar both algebraically and geometrically.

• Be able to find and work with vector equations of lines.

• Be able to graph position vector valued functions.

• Be able to find derivatives and antiderivatives of vector valued functions.

• Know what position vectors, velocity vectors, speed, and acceleration vectors represent,

• Be able to use vectors in application problems talking about position, velocity, speed, and/or accel-eration.

Vectors Review Worksheet 1 303

Consider the two vectors −→v = 〈4, 3〉 and −→w = 〈−2, 1〉.Draw correct tip-to tail diagrams (in the proper order) for each of the following vectors on the grid provided.Be sure to include properly labeled and scaled axes. Get feedback on your graphs.

−→w +−→v

−→v + 2−→w

−→v −−→w

1

2−→v − 3−→w

Vectors Review Worksheet 2 304

The path of an object is given by the position vector function

−→r (t) =

⟨t2

4,t3

8− t⟩

Graph this function on the axes below.

• Use proper labeling:

– Labeling some points with appropriate t = value .

– Use arrows to the direction of increasing t.

• Include the graph of the position, velocity and acceleration vectors at the time t = −2.

• Get feedback on your graph.

Appendix Algebraic Rules 305

Basic Properties

a+ b = b+ a

a+ 0 = a

a− b = a+ (−b)

a− a = a+ (−a) = 0

a(b+ c) = ab+ ac

ab = ba

a · 1 = a

a/b = a · (1/b)

a/a = a · (1/a) = 1

a · 0 = 0

Fraction Properties

a

b± c

d=ad± bcbd

a

b= a · 1

b

a

bc=

1

b· ac

=a

c· 1

b

a

bc

=a

b· 1

c=

a

bc

a

bc

d

=a

b· dc

=ad

bc

a

b· cd

=ac

bd

a+ b

c=

1

c(a+ b)

ab

c= a · b

c=a

c· b

a

b

c

= a · cb

=ac

b

Exponential Properties

a0 = 1

an · am = an+m

(an)m = anm

an

bm= an−m

1

an= a−n

n√a = a1/n

n√am = am/n(ab

)n=an

bn

Logarithmic Properties

ln(1) = 0

ln(ab) = ln(a) + ln(b)

ln(ab

)= ln(a)− ln(b)

ln(an) = n ln(a)

eln(x) = ln(ex) = x

logb(x) =ln(x)

ln(b)

Useful Identities

(a+ b)2 = a2 + 2ab+ b2

(a+ b)3 = a3 + 3a2b+ 3ab2 + b3

a2 − b2 = (a− b)(a+ b)

a3 − b3 = (a− b)(a2 + ab+ b2)

Lines

Slope: m =y2 − y1x2 − x1

Slope-Intercept Form: y = mx+ b

Point-Slope Form: y − y0 = m(x− x0)

Quadratics

Quadratic Function: y = ax2 + bx+ c

Vertex Form: y = a(x− h)2 + k

Quadratic Equation: ax2 + bx+ c = 0

Quadratic Formula: x =−b±

√b2 − 4ac

2a

Interval Notation

Open Intervals:(a, b) = {x : a < x < b}(a,∞) = {x : x > a}(−∞, b) = {x : x < b}

Closed Intervals:[a, b] = {x : a ≤ x ≤ b}[a,∞) = {x : x ≥ a}(−∞, b] = {x : x ≤ b}

Half-Open Intervals:(a, b] = {x : a < x ≤ b}[a, b) = {x : a ≤ x < b}

Appendix Trigonometric Rules 306

Right Triangle Definition

sin(θ) =Opposite

Hypotenuse

cos(θ) =Adjacent

Hypotenuse

tan(θ) =Opposite

Adjacent

csc(θ) =Hypotenuse

Opposite

sec(θ) =Hypotenuse

Adjacent

cot(θ) =Adjacent

Opposite

Special Triangles

Basic Trigonometric Identities

tan(θ) =sin(θ)

cos(θ)

sec(θ) =1

cos(θ)

sin2(θ) + cos2(θ) = 1

cot(θ) =1

tan(θ)

csc(θ) =1

sin(θ)

tan2(θ) + 1 = sec2(θ)

Unit Circle Definition

sin(θ) =y

1= y

cos(θ) =x

1= x

tan(θ) =y

x

csc(θ) =1

y

sec(θ) =1

x

cot(θ) =x

y

Unit Circle

More Trigonometric Identities

sin(θ + φ) = sin(θ) cos(φ)− sin(φ) cos(θ)

sin(2θ) = 2 sin(θ) cos(θ)

sin2(θ) =1

2− 1

2cos(2θ)

sin(θ) sin(φ) =1

2cos(θ − φ)− 1

2cos(θ + φ)

sin(θ) cos(φ) =1

2sin(θ − φ) +

1

2sin(θ + φ)

cos(θ + φ) = cos(θ) cos(φ)− sin(θ) sin(φ)

cos(2θ) = cos2(θ)− sin2(θ)= 2 cos2(θ)− 1= 1− 2 sin2(θ)

cos2(θ) =1

2+

1

2cos(2θ)

cos(θ) cos(φ) =1

2cos(θ − φ) +

1

2cos(θ + φ)

Appendix Calculus Rules 307

Elementary Derivatives

d

dx

(xn)

= nxn−1

d

dx

(lnx)

=1

x

d

dx

(ex)

= ex

d

dx

(sinx

)= cosx

d

dx

(cosx

)= − sinx

Elementary Derivatives with Linear Insides

d

dx

((ax+ b)n

)= a · n(ax+ b)n−1

d

dx

(ln(ax+ b)

)= a · 1

(ax+ b)

d

dx

(eax+b

)= a · eax+b

d

dx

(sin(ax+ b)

)= a · cos(ax+ b)

d

dx

(cos(ax+ b)

)= −a · sin(ax+ b)

Derivative Rules

Linearity Rule

d

dx

(af(x) + bg(x)

)= af ′(x) + bg′(x)

Product Rule

d

dx

(f(x)g(x)

)= f ′(x)g(x) + f(x)g′(x)

Chain Ruled

dx

(f(g(x)

))= f ′

(g(x)

)g′(x)

Quotient Rule

d

dx

(f(x)

g(x)

)=g(x)f ′(x)− f(x)g′(x)(

g(x))2

Elementary Antiderivatives

∫xn dx =

1

n+ 1xn+1 + C; n 6= −1∫

1

xdx =

∫x−1 dx = ln |x|+ C∫

ex dx = ex + C∫sinx dx = − cosx+ C∫cosx dx = sinx+ C

Antiderivatives with Linear Insides

∫(ax+ b)n dx =

1

a· 1

n+ 1(ax+ b)n+1 + C∫

1

ax+ bdx =

1

aln |ax+ b|+ C∫

eax+b dx =1

aeax+b + C∫

sin(ax+ b) dx = −1

acos(ax+ b) + C∫

cos(ax+ b) dx =1

asin(ax+ b) + C

Linearity Rule for Antiderivatives

∫ (af(x) + bg(x)

)dx = a

∫f(x) dx+ b

∫g(x) dx

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