## 550 AP Calculus AB & BC Practice Questions

**Editorial**

Rob Franek, Senior VP, Publisher

Mary Beth Garrick, Director of Production

Selena Coppock, Senior Editor

Calvin Cato, Editor

Kristen O’Toole, Editor

Meave Shelton, Editor

Alyssa Wolff, Editorial Assistant

**Random House Publishing Team**

Tom Russell, Publisher

Alison Stoltzfus, Publishing Director

Ellen L. Reed, Production Manager

Dawn Ryan, Managing Editor

Erika Pepe, Associate Production Manager

Kristin Lindner, Production Supervisor

Andrea Lau, Designer

The Princeton Review

111 Speen Street, Suite 550

Framingham, MA 01701

E-mail: editorialsupport@review.com

Copyright © 2013 by TPR Education IP Holdings, LLC

Cover art © Jonathan Pozniak

All rights reserved. Published in the United States by Random House LLC, New York, a Penguin Random House Company, and in Canada by Random House of Canada Limited, Toronto, Penguin Random House Companies.

The Princeton Review is not affiliated with Princeton University.

AP and Advanced Placement Program are registered trademarks of the College Board, which does not sponsor or endorse this product.

eBook ISBN: 978-0-8041-2446-1

Trade Paperback ISBN: 978-0-8041-2445-4

The Princeton Review is not affiliated with Princeton University.

Editor: Calvin S. Cato

Production Editor: Jesse Newkirk

Production Coordinator: Deborah A. Silvestrini

v3.1

# Acknowledgments

The Princeton Review would like to give a very special thanks to Bikem Polat and Chris Knuth for their hard work on the creation of this title. Their intimate subject knowledge and thorough fact-checking has made this book possible. In addition, The Princeton Review thanks Jesse Newkirk for his hard work in copy editing the content of this title.

# About the Authors

Bikem Ayse Polat has been teaching and tutoring through The Princeton Review since 2010. She came to TPR as an undergraduate at the University of California, San Diego from which she graduated in 2011 with two Bachelors of Science degrees in Bioengineering: Pre-Med and Psychology. After graduation, she moved to Cincinnati, OH for a year where she certified to tutor online and was promoted to Master Tutor. She is currently residing in Philadelphia, PA where she is a graduate student at Temple University, working on her Ph.D. in Urban Education. She is certified to teach and tutor numerous Advanced Placement courses as well as the PSAT, SAT, ACT, LSAT, GRE, and MCAT. In her spare time, Bikem enjoys spending time with her dog, Gary, and her nieces, Laila and Serra.

Chris Knuth have been with the Princeton Review for 9 years and has been teaching Math since the 5th grade through various tutoring organizations and classes. He is certified to teach various Advanced Placement courses as well as SAT, ACT, GMAT, GRE, LSAT, DAT, OAT, PSAT, SSAT, and ISEE. He considers Calculus to be his favorite level of Math. Chris would like to thank his family for their unwavering support and always being the “wind in my sail” and give a very special thank you to the teachers who made math fun, Gai Williams and Anita Genduso.

# Contents

*Cover*

*Title Page*

*Copyright*

*Acknowledgments*

*About the Authors*

**Part I: Using This Book to Improve Your AP Score**

Preview: Your Knowledge, Your Expectations

Your Guide to Using This Book

How to Begin

AB Calculus Diagnostic Test

AB Calculus Diagnostic Test Answers and Explanations

Answer Key

Explanations

BC Calculus Diagnostic Test

BC Calculus Diagnostic Test Answers and Explanations

Answer Key

Explanations

**Part II: About the AP Calculus Exams**

AB Calculus vs BC Calculus

The Structure of the AP Calculus Exams

Overview of Content Topics

How AP Exams Are Used

Other Resources

Designing Your Study Plan

**Part III: Test-Taking Strategies for the AP Calculus Exams**

How to Approach Multiple Choice Questions

How to Approach Free Response Questions

Derivatives and Integrals That You Should Know

Prerequisite Mathematics

**Part IV: Drills**

**Part V: Practice Exams**

# Part I

# Using This Book to Improve Your AP Score

• Preview: Your Knowledge, Your Expectations

• Your Guide to Using This Book

• How to Begin

• AB Calculus Diagnostic Test

• AB Calculus Diagnostic Test Answers and Explanations

• BC Calculus Diagnostic Test

• BC Calculus Diagnostic Test Answers and Explanations

## PREVIEW: YOUR KNOWLEDGE, YOUR EXPECTATIONS

Your route to a high score on the AP Calculus Exam depends a lot on how you plan to use this book. Start thinking about your plan by responding to the following questions.

- Rate your level of confidence about your knowledge of the content tested by the AP Calculus Exam:
A. Very confident—I know it all

B. I’m pretty confident, but there are topics for which I could use help

C. Not confident—I need quite a bit of support

D. I’m not sure

- If you have a goal score in mind, circle your goal score for the AP Calculus Exam:
- What do you expect to learn from this book? Circle all that apply to you.
A. A general overview of the test and what to expect

B. Strategies for how to approach the test

C. The content tested by this exam

D. I’m not sure yet

## YOUR GUIDE TO USING THIS BOOK

This book is organized to provide as much—or as little—support as you need, so you can use this book in whatever way will be most helpful for improving your score on the AP Calculus Exam.

- The remainder of
**Part One**will provide guidance on how to use this book and help you determine your strengths and weaknesses. **Part Two**of this book will- provide information about the struxcture, scoring, and content of the AP Calculus Exam.
- help you to make a study plan.
- point you towards additional resources.

**Part Three**of this book will explore various strategies:- how to attack multiple choice questions
- how to write a high-scoring free response answer
- how to manage your time to maximize the number of points available to you

**Part Four**of this book contains practice drills covering all of the AB Calculus and BC Calculus concepts you will find on the exams.**Part Five**of this book contains practice tests.

You may choose to use some parts of this book over others, or you may work through the entire book. This will depend on your needs and how much time you have. Let’s now look how to make this determination.

## HOW TO BEGIN

**Take a Test**Before you can decide how to use this book, you need to take a practice test. Doing so will give you insight into your strengths and weaknesses, and the test will also help you make an effective study plan. If you’re feeling test-phobic, remind yourself that a practice test is a tool for diagnosing yourself—it’s not how well you do that matters but how you use information gleaned from your performance to guide your preparation.

So, before you read further, take the AP Calculus AB Diagnostic Test starting at this page of this book or take the AP Calculus BC Diagnostic Test starting on this page. Be sure to do so in one sitting, following the instructions that appear before the test.

**Check Your Answers**Using the answer key on this page (for Calculus AB) or this page (for Calculus BC), count how many multiple choice questions you got right and how many you missed. Don’t worry about the explanations for now, and don’t worry about why you missed questions. We’ll get to that soon.

**Reflect on the Test**After you take your first test, respond to the following questions:

- How much time did you spend on the multiple choice questions?
- How much time did you spend on each free response question?
- How many multiple choice questions did you miss?
- Do you feel you had the knowledge to address the subject matter of the essays?
- Do you feel you wrote well organized, thoughtful essays?
- Circle the content areas that were most challenging for you and draw a line through the ones in which you felt confident/did well.
- Functions, Graphs, and Limits
- Differential Calculus
- Integral Calculus
- Polynomial Approximations and Series (for BC Calculus Students)
- Applications of Derivatives
- Applications of Integrals

**Read Part Two and Complete the Self-Evaluation**As discussed in the Goals section above, Part Two will provide information on how the test is structured and scored. It will also set out areas of content that are tested.

As you read Part Two, re-evaluate your answers to the questions above. At the end of Part Two, you will revisit and refine the questions you answered above. You will then be able to make a study plan, based on your needs and time available, that will allow you to use this book most effectively.

**Engage with the Drills as Needed**Notice the word

*engage*. You’ll get more out of this book if you use it intentionally than if you read it passively, hoping for an improved score through osmosis.The drills are designed to give you the opportunity to assess your mastery of calculus concepts through test-appropriate questions.

**Take Test 2 and Assess Your Performance**Once you feel you have developed the strategies you need and gained the knowledge you lacked, you should take one of the practice exams at the end of this book. You should do so in one sitting, following the instructions at the beginning of the test. When you are done, check your answers to the multiple choice sections.

Once you have taken the test, reflect on what areas you still need to work on, and revisit the drills in this book that address those topics. Through this type of reflection and engagement, you will continue to improve.

**Keep Working**As you work through the drills, consider what additional work you need to do and how you will change your strategic approach to different parts of the test.

If you do need more guidance, there are plenty of resources available to you. Our

*Cracking the AP Calculus AB & BC Exams*guide gives you a comprehensive review of all the calculus topics you need to know for the exam and offers 5 practice tests (3 for AB and 2 for BC). In addition, you can go to the AP Central website for more information about exam schedules and calculus concepts.

# AB Calculus Diagnostic Test

(*Click here to download a PDF of Diagnostic Test*)

**AP**^{®} Calculus AB Exam

^{®}Calculus AB Exam

**DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.**

**Instructions**

Section I of this examination contains 45 multiple-choice questions. Fill in only the ovals for numbers 1 through 45 on your answer sheet.

CALCULATORS MAY NOT BE USED IN THIS PART OF THE EXAMINATION.

Indicate all of your answers to the multiple-choice questions on the answer sheet. No credit will be given for anything written in this exam booklet, but you may use the booklet for notes or scratch work. After you have decided which of the suggested answers is best, completely fill in the corresponding oval on the answer sheet. Give only one answer to each question. If you change an answer, be sure that the previous mark is erased completely. Here is a sample question and answer.

Sample Question

Chicago is a

(A) state

(B) city

(C) country

(D) continent

(E) village

Sample Answer

Use your time effectively, working as quickly as you can without losing accuracy. Do not spend too much time on any one question. Go on to other questions and come back to the ones you have not answered if you have time. It is not expected that everyone will know the answers to all the multiple-choice questions.

**About Guessing**

Many candidates wonder whether or not to guess the answers to questions about which they are not certain. Multiple choice scores are based on the number of questions answered correctly. Points are not deducted for incorrect answers, and no points are awarded for unanswered questions. Because points are not deducted for incorrect answers, you are encouraged to answer all multiple-choice questions. On any questions you do not know the answer to, you should eliminate as many choices as you can, and then select the best answer among the remaining choices.

CALCULUS AB

SECTION I, Part A

Time—55 Minutes

Number of questions—28

A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION

Directions: Solve each of the following problems, using the available space for scratchwork. After examining the form of the choices, decide which is the best of the choices given and fill in the corresponding oval on the answer sheet. No credit will be given for anything written in the test book. Do not spend too much time on any one problem.

In this test: Unless otherwise specified, the domain of a function *f* is assumed to be the set of all real numbers *x* for which *f*(*x*) is a real number.

1.

(A) 1

(B) 2

(C) 0

(D) nonexistent

(E) 2π

2.

(A)

(B) 0

(C) ∞

(D) 3

(E) The limit does not exist.

3. At what point does the following function have a removable discontinuity?

*f*(*x*) =

(A) (–5,–1)

(B) (–2,–1)

(C) (–2,1)

(D) (1,1)

(E) (1,–1)

4. Which of the following functions is NOT continuous at *x* = –3?

(A) *f*(*x*) =

(B) *g*(*x*) =

(C) *h*(*x*) =

(D) *j*(*x*) =

(E) *k*(*x*) = (*x* + 3)^{2}

5. Which of the following functions is continuous at *x* = –3?

(A) *f*(*x*) =

(B) *g*(*x*) =

(C) *h*(*x*) =

(D) *j*(*x*) = (*x* − 2)^{}

(E) *k*(*x*) = ()^{2}

6. What is ?

(A)

(B) 0

(C)

(D) 1

(E) The limit does not exist.

7. If *f*(*x*) = (2*x*^{3} + 33)( − 2*x*), then *f*′(*x*) =

(A) (2*x*^{3} + 33) + 6*x*^{2}( − 2*x*)

(B) (2*x*^{3} + 33) + 6*x*^{3}( − 2*x*)

(C) (2*x*^{3} + 33) + 6*x*^{2}( − 2*x*)

(D) (2*x*^{3} + 33) + 6*x*^{2}( − 2*x*)

(E) (2*x*^{3} + 33) + 6*x*^{3}( − 2*x*)

8. If *y* = , then =

(A)

(B)

(C)

(D)

(E)

9. Find the second derivative of *x*^{2}*y*^{2} = 2 at (2,1).

(A) 1

(B) –2

(C)

(D) 2

(E) −

10. If the line *y = ax*^{2} + *bx* + *c* goes through the point (2,1) and is normal to *y* = *x* + 2 at the point (0,2), then *a* = ?

(A) −

(B)

(C)

(D) −

(E) 2

11. If *f*(*x*) = 2*g*(*x*) and if *h*(*x*) = *x*^{3}, then *f*(*h*(*x*)) =

(A) 6*x*^{2}*g*(*x*^{3})

(B) 2*g*(*x*^{3})

(C) 2*x*^{2}*g*(*x*^{3})

(D) 6*g*(*x*^{3})

(E) 2*x*^{3}*g*(*x*^{3})

12. Which of the following statements about the function given by *f*(*x*) = *x*^{5} − 2*x*^{3} is true?

(A) The function has no relative extrema.

(B) The graph has one point of inflection and two relative extrema.

(C) The graph has three points of inflection and one relative extremum.

(D) The graph has three points of inflection and two relative extrema.

(E) The graph has two points of inflection and two relative extrema.

13. ∫(2*x* − 5)^{3} *dx* =

(A) + *C*

(B) + *C*

(C) + *C*

(D) + *C*

(E) + *C*

14.

(A) 4*x*^{20} + *C*

(B) + *C*

(C) + *C*

(D) + *C*

(E) + *C*

15. Find the average value of *f*(*x*) = 3*x*^{2} sin *x*^{3} on the interval [0, ].

(A)

(B) 2

(C)

(D)

(E) 1

16. Find *t*^{2} + 4*t dt*.

(A) 9*x*^{4} + 12*x*^{2}

(B) 6*x*(9*x*^{3} + 12*x*^{2})

(C) 6*x*^{2}(9*x*^{4} + 12*x*^{2})

(D) 90*x*^{3}

(E) 54*x*^{5} + 72*x*^{3}

17.

(A) −ln|cos *x*| + *C*

(B) ln|sin *x*| + *C*

(C) ln|sec *x* + tan *x*| + *C*

(D) −ln|csc *x* + cot *x*| + *C*

(E) −ln|sin *x*| + *C*

18. What is the area between *y = x*^{3} and *y = x*?

(A) 0

(B) 1

(C)

(D)

(E)

19. Find the volume of the region bounded by *y* = (*x* – 5)^{3}, the *x*-axis, and the line *x* = 10 as it is revolved around the line *x* = 2. Set up, but do not evaluate the integral.

(A) 2π *x*(*x* − 5)^{3} *dx*

(B) 2π (*x* − 2)(*x* − 5)^{3} *dx*

(C) 2π *x*(*x* − 5)^{3} *dx*

(D) 2π (*x* − 2)(*x* − 5)^{3} *dx*

(E) 2π (*x* − 2)(*x* − 5)^{3} *dx*

20. ∫3*x*^{2}(*x*^{3} − 3)^{7} *dx*

(A) 8(*x*^{3} – 3)^{8} + *C*

(B) + *C*

(C) (*x*^{3} – 3)^{8} + *C*

(D) *x*^{3} + *C*

(E) *x*^{3}(*x*^{3} – 3)^{8} + *C*

21. Find the equation for the normal line to *y* = 3*x*^{2} – 6*x* at (2,0).

(A) *y* = −6*x* −

(B) *y* = −*x* +

(C) *y* = *x* =

(D) *y* = 6*x* + 3

(E) *y* = −*x* −

22.

(A) 0

(B) –1

(C) −∞

(D) ∞

(E) 1

23. (2 − *t*^{2})*dt* =

(A) 6*x*^{2} + 3*x*^{8}

(B) 6*x*^{2} – 9*x*^{4}

(C) 2 – *x*^{2}

(D) 6*x*^{2} – 3*x*^{8}

(E) 2 – *x*^{6}

24.

(A) + *C*

(B) + *C*

(C) + *C*

(D) + *C*

(E) + *C*

25. Find for *f*′ (*x*) = *x*^{3} + 2*x* when *x* = 1.

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6

26. (*x*^{3} + 2*x*^{2} − 14) =

(A) 6*x*

(B) 6*x* + 4

(C) 3*x*^{2} + 4*x*

(D) 3*x*^{2}

(E) 4*x*

27. Find the absolute maximum on the interval [–2, 2] for the curve *y* = 4*x*^{5} – 10*x*^{2} – 8.

(A) –2

(B) –1

(C) 0

(D) 1

(E) 2

28. Find at (–1,–2) if 3*x*^{3} – 2*x*^{2} + *x* = *y*^{3} + 2*y*^{2} + 3*y*.

(A)

(B) 2

(C) −

(D) –2

(E) 0

**END OF PART A, SECTION IIF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON PART A ONLY.DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO.**

CALCULUS AB

SECTION I, Part B

Time—50 Minutes

Number of questions—17

A GRAPHING CALCULATOR IS REQUIRED FOR SOME QUESTIONS ON THIS PART OF THE EXAMINATION

Directions: Solve each of the following problems, using the available space for scratchwork. After examining the form of the choices, decide which is the best of the choices given and fill in the corresponding oval on the answer sheet. No credit will be given for anything written in the test book. Do not spend too much time on any one problem.

In this test:

1. The **exact** numerical value of the correct answer does not always appear among the choices given. When this happens, select from among the choices the number that best approximates the exact numerical value.

2. Unless otherwise specified, the domain of a function *f* is assumed to be the set of all real numbers *x* for which *f*(*x*) is a real number.

29. If *f*(*x*) = *x*^{−3} + 3 + 5π − *e*^{2}, then *f*′(*x*) =

(A) − 3*x*^{4}

(B) − 3*x*^{−4}

(C) 3*x*^{−4} +

(D)

(E) 3*x*^{2} +

30. Find the value of *c* that satisfies Rolle’s theorem for *f*(*x*) = 2*x*^{4} − 16*x* on the interval [0,2].

(A) 2

(B) (−2)^{−}

(C) 2^{}

(D) (−2)^{}

(E) 2^{}

31. Find the absolute maximum of *y* = *x*^{3} − *x*^{2} − 7*x* on the interval [–2,2].

(A)

(B)

(C) 0

(D)

(E) −

32. Approximate the area under the curve *y* = *x*^{2} + 2 from *x* = 1 to *x* = 2 using four right-endpoint rectangles.

(A) 4.333

(B) 3.969

(C) 4.719

(D) 4.344

(E) 4.328

33. Approximate the area under the curve *y* = *x*^{2} + 2 from *x* = 1 to *x* = 2 using four inscribed trapezoids.

(A) 4.333

(B) 3.969

(C) 4.719

(D) 4.344

(E) 4.328

34. Evaluate .

(A) 1

(B) 2

(C) –1

(D)

(E) 0

35. Suppose *F*(*x*) = . What is the change in *F*(*x*) as *t* increases from 1 to 4.

(A) 72

(B) 71.25

(C) 24.75

(D) 6

(E) 0.75

36. In the *xy*-plane, 2*x* + *y* = *k* is tangent to the graph of *y* = 2*x*^{2} – 8*x* + 14. What is the value of *k*?

(A)

(B)

(C) 5

(D)

(E)

37. The function *f* is continuous on the closed interval [0,4] and twice differentiable over the open interval, (0, 4). If *f*′(2) = –5 and *f*″(*x*) > 0 over the interval (0, 4), which of the following could be a table of values for *f*?

(A)

(B)

(C)

(D)

(E)

38. When is the particle whose path is described by *x*(*t*) = 2*t*^{3} − *t*^{2} + 9*t* − 16, from *t* > 0, slowing down?

(A) 0 < *t* < 3

(B) < *t* < 3

(C) < *t* <

(D) *t* > 3

(E) < *t* < 3

39. What is the are enclosed by the curve *f*(*x*) = 4*x*^{2} − *x*^{4} and the *x*-axis.

(A)

(B)

(C)

(D)

(E) 0

40. Which of the following is an asymptote for the curve *y* = ?

(A) *x* = 7

(B) *y* = 3

(C) *y* = –7

(D) *x* = 3

(E) *x* = –3

41.

(A) + *C*

(B) 2 ln|*x*^{2} − 7| + *C*

(C) ln|*x*^{2} − 7| + *C*

(D) ln|*x*^{2} − 7| + *C*

(E) ln|*x*| + *C*

42. What is the area between the curves *y* = *x*^{3} – 2*x*^{2} – 5*x* + 6 and *y* = *x*^{2} – *x* – 6 from *x* = –2 to *x* = 3?

(A) 30

(B)

(C) 32

(D)

(E) 34

43. Find the average value of *f*(*x*) = (3*x* − 1)^{3} on the interval from *x* = –1 to *x* = 3.

(A) 10.667

(B) 12

(C) 9.333

(D) 15

(E) 18

44. The curve *y* = *ax*^{2} + *bx* + *c* passes through the point (1,5) and is normal to the line –*x* + 5*y* = 15 at (0,3). What is the equation of the curve?

(A) *y* = 7*x*^{2} – 0.2*x* + 3

(B) *y* = 2.2*x*^{2} – 0.2*x* + 3

(C) *y* = 7*x*^{2} + 5*x* + 3

(D) *y* = 7*x*^{2} – 5*x* + 3

(E) *y* = 5*x*^{2} – 7*x* + 3

45.

(A) –1

(B) 0

(C) 1

(D) 8

(E) ∞

**STOPEND OF PART B, SECTION IIF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON PART B ONLY.DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.**

SECTION II

GENERAL INSTRUCTIONS

You may wish to look over the problems before starting to work on them, since it is not expected that everyone will be able to complete all parts of all problems. All problems are given equal weight, but the parts of a particular problem are not necessarily given equal weight.

A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROBLEMS OR PARTS OF PROBLEMS ON THIS SECTION OF THE EXAMINATION.

- You should write all work for each part of each problem in the space provided for that part in the booklet. Be sure to write clearly and legibly. If you make an error, you may save time by crossing it out rather than trying to erase it. Erased or crossed-out work will not be graded.
- Show all your work. You will be graded on the correctness and completeness of your methods as well as your answers. Correct answers without supporting work may not receive credit.
- Justifications require that you give mathematical (noncalculator) reasons and that you clearly identify functions, graphs, tables, or other objects you use.
- You are permitted to use your calculator to solve an equation, find the derivative of a function at a point, or calculate the value of a definite integral. However, you must clearly indicate the setup of your problem, namely the equation, function, or integral you are using. If you use other built-in features or programs, you must show the mathematical steps necessary to produce your results.
- Your work must be expressed in standard mathematical notation rather than calculator syntax. For example,
*x*^{2}*dx*may not be written as fnInt (X^{2}, X, 1, 5). - Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point.
- Unless otherwise specified, the domain of a function
*f*is assumed to be the set of all real numbers*x*for which*f*(*x*) is a real number.

SECTION II, PART A

Time—30 minutes

Number of problems—2

**A graphing calculator is required for some problems or parts of problems.**

During the timed portion for Part A, you may work only on the problems in Part A.

On Part A, you are permitted to use your calculator to solve an equation, find the derivative of a function at a point, or calculate the value of a definite integral. However, you must clearly indicate the setup of your problem, namely the equation, function, or integral you are using. If you use other built-in features or programs, you must show the mathematical steps necessary to produce your results.

1. A cylindrical drum is filling with water at a rate of 25*π* in^{3}/sec.

(a) If the radius of the cylinder is 1/3 the height, write an expression for the volume of water in terms of the height at any instance.

(b) At what rate is the height changing when the height is 10 in?

(c) What is the height of the water when it is increasing at a rate of 12 in/sec?

2. The function *f* is defined by *f*(*x*) = (9 − *x*^{2})^{} for −3 ≤ *x* ≤ 3.

(a) Find *f*′(*x*).

(b) Write an equation for the line tangent to the graph of *f* at *x* = –2.

(c) Let *g* be the function defined by *g*(*x*) = . Is *g* continuous at *x* = –2? Use the definition of continuity to explain your answer.

(d) Find the value of .

SECTION II, PART B

Time—1 hour

Number of problems—4

**No calculator is allowed for these problems.**

During the timed portion for Part B, you may continue to work on the problems in Part A without the use of any calculator.

3. A particle moves with velocity *v*(*t*) = 9*t*^{2} + 18*t* – 7 for *t* ≥ 0 from an initial position of *s*(0) = 3.

(a) Write an equation for the position of the particle.

(b) When is the particle changing direction?

(c) What is the total distance covered from *t* = 2 to *t* = 5?

4. Let *f* be the function given by *f*(*x*) = –2*x*^{4} + 6*x*^{2} + 2.

(a) Find the equation for the line normal to the graph at (1,6).

(b) Find the *x* and *y* coordinates of the relative maximum and minimum points.

(c) Find the *x* and *y* coordinates of the points of inflection.

5. Consider the curve given by *x*^{3}*y*^{2} – 5*x* + *y* = 3.

(a) Find .

(b) Find .

(c) Find the equation of the normal lines at each of the two points on the curve whose *x*-coordinate is –1.

6. Let *f* be the function given by *f*(*x*) = 3*x*^{3} – 6*x*^{2} + 4*x*.

(a) Find an equation for the normal line at *x* = 2.

(b) Where are the relative maxima and minima of the curve, if any exist? Verify your answer.

(c) Where are the points of inflection? Verify your answer. If there are none, explain why.

**STOPEND OF EXAM**

# AB Calculus Diagnostic Test Answers and Explanations

## ANSWER KEY

### Section I

1. C

2. A

3. B

4. D

5. E

6. E

7. A

8. E

9. C

10. B

11. A

12. D

13. A

14. B

15. A

16. E

17. A

18. E

19. B

20. B

21. B

22. E

23. D

24. B

25. D

26. B

27. E

28. A

29. D

30. E

31. D

32. C

33. D

34. E

35. B

36. D

37. E

38. B

39. A

40. B

41. D

42. B

43. B

44. D

45. B

## EXPLANATIONS

### Section I

1. **C** Use the double angle formula for sine, sin 2*θ* = 2 sin*θ* cos*θ*. to rewrite the limit and then solve:

2. **A** If the highest power of *x* in a rational expression is the same in both the numerator and the denominator then the limit as *x* approaches infinity is the coefficient of the highest term in the numerator divided by the coefficient of the highest term in the denominator. The coefficient of the highest term in the numerator is 1 and the coefficient of the highest term in the denominator is 3, so the limit is .

3. **B** A removable discontinuity occurs when a rational expression has common factors in the numerator and the denominator. The reduced function has the factor (*x* + 2) in the numerator and denominator, hence there is a removable discontinuity when *x* = –2. The *y*-coordinate of the discontinuity is found by plugging *x* = –2 into the reduced function, *f*(*x*) = . Thus, the point where a removable discontinuity exists is (–2,–1).

4. **D** There are three conditions a function must fulfill for it to be continuous at a point *x* = *c*: 1) *f*(*c*) exists, 2) *f*(*x*) exists, and 3) *f*(*x*) = *f*(*c*). All of the functions above satisfy all three conditions, except *j*(*x*). *j*(–3) does not exist (and the other two conditions are not met either).

5. **E** There are three conditions a function must fulfill for it to be continuous at a point *x* = *c*: *f*(*c*) exists, *f*(*x*) = exists, and *f*(*x*) = *f*(*c*). Answer choice E is the only one that satisfies all three. *F*(*c*) does not exist in answer choice D. *f*(*x*) does not exist in answer choices A or C. *f*(*x*) ≠ *f*(*c*) in answer choice B.

6. **E** Recall the definition of the derivative says: = *f*′(*x*) and the derivative of sec *x* is tan *x* sec *x*. Thus, = = = . Therefore, the limit does not exist.

7. **A** Using the Product Rule, , take the derivative of *f* (*x*) and you get A.

8. **E** Using the Chain Rule, , take the derivative of *f*(*x*) and you get E.

9. **C** First, use implicit differentiation to find :

2*x*^{2}*y* + 2*xy*^{2} = 0

Isolate and simplify:

Next, take the second derivative via implicit differentiation:

Plug in :

There is no need to simplify, just plug in the point (2,1):

10. **B** Start by plugging (0,2) into *y* = *ax*^{2} + *bx* + *c*. Then, *c* = 2. Next, take the derivative of *y* = *ax*^{2} + *bx* + *c*: = 2*ax* + *b*. Given that *y* = *x* + 2 is normal to *y* = *ax*^{2} + *bx* + *c*, the slope of *y* = *ax*^{2} + *bx* + *c,* or , is –3. Thus, by plugging *x* = 0 into = 2*ax* + *b*, *b* = –3. Finally, plug the point (2,1) and the values of *c* and *b* into *y* = *ax*^{2} + *bx* + *c*. Therefore, *a* = .

11. **A** Plug *h*(*x*) in as *x* into *g*(*x*) and take the derivative using the Chain Rule, so (2*g*(*x*^{3})) = 6*x*^{2}*g*(*x*^{3}).

12. **D** In order to determine the number of relative extrema, the derivative of *f*(*x*) must be set equal to zero and the critical points found. When this is done, *f*′(*x*) = 6*x*^{4} − 6*x*^{2} and the critical points are located at *x* = –1, *x* = 0, and *x* = 1. Therefore, there are three possible relative extrema for this curve, and answer choices A and C can be eliminated. Next, in order to determine the points of inflection, set the second derivative of *f*(*x*) equal to zero and solve. When this is done, *f*″(*x*) = 24*x*^{3} − 12*x* and the points of inflection are located at *x* = 0, *x* = − and *x* = . Thus, there are three points of inflection and the answer is D.

13. **A** Use *u*-substitution. Here, *u* = 2*x* – 5 and *du* = 2 *dx*. Then, . Replace *u* for the final solution: + *C*.

14. **B** Use *u*-substitution. Here *u* = *x*^{3} − 6, and *du* = 4*x*^{2} *dx*. Then, . Replace *u* for the final solution: + *C*.

15. **A** Use the mean value theorem for integrals, *f*(*c*) = . Thus, for this problem, .

16. **E** Use the Second Fundamental Theorem of Calculus: . Thus, for this problem, = 6*x*((3*x*^{2})^{2} + 4(3*x*^{2})) = 54*x*^{5} + 72*x*^{3}.

17. A First, rewrite the integral: = ∫ tan *x dx*. You can either derive the integral from using *u*-substitution, or you should have memorized that ∫tan *x dx* = −ln|cos *x*| + *C*.

18. **E** Using vertical slices, the area between the curves follows the general formula: . First, it is important to determine where these two curves intersect and which curve is “on top.” The two curves intersect at *x* = –1, *x* = 0, and *x* = 1. Over the interval [–1, 0], the curve *y* = *x*^{3} is “on top.” Therefore, the integral is . Over the interval [0, 1], the curve *y* = *x* is “on top.” Therefore, the integral is . Thus the area between the curves is the sum of these two areas: .

19. **B** Since you are not told which method to use to find the volume you must decide, a big hint is the answer choices. However, if you didn’t have this hint, then you can use the rule of thumb that it is TYPICALLY (but not always) better to use cylindrical shells, if the region is bound by more than two curves (including an axis) or if one or more curves are given as *y* = and the others are given as *x* =. Both conditions are satisfied in this problem, so cylindrical shells is probably best. The general formula for cylindrical shells is 2π *x*[*f*(*x*) − *g*(*x*)]*dx*. First, the points of intersection between all these curves must be found, where the region is bound, to establish the limits of integration. The bounds are *x* = 5 and *x* = 10. Next, determine which curve is “on top” or “more positive.” In this case, the curves in question are *y* = (*x* – 5)^{3} and *y* = 0. Since *y* = (*x* – 5)^{3} is always more positive than *y* = 0, *y* = (*x* – 5)^{3} = *f*(*x*) and *y* = 0 = *g*(*x*). Finally, the general formula is for a region that is rotated about the *y*-axis, or *x* = 0. Since our curve is shifted to be rotated around *x* = 2, the radius of the cylinder, *x*, is now *x* – 2 to account for the shift. Thus, the final integral is: 2π (*x* − 2)(*x* − 5)^{3} *dx*.

20. **B** Use *u*-substitution in which *u* = *x*^{3} – 3 and *du* = 3*x*^{2} *dx*. Thus, the integral is:

21. **B** In order to determine the equation for the normal line, take the derivative with respect to *x* at (2,0): = 6*x* − 6 which is 6 at (2,0). Since this is the slope of the tangent line, and the slope of the normal line is the opposite reciprocal, the slope of the normal line is −. Plug this information into the point-slope formula of a line and simplify to slope-intercept form: *y* − 0 = −(*x* − 2) or *y* = −*x* + .

22. **E** Either use L’Hôpital’s rule or recall that and . In this case, can be rewritten as .

23. **D** Following the Second Fundamental Theorem of Calculus,

(2 − *t*^{2})*dt* = [2 − (*x*^{3})^{2}](3*x*^{2}) = (2 − *x*^{6})(3*x*^{2}) = 6*x*^{2} − 3*x*^{8}.

24. **B** Use *u*-substitution and recognize that the solution will be an inverse sine function. . Thus, the final solution is: + *C*.

25. **D** *f*′(*x*) = 3*x*^{2} + 2 and *f*′(1) = 5.

26. **B** The first derivative is = 3*x*^{2} + 4*x*. The second derivative is = 6*x* + 4.

27. **E** For an absolute maximum, find the first derivative and set it equal to zero to determine the critical points. = 20*x*^{4} − 20*x* = 0. The critical points are at *x* = 0 and *x* = 1. Find the second derivative to determine which of these points are maxima and which are minima. Maxima are located where the second derivative is negative; minima where the second derivative is positive. = 80*x*^{3} − 20. From here, *x* = 0 corresponds with maximum points. Finally, plug in this value into the original equation to determine which has a higher *y*-value. At *x* = 0, *y* = –8. However, you must remember when determining absolute maxima and minima on a closed interval to always check the endpoints. So, plug in *x* = –2 and *x* = 2 into the original equation and determine if they have *y*-values that are more positive than at *x* = 0. At *x* = –2, *y* = –176 and at *x* = 2, *y* = 80. Therefore, *x* = 2 is the absolute maximum.

28. **A** Using implicit differentiation, find the first derivative of the equation and solve for : Next, determine the second derivative, but don’t simplify the equation: . Finally evaluate the first and second derivative at (–1,–2). The first derivative at (–1,–2) is 2. The second derivative at (–1,–2) is .

29. **D** Using the Power and Addition Rules, take the derivative of *f*(*x*) and you get D. Remember that π and *e* are constants.

30. **E** Rolle’s Theorem states that if *y* = *f*(*x*) is continuous on the interval [*a, b*], and is differentiable everywhere on the interval (*a, b*), and if *f*(a) = *f*(*b*) = 0, then there is at least one number *c* between *a* and *b* such that *f*′(*c*) = 0. *f* (*x*) = 0 at both *x* = 0 and *x* = 2. Then, solve *f*′(*c*) = 8*x*^{3} − 16 = 0. *c* = 2^{}.

31. **D** An absolute maximum or minimum occurs when the derivative of a function is zero or where the derivative fails to exist or at an endpoint. First, find the derivative of *y*, set it equal to zero and solve for *x*. = 5*x*^{2} − 2*x* − 7 = 0, then *x* = –1 and *x* = . Determine the *y*-values corresponding to each of these *x*-values and at the endpoints, *x* = –2 and *x* = 2. The resulting points are , , , and . The maximum is occurs at .

32. **C** The formula for the area under a curve using right-endpoint rectangles is: *A* = (*y*_{1} + *y*_{2} + *y*_{3} +…*y _{n}*), where

*a*and

*b*are the

*x*-values that bound the area and

*n*is the number of rectangles. Since we are interested in the right-endpoints, the

*x*-coordinates are

*x*

_{1}= ,

*x*

_{2}= ,

*x*

_{3}= , and

*x*

_{4}= 2. The

*y*-coordinates are found by plugging these values into the equation for

*y*, so

*y*

_{1}= 3.5625,

*y*

_{2}= 4.25,

*y*

_{3}= 5.0625, and

*y*

_{4}= 6. Then,

*A*= (3.5625 + 4.25 + 5.0625) = 4.71875.

33. **D** The formula for the area under a curve using inscribed trapezoids is: *A* = (*y*_{0} + 2*y*_{1} + 2*y*_{2} +…+ 2*y _{n}*

_{−1}+

*y*), where

_{n}*a*and

*b*are the

*x*-values that bound the area and

*n*is the number of rectangles. The

*x*-coordinates are

*x*

_{0}= 1,

*x*

_{1}= ,

*x*

_{2}= ,

*x*

_{3}= , and

*x*

_{4}= 2. The

*y*-coordinates are found by plugging these values into the equation for

*y*, so

*y*

_{0}= 3,

*y*

_{1}= 3.5625,

*y*

_{2}= 4.25,

*y*

_{3}= 5.0625, and

*y*

_{4}= 6. Then,

*A*= (3 + 2(3.5625) + 2(4.25) + 2(5.0625) + 6) = 4.34375.

34. **E** Use the Fundamental Theorem of Calculus: *f*(*x*) *dx* = *F*(*b*) − *F*(*a*) and *u*-substitution. For this problem, *u* = *x*^{2} and *du* = 2*x dx.* Then:

35. **B** This is an accumulation problem, so use the Second Fundamental Theorem of Calculus, *f*(*t*) *dt* = *f*(*x*). Thus, *F*(*x*) = = 72 − 0.75 = 71.25.

36. **D** First, rewrite 2*x* + *y* = *k* in slope-intercept form: *y* = –2*x* + *k*; thus, the slope of the tangent, or the first derivative of *y*, is –2. Next, take the derivative of *y* = 2*x*^{2} – 8*x* + 14 and set it equal to –2: = 4*x* − 8 = −2. Use this equation to solve for the *x*-value that corresponds to the first derivative of –2; the *x*-coordinate is . Use this *x*-coordinate to solve for the corresponding *y*-coordinate; plug *x* = into the equation for the curve *y*. The *y*-coordinate equals . Finally, go back to the equation for the tangent line and plug in the *x* and *y* values found. Solve for *k: k* = .

37. **E** Since, *f*′(*x*) < 0, the curve is decreasing, which is true in all answer choices. In addition, because *f*″(*x*) > 0, the curve is concave up. The best way to determine which curve is concave up is to make a quick sketch of the graphs from the data points. When this is done, E is concave up.

38. **B** The particle slows down when the velocity and acceleration have different signs. First, take the first derivative of *x*(*t*) and set it equal to zero to solve for the times when the velocity is changing sign: *x*′(*t*) = 6*t*^{2} − 21*t* + 9 = 0 when *t* = and *t* = 3. Next, take the second derivative of *x*(*t*), and determine when that is changing sign: *x*″(*t*) = 12*t* − 21 = 0 when *x* = . Since there are three different values for which either the velocity or acceleration equals zero, there are four intervals to check the signs of both the velocity and acceleration:

Since the velocity and acceleration have different signs over the intervals and 0 < *t* < and < *t* < 3, the correct answer is B.

39. **A** First, determine where *f*(*x*) and the *x*-axis intercept, i.e., solve *f*(*x*) = 0 for *x.* Thus, *x* = 0, *x* = –2, and *x* = 2. In order to determine the area under the curve, we must set up and solve two integrals: .

40. **B** The line *y* = *c* is a horizontal asymptote of the graph of *y* = *f*(*x*) if the limit of the function as *x* approaches positive and negative infinity equals *c*. Similarly, the line *x* = *k* is a vertical asymptote of the graph of *y* = *f*(*x*) if the limit of the function as *x* approaches *k* from the left and right is positive or negative infinity. First, check for a horizontal asymptote, = 3 and = 3, so there is a horizontal asymptote at *y* = 3. Next, check for a vertical asymptote; always check the point where the denominator is undefined, in this case, *x* = –7: and . Thus, there is a vertical asymptote at *x* = –7.

41. **D** Use *u*-substitution in which *u* = *x*^{2} – 7 and *du* = 2*x dx*. Thus, the integral is:

= ln |*x*^{2} − 7| + *C*

42. **B** The formula for the area between two curves is (*f*(*x*) − *g*(*x*)) *dx*, where *a* and *b* are the *x*-coordinates that bind the region and *f*(*x*) is the more positive curve. Be careful to check if the curves cross the *x*-axis because multiple integrals will be required if this happens. In this case, the curves intersect on the *x*-axis at *x* = 2. Therefore, the area between the curves will be:

43. **B** Using the MVTI, = 12.

44. **D** Plug in the given point, (0,3), into the equation for the curve, *y* = *ax*^{2} + *bx* + *c*, thus *c* = 3. Next, rewrite the equation for the normal line in slope-intercept form, *y* = *x* + 3. Since this line is normal to the curve, the slope of the tangent line is the opposite reciprocal to the slope of the normal line. The slope of the tangent line is –5 to evaluate it at (0,3). Take the derivative of *y* and set it equal to –5: = 2*ax* + *b* and = 2(*a*)(0) + *b* = −5. Therefore, *b* = –5. Finally, solve for *a* by plugging the second point (1,5) into the equation for the curve. Also, plug in *b* and *c*, so: 5 = *a* + *b* + *c* or *a* = 7, when *b* = –5 and *c* = 3. Finally, the equation of the curve is *y* = 7*x*^{2} – 5*x* + 3.

45. **B** ((*x*^{3} + 2*x*^{2} − 3)(*x*^{−2} + 7*x*)) = (*x*^{3} + 2*x*^{2} − 3) • (*x*^{−2} + 7*x*) = 0 • 8 = 0

### Section II

1. A cylindrical drum is filling with water at a rate of 25*π* in^{3}/sec.

(a) If the radius of the cylinder is the height, write an expression for the volume of water in terms of the height at any instance.

(a) *V* = π*r*^{2}*h* and *r* = . Thus, the volume can be found by solving *V* = .

(b) At what rate is the height changing when the height is 10 in?

(b) The rate the height is changing can be found by taking the first derivative with respect to time. . Plug in the values given and solve for in/sec.

(c) What is the height of the water when it is increasing at a rate of 12 in/sec?

(c) Use the derivative from part (b) and plug in the values to solve for *h*, so *h* = in.

2. The function *f* is defined by *f*(*x*) = (*9* − *x*^{2})^{} for −3 ≤ *x* ≤ 3.

(a) Find *f*′(*x*).

(a) *f*′(*x*) = (9 − *x*^{2})^{}(2*x*) = 3*x*(9 − *x*^{2})^{}

(b) Write an equation for the line tangent to the graph of *f* at *x* = –2.

(b) Use the equation for *f*′(*x*) from part (a) to find the slope of the tangent at *x* = –2: *f*′(−2) = −6. Determine the *y*-coordinate that corresponds with *x* = –2 by plugging it into *f*(*x*): *f*(−2) = 5. Finally, the equation for the tangent line is: *y* − 5 = −6(*x* + 2).

(c) Let *g* be the function defined by *g*(*x*) = . Is *g* continuous at *x* = –2? Use the definition of continuity to explain your answer.

(c) The definition of continuity states a function is continuous if three conditions are met: *f*(*c*) exists, *f*(*x*) exists, *f*(*x*) = *f*(*c*). The first condition is met, *f*(–2) exists, from part (b). The second condition is violated; the left and right hand limits as *x* approaches –2 are not equal, so the limit does not exist and the function is not continuous: *g*(*x*) = 5 and *g*(*x*) = 5.

(d) Find the value of

(d) Solve using *u*-substitution, in which *u* = 9 – *x*^{2} and *du* = 2*x dx*. Thus, *dx* = 145.8.

3. A particle moves with velocity *v*(*t*) = 9*t*^{2} + 18*t* – 7 for *t* ≥ 0 from an initial position of *s*(0) = 3.

(a) Write an equation for the position of the particle.

(a) The position function of the particle can be determined by integrating the velocity with respect to time, thus *s*(*t*) = ∫*v*(*t*) *dt*. For this problem, *s*(*t*) = ∫(9*t*^{2} + 18*t* − 7) *dt* = 3*t*^{3} + 9*t*^{2} − 7*t* + *C*. Since we are given the initial position, *s*(0) = 3, plug that in to solve for *C.* Thus, *C* = 3 and the equation for the position of the particle is *s*(*t*) = 3*t*^{3} + 9*t*^{2} – 7*t* + 3.

(b) When is the particle changing direction?

(b) The particle changes direction when the velocity is zero, but the acceleration is not. In order to determine when those times are, set the velocity equal to zero and solve for *t*. *v*(*t*) = 9*t*^{2} + 18*t* – 7 = 0 when *t* = and *t* = . Since, the time range in question is *t* ≥ 0, we can ignore *t* = − . Then, take the derivative of the velocity function to find the acceleration function, as (*v*(*t*)) = *a*(*t*). For the given *v*(*t*), *a*(*t*) = 18*t* + 18. Check that the acceleration at time *t* = is not zero by plugging into the acceleration function: *a*(*t*) = 30. Therefore, the particle is changing direction at *t* = because *v*(*t*) = 0 and *a*(*t*) ≠ 0.

(c) What is the total distance covered from *t* = 2 to *t* = 5?

(c) The distance covered is found by using the position function found in part (a). Determine the position at *t* = 2 and subtract it from the position at *t* = 5. From part (b), we know that the object does not change direction over this time interval, so we do not need to find the time piecewise. Thus, *s*(5) – *s*(2) = 568 – 49 = 519.

4. Let *f* be the function given by *f*(*x*) = −2*x*^{4} + 6*x*^{2} + 2.

(a) Find the equation for the line normal to the graph at (1, 6).

(a) The line normal to the graph will have a slope that is the opposite reciprocal of the tangent line at that point. Therefore, begin by finding the slope of the tangent line, i.e., the first derivative. *f*′(*x*) = −8*x*^{3} + 12*x* and *f*′(1) = 4. The slope of the normal line is –. So, the equation of the normal line is: *y* − 6 = −(*x* − 1).

(b) Find the *x* and *y* coordinates of the relative maximum and minimum points.

(b) The relative maximum and minimum will occur at the points when the first derivative is zero or undefined. In this case, set the first derivative to zero and solve for *x*: *f*′(*x*) = −8*x*^{3} + 12*x* = 0, and *x* = 0, *x* = , and *x* = − . To determine which of these is a relative maximum and which is a relative minimum, find the second derivative at each of these critical points. *f*″(*x*) = −24*x*^{2} + 12, and *f*″(0) = 12, = − 24, and = − 24. By the second derivative test, *x* = 0 corresponds with a relative minimum because *f*″(0) > 0 and *x* = , and *x* = − correspond with relative maximums because and < 0. To determine the *y* coordinates of these points, plug them back into *f*(*x*): *f*(0) = 0, and . So, a relative minimum occurs at (0,2) and relative maximums occur at and .

(c) Find the *x* and *y* coordinates of the points of inflection.

(c) Points of inflection occur when the second derivative equals zero. Take the second derivative from part (b) and solve for the *x* values when *f*″(*x*) = 24*x*^{2} + 12 = 0. So, *x* = and *x* = − . To determine the *y*-coordinates for these points of inflection, determine *f*(*x*) at each of these points: and . So the points of inflection occur at and .

5. Consider the curve given by *x*^{3}*y*^{2} – 5*x* + *y* = 3.

(a) Find .

(a) Use implicit differentiation: 2*x*^{3}*y* + 3*x*^{2}*y*^{2} − 5 + = 0. Simplify and isolate : .

(b) Find .

(b) Use the derivative from part (a) and differentiate again using implicit differentiation: . Replace the value for from part (a) for the final solution:

No need to simplify.

(c) Find the equation of the normal lines at each of the two points on the curve whose *x*-coordinate is –1.

(c) From the original equation, at *x* = –1, *y* = –2 or *y* = 1. Plug those values into from part (a) to get the slope of the tangents at those points: − and –2, respectively. The slopes of the normal lines are the opposite reciprocals of those slopes, so and , respectively. The equations for the normal are then found by using the point-slope formula. The equations are *y* + 2 = (*x* + 1) and *y* − 1 = (*x* + 1).

6. Let *f* be the function given by *f*(*x*) = 3*x*^{3} – 6*x*^{2} + 4*x*.

(a) Find an equation for the normal line at *x* = 2.

(a) First determine what *f*(2) equals: *f*(2) = 8. Next, take the first derivative of *f*(*x*): *f*′(*x*) = 9*x*^{2} − 12*x* + 4. At *x* = 2, *f*′(*x*)(*x*) = 16. The slope of the normal line would then be −. The equation of the normal line is *y* − 8 = − (*x* − 2).

(b) Where are the relative maxima and minima of the curve, if any exist? Verify your answer.

(b) Relative maxima and minima exist where the first derivative is zero or does not exist. Set the first derivative equal to zero and solve or *x*, *f*′(*x*) = 0 at *x* = . To determine whether this point is a maximum or minimum, check the second derivative. If the second derivative is negative, the point is a maximum. If the second derivative is positive, the point is a minimum. However, if the second derivative is zero, the point is a point of inflection. At *x* = , *f*″(*x*) = 0, so the point is a point of inflection. This curve has no maxima or minima.

(c) Where are the points of inflection? Verify your answer. If there are none, explain why.

(c) It was discovered in part (b) that the point of inflection is at *x* = . Plug that into *f*(*x*) to find the corresponding *y*-value, which is . Thus, the point of inflection is located at . Justification is the same as the justification in part (b).

# BC Calculus Diagnostic Test

(*Click here to download a PDF of Diagnostic Test*)

## AP^{®} Calculus BC Exam

**DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.**

**Instructions**

Section I of this examination contains 45 multiple-choice questions. Fill in only the ovals for numbers 1 through 45 on your answer sheet.

CALCULATORS MAY NOT BE USED IN THIS PART OF THE EXAMINATION.

Indicate all of your answers to the multiple-choice questions on the answer sheet. No credit will be given for anything written in this exam booklet, but you may use the booklet for notes or scratch work. After you have decided which of the suggested answers is best, completely fill in the corresponding oval on the answer sheet. Give only one answer to each question. If you change an answer, be sure that the previous mark is erased completely. Here is a sample question and answer.

Sample Question

Chicago is a

(A) state

(B) city

(C) country

(D) continent

(E) village

Sample Answer

Use your time effectively, working as quickly as you can without losing accuracy. Do not spend too much time on any one question. Go on to other questions and come back to the ones you have not answered if you have time. It is not expected that everyone will know the answers to all the multiple-choice questions.

**About Guessing**

Many candidates wonder whether or not to guess the answers to questions about which they are not certain. Multiple choice scores are based on the number of questions answered correctly. Points are not deducted for incorrect answers, and no points are awarded for unanswered questions. Because points are not deducted for incorrect answers, you are encouraged to answer all multiple-choice questions. On any questions you do not know the answer to, you should eliminate as many choices as you can, and then select the best answer among the remaining choices.

CALCULUS BC

SECTION I, Part A

Time—55 Minutes

Number of questions—28

A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION

Directions: Solve each of the following problems, using the available space for scratchwork. After examining the form of the choices, decide which is the best of the choices given and fill in the corresponding oval on the answer sheet. No credit will be given for anything written in the test book. Do not spend too much time on any one problem.

In this test: Unless otherwise specified, the domain of a function *f* is assumed to be the set of all real numbers *x* for which *f*(*x*) is a real number.

1.

(A) 0

(B) 1

(C) 3

(D) 5

(E) The limit does not exist.

2.

(A) 0

(B) 1

(C) 2*π*

(D) ∞

(E) The limit does not exist.

3.

(A)

(B)

(C)

(D)

(E)

4.

(A) –1

(B) 0

(C) 1

(D) ∞

(E) The limit does not exist.

5. What type(s) of discontinuity/ies does the function, *f*(*x*) = , have?

(A) jump

(B) point

(C) essential

(D) jump and removable

(E) essential and removable

6. Find the fourth derivative of *f*(*x*) = .

(A)

(B)

(C)

(D)

(E)

7.

(A) tan 5*x*

(B) sec 5*x*

(C) csc 5*x*

(D) sec 5*x* tan 5*x*

(E) 5sec 5*x* tan 5*x*

8. What is the volume of the solid formed by rotating the region between the curves *y* = 6*x*^{2} – *x* and *y* = *x*^{2} – 6*x* about the *y*-axis?

(A)

(B) 5*π*

(C)

(D) *π*

(E)

9. ∫*x*^{3} ln 2*x dx*

(A) (4 ln 2*x* − 1) + *C*

(B) (ln 2*x* + 1) + *C*

(C) *x*^{4}(ln 2*x* − 1) + *C*

(D) (ln 2*x* − 1) + *C*

(E) (4 ln 2*x* + 1) + *C*

10.

(A) 3 ln |*x* + 3| − ln |2*x* − 1| + *C*

(B) ln |*x* + 3| − 3 ln |2*x* − 1| + *C*

(C) ln |*x* + 3| − ln |2*x* − 1| + *C*

(D) 3 ln |*x* + 3| + ln |2*x* − 1| + *C*

(E) −ln |*x* + 3| + ln |2*x* − 1| + *C*

11.

The polynomial function *f* has selected values of its second derivative, *f*″, given in the table above. Which of the following could be false?

(A) The graph of *f* changes concavity in the interval (–2, 1).

(B) There is a point of inflection on the graph of *f* at *x* = 0.

(C) The graph of *f* has a point of inflection at *x* = 1.5.

(D) The graph of *f* is concave down at *x* = 1.

(E) The graph of *f* changes concavity in the interval (1, 2).

12. If *y* = , =

(A)

(B)

(C)

(D)

(E)

13. If *y* = *x*^{3} + 3*x*^{2} + 5, then =

(A) 3*x*^{3} + 6*x*^{2} + 5

(B) 3*x*^{2} + 6*x*

(C) + *x*^{3} + 5*x*

(D) 15*x*^{5}

(E) 3*x*^{2} + 6*x* + 5

14. If *f*(*x*) = , then *f*′(*x*) =

(A)

(B)

(C)

(D)

(E)

15. If *f*(*x*) = 2 sin(cos *x*), then *f*′(*x*) =

(A) −2 sin *x* · cos (cos *x*)

(B) −2 cos *x* · sin (sin *x*)

(C) 2 sin *x* · cos *x*

(D) − 2 sin *x* · cos *x*

(E) 2 sin *x* · cos (cos *x*)

16. Find if *x*^{3} + 3*x*^{2}*y* + 3*xy*^{2} + *y*^{3} = 27.

(A) 27

(B)

(C) 1

(D)

(E) –1

17. Water is filling a conical cup at a rate of *π* in^{3}/sec. If the cup has a height of 18 in and a radius of 6 in, how fast is the water level rising when the water is 6 in deep?

(A) in/s

(B) 6 in/s

(C) in/s

(D) in/s

(E) in/s

18. Find the derivative of log_{g}(*x*^{2} + 2)^{3}

(A)

(B)

(C)

(D)

(E)

19. What curve is represented by *x* = cos^{2} *t* and *y* = sin^{2} *t*?

(A) *y* = *x* + 1

(B) *y* = –*x* + 1

(C) *y* = –*x* – 1

(D) *y* = *x* – 1

(E) *y* = *x*

20. Given the position function *x*(*t*) = *t*^{3} – 18*t*^{2} – 84*t* + 11 for *t* ≥ 0, for what values of *t* is speed increasing?

(A) 0 ≤ *t* ≤ 14

(B) 0 ≤ *t* ≤ 6 and *t* > 14

(C) 0 ≤ *t* ≤ 6

(D) *t* > 6

(E) 6 < *t* < 14

21. Find the derivative of *f*(*x*) = *e*^{sin2}^{x}

(A) 2*e*^{sin2}* ^{x}* sin

*x*cos

*x*

(B) *e*^{sin2}^{x}

(C) *e*^{sin2}* ^{x}* sin

^{2}

*x*

(D) 2*e*^{sin2}* ^{x}* sin

*x*

(E) *e*^{sin2}* ^{x}* sin

*x*cos

*x*

22. Given:

Use a left-hand Riemann sum with the six subintervals indicated by the data to approximate *f*(*x*)*dx*.

(A) 63

(B) 99

(C) 40

(D) 64

(E) 100

23.

(A) 2 ln |*x*^{2} + 1| + *C*

(B) ln |3*x*^{3} + *x*| + *C*

(C) ln |3*x*^{3} + *x*| + *C*

(D) ln |*x*^{2} + 1| + *C*

(E) 2 ln |3*x*^{3} + *x*| + *C*

24. ∫*e ^{x}* sin

*x dx*=

(A) + *C*

(B) + *C*

(C) + *C*

(D) + *C*

(E) + *C*

25. Which of the following integrals converges?

I.

II.

III.

(A) I

(B) II

(C) III

(D) I & II

(E) I, II, & III

26. Find the area of the region in the plane enclosed by *r* = 5 + 2 cos *θ*.

(A) 23*π*

(B) 24*π*

(C) 25*π*

(D) 26*π*

(E) 27*π*

27. If and *y* = –1 when *x* = 0, then the equation for the curve is

(A) *y*^{3} = 3sin *x* – 1

(B) *y*^{3} = 3cos *x* – 1

(C) *y* = sin *x* – 1

(D) *y* = 3sin *x +* 1

(E) *y*^{3} = sin *x*

28. Which of the following series converges?

I.

II.

III.

(A) I

(B) II

(C) III

(D) I & II

(E) I & III

**END OF PART A, SECTION IIF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON PART A ONLY.DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO.**

CALCULUS BC

SECTION I, Part B

Time—50 Minutes

Number of questions—17

A GRAPHING CALCULATOR IS REQUIRED FOR SOME QUESTIONS ON THIS PART OF THE EXAMINATION

In this test:

1. The **exact** numerical value of the correct answer does not always appear among the choices given. When this happens, select from among the choices the number that best approximates the exact numerical value.

2. Unless otherwise specified, the domain of a function *f* is assumed to be the set of all real numbers *x* for which *f*(*x*) is a real number.

29. A spherical balloon is losing air at a rate of –24*π* in^{3}/sec. How fast is the balloon surface area of the balloon shrinking when the radius of the balloon is 2 in?

(A) –20*π* in^{2}/sec

(B) –24*π* in^{2}/sec

(C) –48*π* in^{2}/sec

(D) –12*π* in^{2}/sec

(E) –60*π* in^{2}/sec

30. Find .

(A)

(B)

(C)

(D)

(E)

31. When are the horizontal and vertical components of the velocity of a curve whose motion is given by *x* = *t*^{2} + 6 and *y* = 2*t*^{3} – *t*^{2} + *t* equal?

(A) *t* = –1

(B) *t* = 2

(C) *t* =

(D) *t* = 6

(E) *t* = –2

32. Find the derivative of *x*^{2} + 3*x*^{2}*y* + 3*xy*^{2} + *y*^{3} = 2 at (3,2).

(A) 7

(B)

(C)

(D) −

(E) −

33. A kid on a bike is riding home in the woods on a straight path that is 50 m from the nearest point on the road. His home is 1500 m from the nearest point on the road. If the kid rides at 3 m/s in the woods and 5 m/s on the road, how far from his house should the kid cross to the road to get home in the shortest time?

(A) 37.5 m

(B) 1500 m

(C) 300.5 m

(D) 1200 m

(E) 1462.5 m

34. A shoe company determined that its profit equation (in millions of dollars) is given by *P* = 2*x*^{3} – 105*x*^{2} + 1500*x* – 1200, where *x* is the number of thousands of pairs of shoes sold and 0 ≤ *x* ≤ 50. Optimize the manufacturer’s profit.

(A) $5.3 billion

(B) $61.3 billion

(C) $1.925 billion

(D) $1.2 billion

(E) $65 billion

35. If the function *f*(*x*) = *x*^{4} has an average value of 5 on the closed interval [0, *k*], then *k* =

(A) 5

(B)

(C) 1

(D)

(E)

36. ∫3*x*(7^{3x2+2})*dx* =

(A) + *C*

(B) + *C*

(C) + *C*

(D) + *C*

(E) + *C*

37. A rectangle with one side on the *x*-axis has its upper vertices on the graph of *y* = cos *x* + 1. What is the minimum area between the rectangle and the graph of *y* = cos *x* + 1 on the interval (−*π* ≤ *x* ≤ *π*)?

(A) 6.283

(B) 2.988

(C) 1.307

(D) 3.296

(E) 5.022

38.

(A) cot^{–1}(*x* + 3) + *C*

(B) sin^{–1}(*x* + 3) + *C*

(C) sec^{–1}(*x* + 3) + *C*

(D) tan^{–1}(*x* + 3) + *C*

(E) cos^{–1}(*x* + 3) + *C*

39. If the path of the particle is given by *x*(*t*) = 2*t*^{2} – 3*t* + 1, how far does the particle travel between *t* = 0 and *t* = 4?

(A) 20

(B) 21

(C)

(D) 22

(E)

40. Given the following values for *x* and *f*(*x*), what is the area under *f*(*x*). Use a left-hand Riemann sum to approximate.

(A) 46

(B) 57

(C) 116

(D) 207

(E) 253

41. Find the length of the curve given by *y* = (*x*^{2} − 2)^{} from *x* = 0 to *x* = 4.

(A) 12

(B) 16

(C) − 4

(D)

(E) + 4

42. Find (*t* − *t*^{4})*dt*.

(A) *x* – *x*^{4} + 1

(B) *x*^{4} – *x* + 1

(C) *x*^{4} – *x*

(D) *x* – *x*^{4} – 1

(E) *x* – *x*^{4}

43. A cylindrical pool is filling at a rate of 96*π* . If the radius of the pool is 4 feet, how fast is the height changing?

(A) 3 ft/hr

(B) 4 ft/hr

(C) 5 ft/hr

(D) 6 ft/hr

(E) 7 ft/hr

44. A child jumps on a trampoline and rises at a rate of 10 ft/min. The child’s mother is watching from the patio 40 ft away. How fast, in rad/sec, is the angle of elevation between the trampoline and the mother’s line of sight of her child increasing when the child is 30 ft in the air?

(A) rad/sec

(B) rad/sec

(C) rad/sec

(D) rad/sec

(E) rad/sec

45. Find the derivative of *y* = .

(A)

(B)

(C)

(D)

(E)

**STOPEND OF PART B, SECTION IIF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON PART B ONLY.DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.**

SECTION II

GENERAL INSTRUCTIONS

You may wish to look over the problems before starting to work on them, since it is not expected that everyone will be able to complete all parts of all problems. All problems are given equal weight, but the parts of a particular problem are not necessarily given equal weight.

A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROBLEMS OR PARTS OF PROBLEMS ON THIS SECTION OF THE EXAMINATION.

- You should write all work for each part of each problem in the space provided for that part in the booklet. Be sure to write clearly and legibly. If you make an error, you may save time by crossing it out rather than trying to erase it. Erased or crossed-out work will not be graded.
- Show all your work. You will be graded on the correctness and completeness of your methods as well as your answers. Correct answers without supporting work may not receive credit.
- Justifications require that you give mathematical (noncalculator) reasons and that you clearly identify functions, graphs, tables, or other objects you use.
- You are permitted to use your calculator to solve an equation, find the derivative of a function at a point, or calculate the value of a definite integral. However, you must clearly indicate the setup of your problem, namely the equation, function, or integral you are using. If you use other built-in features or programs, you must show the mathematical steps necessary to produce your results.
- Your work must be expressed in standard mathematical notation rather than calculator syntax. For example,
*x*^{2}*dx*may not be written as fnInt (X^{2}, X, 1, 5). - Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point.
- Unless otherwise specified, the domain of a function
*f*is assumed to be the set of all real numbers*x*for which*f*(*x*) is a real number.

SECTION II, PART A

Time—30 minutes

Number of problems—2

**A graphing calculator is required for some problems or parts of problems.**

During the timed portion for Part A, you may work only on the problems in Part A.

On Part A, you are permitted to use your calculator to solve an equation, find the derivative of a function at a point, or calculate the value of a definite integral. However, you must clearly indicate the setup of your problem, namely the equation, function, or integral you are using. If you use other built-in features or programs, you must show the mathematical steps necessary to produce your results.

1. Let *y* be the function satisfying *f*′(*x*) = −*x*(1 + *f*(*x*)) and *f*(0) = 5.

(a) Use Euler’s method to approximate *f*(1) with a step size of 0.25.

(b) Find an exact solution for *f*(*x*) when *x* = 1.

(c) Evaluate −*x*(1 + *f*(*x*))*dx*.

2. Let R be the region enclosed by the graphs of *y* = *x*^{2} – *x* – 6 and *y* = *x*^{3} – 2*x*^{2} – 5*x* + 6 and the lines *x* = –2 and *x* = 2.

(a) Find the area of R.

(b) The horizontal line *y* = 0 splits R into two parts. Find the area of the part of R above the horizontal line.

(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the *x*-axis is an equilateral triangle. Find the volume of this solid.

(d) What is the volume of the solid generated by the region R when it is revolved about the line *x* = –3.

SECTION II, PART B

Time—1 hour

Number of problems—4

**No calculator is allowed for these problems.**

During the timed portion for Part B, you may continue to work on the problems in Part A without the use of any calculator.

3. The derivative of a function *f* is *f*′(*x*) = (2*x* + 6)*e ^{−x}* and

*f*(2) = 15.

(a) The function has a critical point at *x* = –3. Is there a relative maximum, minimum, or neither at this point on *f*? Justify your response.

(b) On what interval, if any, is the graph of *f* both increasing and concave down? Explain your reasoning.

(c) Find the value of *f*(5).

4. Consider the equation *x*^{3} – 2*x*^{2}*y* + 3*xy*^{2} – 4*y*^{3} = 10.

(a) Write an equation for the slope of the curve at any point.

(b) Find the equation of the normal line to the curve at the point *x* = 1.

(c) Find at *x* = 1.

5. Given that *f*(*x*) = sin *x*:

(a) Find the 6^{th} degree Maclaurin series.

(b) Use the polynomial to estimate sin 0.2.

(c) Estimate the remainder of the approximation.

6. Two particles travel in the *xy*-plane for time *t* ≥ 0. The position of particle A is given by *x* = 2*t* – 3 and *y* = (2*t* + 1)^{2} and the position of particle B is given by *x* = *t* – 1 and *y* = *t* + 23.

(a) Find the velocity vector for each particle at *t* = 3.

(b) Set up, but do not evaluate, an integral expression for the distance traveled by particle A from *t* = 3 to *t* = 5.

(c) At what time do the two particles collide? Justify your answer.

**STOPEND OF EXAM**

# BC Calculus Diagnostic Test Answers and Explanations

## ANSWER KEY

### Section I

1. D

2. A

3. B

4. C

5. E

6. E

7. D

8. E

9. A

10. E

11. C

12. A

13. B

14. C

15. A

16. E

17. A

18. B

19. B

20. B

21. A

22. D

23. E

24. A

25. A

26. E

27. A

28. D

29. B

30. A

31. C

32. D

33. E

34. B

35. B

36. A

37. B

38. D

39. E

40. E

41. C

42. E

43. D

44. B

45. C

## EXPLANATIONS

### Section I

1. **D** .

2. **A** .

3. **B**

4. **C** .

5. **E** Jump discontinuities exist when the left and right hand limits of the function are not equal at a certain value. A point discontinuity exists when the limit of the function at a certain value does not equal the function at that value. An essential discontinuity is a vertical asymptote. A removable discontinuity occurs where a canceled factor in a rational expression existed. For this function, *f*(*x*) = , since the factor (*x* – 9) can be factored out, there will be a removable discontinuity at *x* = 9. Also, since the function does not exist at *x* = 3, there is a vertical asymptote there, so there is both a removable and an essential discontinuity on this function’s graph.

6. **E** First, the function can simplify to *f*(*x*) = . Use the quotient rule and be prepared to simplify to make the multiple derivatives easier to evaluate. The first derivative is *f*′(*x*) = . The second derivative is *f*″(*x*) = . The third derivative is *f*′″(*x*) = . Finally, the fourth derivative is *f*^{(4)}(*x*) = .

7. **D** Use *u*-substitution and remember your derivatives of trig functions.

= (5)(sec 5*x* tan 5*x*) = sec 5*x* tan 5*x*.

8. **E** First, determine where the two curves intersect. Set them equal to each other and solve for *x*. Thus, *x* = 0 and *x* = –1 at the points of intersection that bind the area. (Determine which curve is “on top” in order to solve properly.) Next, integrate using cylindrical shells from *x* = 0 to *x* = –1: 2*π* *x*(6*x*^{2} − *x* − (*x*^{2} − 6*x*)) *dx* = . The washer method would have worked, but you would have had to convert the equations into “*x* =” form. Recall, it is generally best to use the cylindrical shells method when the axis of rotation is in “*x* =” form and the equations in “*y* =” form. When everything is in “*y* =” form the washer method is generally better.

9. **A** Since *u*-substitution will not work, you must integrate by parts. Recall, the formula is ∫*u dv* = *uv* − ∫*v du*. For this problem, *u* = ln 2*x*, *du* = , *dv* = *x*^{3} *dx*, and *v* = . (If you are unsure of the derivative of ln 2*x*, remember the product rule of logarithms and then differentiate.) Now, the integral is ∫*x*^{5} ln 2*x dx* = ln 2*x* − ∫*x*^{3} *dx* = (4ln(2*x*) − 1) + *C*.

10. **E** Use partial fractions to solve this problem. When this is done, the fraction in the integrand becomes . If those two fractions are used as integrands, via *u*-substitution, the solution is −ln|*x* + 3| + ln|2*x* − 1| + *C*.

11. **C** By the second derivative test, when the second derivative equals zero, there is a point of inflection at that point. A point of inflection signifies a change in the concavity of the graph of the original function. In addition, if the second derivative is negative over an interval, the graph of the original function is concave down over that interval. If the second derivative is positive, the graph is concave up.

12. **A** Rather than deal with the chain, product, and quotient rules, use logarithmic differentiation:

13. **B** Using the Power and Addition Rules, take the derivative of *f*(*x*) and you get B. Be careful, C is the integral.

14. **C** Using the Quotient Rule, , take the derivative of *f*(*x*) and you get C.

15. **A** Using the Chain Rule, , and *d* (sin *x*) = cos *x* and *d* (cos *x*) = –sin *x,* the answer is A.

16. **E** Use implicit differentiation to find :

3*x*^{2} + 6*xy* + 3*x*^{2} + 3*y*^{2} + 6*xy* + 3*y*^{2} = 0

Isolate terms containing by moving all terms that don’t contain to the right side of the equals sign:

3*x*^{2} + 6*xy* + 3*y*^{2} = −3*x*^{2} − 6*xy* − 3*y*^{2}

Factor out :

(3*x*^{2} + 6*xy* + 3*y*^{2}) = −3*x*^{2} − 6*xy* − 3*y*^{2}

Isolate and simplify:

= −1

17. **A** The volume of a cone is found from the formula *V* = *πr*^{2}*h*. Because we are told the radius of the cup is 6 in and the height is 18 in, then at any level, the height of the water will be three times the radius, thus, *h* = 3*r*. Using this relationship, *r* can be replaced with in the formula for volume, so *V* = . Then, to determine the rate that the water level is rising, we must differentiate both sides with respect to *t*: . Insert the rate of the cup filling and the water level at the instant of interest, 6 in, and solve for . Thus, in/s.

18. **B** When *y* = log* _{a} u*, . For this problem,

*u*= (

*x*

^{2}+ 2)

^{3}and = 6

*x*(

*x*

^{2}+ 2)

^{2}(from the Chain Rule). Therefore,

*f*′(

*x*) = .

19. **B** Since you are relating a sine function and a cosine function, look for a trig identity that easily relates those. In this case, cos^{2}*t +* sin^{2}*t* = 1. When you replace cos^{2}*t* and sin^{2}*t* with *x* and *y*, respectively, the equation becomes *x* + *y* = 1. Solve for *y* to get *y* = –*x* + 1.

20. **B** Recall that the velocity function is the first derivative of a position function with respect to time and the acceleration function is the second derivative of a position function with respect to time. Further, when the velocity and acceleration of a particle have the same sign, the speed is increasing. Thus, to solve this problem, first the position function must be differentiated with respect to time: *v*(*t*) = 3*t*^{2} – 36*t* – 84. This function is set equal to zero and the critical values, or the times when velocity is equal to 0, are found: *t* = –2 and *t* = 14. As the problem states, we are only interested in *t* ≥ 0, so we can ignore *t* = –2. Next, find the sign of the velocity over the time ranges of interest: 0 ≤ *t* < 14 results in *v*(*t*) < 0, and *t* > 14 results in *v*(*t*) > 0. Now to determine when the speed is increasing, differentiate the velocity with respect to time to get the acceleration function: *a*(*t*) = 6*t* – 36. Determine when the acceleration is 0 (*t* = 6) and then find the sign of the acceleration around that time: 0 ≤ *t* < 6 has *a*(*t*) < 0 and *t* > 6 has *a*(*t*) > 0. The times when both the velocity and the acceleration have the same sign is when 0 ≤ *t* < 6 and when *t* > 14.

21. **A** When *y* = *e ^{u}*, . For this problem,

*u*= sin

^{2}

*x*and = 2 sin

*x*cos

*x*(from the Chain Rule). Therefore,

*f*′(

*x*) = 2

*e*

^{sin2x}sin

*x*cos

*x*.

22. **D** The formula for the area under a curve using a left-hand Reimann sum is: *A* = (*y*_{0} + *y*_{1} + *y*_{1} + *y*_{3} +…+ *y _{n}*), where

*a*and

*b*are the

*x*-values that bound the area and

*n*is the number of rectangles. Since we do not have evenly spaced subintervals, we must find the width of each subinterval individually and multiply it by the left-endpoint

*y*-value, so

*A*= 1(2) + 3(4) + 7(1) + 5(2) + 3(3) + 6(4) = 64.

23. **E** Recall, = ln |*u*| + *C*, so using *u*-substitution, *u* = 3*x*^{3} + *x* and *du* = (9*x*^{2} + 1). Then, = 2 ln |*u*| + *C* = 2 ln |3*x*^{3} + *x*| + *C*.

24. A This is a complicated integral and integration by parts is the best way to go for solving it. Recall, the formula for integration by parts is ∫*udv* = *uv* − ∫*vdu*. For this problem, let *u* = *e ^{x}* and

*dv*= sin

*x dx.*From these,

*du*=

*e*and

^{x}dx*v*= –cos

*x*. When you input these using the integration by parts formula you get: ∫

*e*sin

^{x}*x dx*= −

*e*cos

^{x}*x*+ ∫ cos

*xe*. Since we cannot readily integrate ∫cos

^{x}dx*xe*, we must use integration by parts again. In this case,

^{x}dx*u*=

*e*and

^{x}*dv*= cos

*x dx.*From these,

*du*=

*e*and

^{x}dx*v*= sin

*x*. Insert these in to the equation in place of ∫

*cos xe*and you get: ∫

^{x}dx*e*sin

^{x}*x dx*= −

*e*cos

^{x}*x*+

*e*sin

^{x}*x*− ∫

*e*sin

^{x}*x dx*. We are back to where we started. However, we are in a good position now, because we can add ∫

*e*sin

^{x}*x dx*to both sides and we end up with 2∫

*e*sin

^{x}*x dx*= −

*e*cos

^{x}*x*+

*e*sin

^{x}*x*. In order to solve for ∫

*e*sin

^{x}*x dx*, just divide both sides by 2: ∫

*e*sin

^{x}*x dx*= +

*C*.

25. **A** All the options are improper integrals and they will converge if their limits as they approach infinity exist. So, check them one at a time and POE. I. , so I. converges. II. = *undefined*, so II. diverges. III. = *infinity*, so III. diverges.

26. **E** Area under polar curve is found by using: *A* = *r*^{2} *dθ*. This curve repeats after 2*π*, thus the region is bound by 0 and 2*π*. The area is then found from: *A* = (5 + 2 cos*θ*)^{2}*dθ*. Solve this and the area is found to be 27*π*.

27. **A** Solve the differential equation by separating the variables and integrating. The equation would then become: ∫*y*^{2} *dy* = ∫cos *x dx*. From here, *y*^{3} = 3 sin *x* + *C*. Next, use the initial condition (0, –1) to get the exact equation: *y*^{3} = 3 sin *x* – 1.

28. **D** A series converges to *L* when *a _{n}* =

*L*. There are many tests that can be used to test whether a series will converge. The test depends on the type of series. So, check each series in this problem one at a time and POE. I. is a geometric series. Geometric series converge if |

*r*| < 1 and the general form of the series is

*ar*

^{n−1}. The

*r*in I. is , so the series converges. II. can be tested using the ratio test which states that a series in the form

*a*converges if < 1. Then II. can be tested as follows, . Since is less than 1, the series converges. For III., the integral test can be used. The integral test states

_{n}*a*and

_{n}*f*(

*x*)

*dx*either both converge or both diverge, for

*a*=

_{n}*f*(

*n*). Therefore, evaluate the improper integral . This integral diverges (it equals infinity), so the series diverges.

29. **B** The balloon losing air at a rate of −24*π* in^{3}/sec means the volume is shrinking at that rate, but it does not directly relate to surface area. Therefore, we must find an intermediate rate to relate the two formulas, *V* = *πr*^{3} and *A* = 4*πr*^{2}. The common value is *r*, so the common rate will be . To solve for , differentiate the formula for *V* with respect to *t*: . Insert the values for and *r* and solve for . Thus, in. Next, differentiate the formula for *A* with respect to *t*: . Insert the values for and *r* to solve for . Then, = −24*π* in^{2}/sec.

30. **A** When you insert 0 for x, the limit is , which is indeterminate. Use L’Hôpital’s Rule to evaluate the limit: . Since this limit is also indeterminate, use L’Hôpital’s Rule again: . This limit exists and equals .

31. **C** The horizontal and vertical components of the velocity of a curve are found parametrically as the derivatives with respect to time of the *x* and *y* components of the motion, respectively. For this curve, = 5*t* and = 6*t*^{2} − 2*t* + 1. When those two expressions are set equal to each other and solved for *t*, *t* = 1 and *t* = . Since *t* = 1 is not an answer choice, C is the correct answer.

32. **D** Use implicit differentiation to find :

2*x* + 6*xy* + 3*x*^{2} + 3*y*^{2} + 6*xy* + 3*y*^{2} = 0

Do not rearrange the terms to isolate . Instead, plug in (3,2) immediately, solve for the derivative, and simplify:

2(3) + 6(3)(2) + 3(3)^{2} + 3(2)^{2} + 6(3)(2) + 3(2)^{2} = 0

6 + 36 + 27 + 12 + 36 + 12 = 0

54 + 75 = 0

33. **E**

From the diagram, it is clear *D*_{1} = and *D*_{2} = 1500 − *x*. Since distance = rate • time and *r*_{1} = 3 and *r*_{2} = 5, the times for each leg of the trip are *t*_{1} = and *t*^{2} = . Therefore, the total time to ride home is *T* = + 300. To minimize the time, the derivative of T is taken and set equal to zero. and *x* = ±. The negative value can be ignored. Because the derivative is ugly, it will take too much time to check the second derivative. (Don’t make skipping this a habit, though!) Therefore, the time is minimized when *x* = m. The kid should cross at 1462.5 m from his house.

34. **B** The profit is maximized at the values that make the derivative of the profit equation equal to zero or at the end points of the range. The derivative of the profit equation is = 6*x*^{2} − 210*x* − 1500 = 0. This equation is true when *x* = 10 or 25. Then, those values and the endpoints, *x* = 0 and *x* = 50 are used to solve for *P*. The resulting points are (0, –1200), (10, 5300), (25, 1925), and (50, 61300). Since the profit equation is maximized at *x* = 50, and the profit is in millions of dollars, the final result is $61.3 billion.

35. **B** Use the mean value theorem for integrals, *f*(*c*) = *f*(*x*)*dx*. Thus, for this problem, 5 = *x*^{4} *dx*. Using the Fundamental Theorem of Calculus and solving the equation for *k*, *k* = ±.

36. **A** Recall, ∫*a ^{u} du* = +

*C*. In this problem,

*u*= 3

*x*

^{2}+ 2 and

*du*= 6

*x dx*.

Thus, ∫7* ^{u} du* = +

*C*

37. **B** The area of a rectangle is *A = lw*. In this case, the width is 2*x* and the length cos *x* + 1, so *A* = 2*x* (cos *x* + 1). To minimize the area between the rectangle and the graph, the area of the of the rectangle will be maximized. To maximize the area of the rectangle, you must take the derivative of the area and set it equal to zero: = 2 cos *x* − 2*x* sin *x* + 2 = 0. When you solve this with your calculator, there will be four critical points: −*π*, −1.30654, 1.30654, *π*. Plug the points into the formula to determine the area at each point. At –*π* and *π*, the area is 0. At −1.30654 and 1.30654, the area is 3.29559. (You can confirm this is the maximum by taking the second derivative and checking that it is negative at 1.30654.) Next, determine the area under the curve by using a definite integral. cos *x* + 1*dx* = 2*π*. To determine the area between the rectangle and the curve, subtract the area of the rectangle from the area under the curve: *A*_{between} = 2*π* − 3.29559 = 2.98759.

38. **D** The current state of the integral makes it appear very difficult to solve. However, notice that there is an *x*^{2} term and no square root. So, think inverse tangent! First, complete the square: *x*^{2} + 6*x* + 10 = (*x* + 3)^{2} + 1. Thus, you can rewrite the integral as . Now, *u* = *x* + 3 and *du = dx*. Use the pattern = tan^{−1} *u* + *C*. Therefore, = tan^{−1}(*x* + 3) + *C*.

39. **E** In order to find the distance the particle travels, set the first derivative equal to zero to determine, when, if at all, it changes direction over the time interval: *x*′(*t*) = 4*t* − 3 = 0, hence *t* = . Since the particle changes direction at *t* = , we must find the position of the particle at *t* = 0, *t* = , and *t* = 4. *x*(0) = 1, , and *x*(4) = 21. The total distance the particle travels is found by .

40. **E** Given a table of values, a left-hand Riemann sum can be calculated by multiplying the size of the intervals (i.e., the difference in *x*-values), by the left endpoint *y*-values and summing them all up.

41. **C** Recall *L* = . In this case, = *x*(*x*^{2} − 2)^{} and *L* = .

42. **E** From the Second Fundamental Theorem of Calculus, (*t* − *t*^{4}) *dt* = *x* − *x*^{4}.

43. **D** This is a related rates problem. Since the volume of a cylinder is given by *V* = *πr*^{2}*h*, begin by taking the derivative of both sides of the equation with respect to *t*. Because the radius is constant, treat it as a constant when taking the derivative: = *πr*^{2} . Now, plug in the given values for and *r*: 96*π* = *π*4^{2}. Solve for = 6.

44. **B** Draw a picture of the situation:

The question is asking about *θ* in this right triangle and we are given the measurements of the two legs, so set up your equation using tan: tan *θ* = . As this is a related rates problem, take the first derivative with respect to *t*: . Notice the rate of the child rising, , was given in ft/min while the question asks for the solution in rad/sec. Convert into ft/sec: ft/sec. Also, note that a height, *h,* was given in the problem. Plug that into the original equation for tan to solve for *θ* or directly solve for sec^{2} *θ*: tan *θ* = . Recall, 1 + tan^{2}*θ* = sec^{2}*θ*, so plug in to solve for sec^{2}*θ*: sec^{2}*θ* = . Now, plug in all these values into the derivative of tan and solve for or rad/sec.

45. **C** Use the quotient rule and recall the derivative of the natural log and *e*.

.

### Section II

1. Let *y* be the function satisfying *f*′(*x*) = −*x*(1 + *f*(*x*)) and *f*(0) = 5.

(a) Use Euler’s method to approximate *f*(1) with a step size of 0.25.

(a) There are two equations you need to know for Euler’s method: 1. *x _{n}* =

*x*

_{n−1}+

*h*and 2.

*y*=

_{n}*y*

_{n−1}+

*hy*′

_{n−1}. In this case, there will be four steps (

*n*= 4) and

*h*= 0.25. From the given information,

*y*′ = −

*x*(1 +

*y*), so

*y*′

_{0}= 0. It is also given that

*x*

_{0}= 0 and

*y*

_{0}= 5. Then,

*x*

_{1}= 0.25,

*y*

_{1}= 5, and

*y*′

_{1}= 1.5. Next,

*x*

_{2}= 0.5,

*y*

_{2}= 4.625, and

*y*′

_{2}= –2.8125. Once more,

*x*

_{3}= 0.75,

*y*

_{3}= 3.92188, and

*y*′

_{3}= –3.69141. Finally,

*x*

_{4}= 1 and

*y*

_{4}= 2.999.

(b) Find an exact solution for *f*(*x*) when *x* = 1.

(b) Solve the differential equation: = −*x*(1 + *y*). Then, *y* = *Ce*^{} − 1. With the initial condition that *f*(0) = 5, *f*(*x*) = 6*e*^{} − 1. Therefore, *f*(1) = 6*e*^{−} − 1 ≈ 2.639.

(c) Evaluate −*x*(1 + *f*(*x*))*dx*.

(c) First, remember the Fundamental Theorem of Calculus: *f*′(*x*)*dx* = *f*(*b*) − *f*(*a*). From part (b), *f*(*x*) = 6*e*^{} − 1. Notice that −*x*(1 + *f*(*x*))*dx* is an improper integral. So, evaluate *f*(*a*) and *f*(0) and find the difference. *f*(*a*) = −1 and *f*(0) = 5. Therefore, −*x*(1 + *f*(*x*))*dx* = −6.

2. Let R be the region enclosed by the graphs of *y* = *x*^{2} – *x* – 6 and *y* = *x*^{3} – 2*x*^{2} – 5*x* + 6 and the lines *x* = –2 and *x* = 2.

(a) Find the area of R.

(a) *Area* = [*f*(*x*) − *g*(*x*)]*dx*. We are told the region is bound by *x* = –2 and *x* = 2. Further, *y* = *x*^{3} – 2*x*^{2} – 5*x* + 6 is always more positive or “on top” over the entire region, so *f*(*x*) = *x*^{3} − 2*x*^{2} − 5*x* + 6 and *g*(*x*) = *x*^{2} – *x* – 6. Now, the integral can be set up: *A* = (*x*^{3} − 2*x*^{2} − 5*x* + 6) − (*x*^{2} − *x* − 6)*dx*. When solved, the area is found to be 32 units squared.

(b) The horizontal line *y* = 0 splits R into two parts. Find the area of the part of R above the horizontal line.

(b) The line *y* = 0 divides R so that the area above R is the region enclosed by *y* = 0 and *y* = *x*^{3} – 2*x*^{2} – 5*x* + 6. Now, the limits of integration are when these two curves intersect. These curves intersect at *x* = –2, *x* = 1, and *x* = 3. However, R is bound between *x* = –2 and *x* = 2, so the limits of integration are *x* = –2 and *x* = 1. The same formula for area can be used from part (a), but the specific values input would be the ones described here. Thus, *A* = (*x*^{3} − 2*x*^{2} − 5*x* + 6 − 0)*dx*. Thus, the area of this portion of *R* = .

(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the *x*-axis is an equilateral triangle. Find the volume of this solid.

(c) The volume of the solid is found by integrating the area of the cross-section, so *V* = *A*(*x*)*dx*. In this case, since the cross-section is an equilateral triangle, *A* = (*side*)^{3}, where the side length is the length between the curves, i.e., *f*(*x*) – *g*(*x*). From part (a), we know and *g*(*x*) = *x*^{2} – *x* – 6 and *g*(*x*) = *x*^{2} − *x* − 6 and the limits of integration are *x* = –2 and *x* = 2. The equation for the volume is then *V* = [(*x*^{3} − 2*x*^{2} − 5*x* + 6) − (*x*^{2} − *x* − 6)]*dx* = = 141.467 units cubed.

(d) What is the volume of the solid generated by the region R, when it is revolved about the line *x* = –3.

(d) Since you are not told which method to use to find the volume you must decide. You can use the rule of thumb that it is typically (but not always) better to use cylindrical shells, if the region is bound by more than two curves (including an axis) or if one or more curves are given as *y* = and the others are given as *x* =. Both conditions are satisfied in this problem, so cylindrical shells is probably best. The general formula for cylindrical shells is 2*π* *x*[*f*(*x*) − *g*(*x*)]*dx*. We know from part (a) the limits of integration are *x* = –2 and *x* = 2 and we have established which curve is *f*(*x*) and which is *g*(*x*) in part (a). The only thing left to establish is the radius of the cylinder from the axis of rotation. Since there is a shift in the axis to *x* = –3, the radius is *x* + 3. Thus, the final integral is: *V* = 2*π* (*x* + 3)[(*x*^{3} − 2*x*^{2} − 5*x* + 6) − (*x*^{2} − *x* − 6)]*dx* = = 542.688 units cubed.

3. The derivative of a function *f* is *f*′(*x*) = (2*x* + 6)*e ^{−x}* and

*f*(2) = 15.

(a) The function has a critical point at *x* = –3. Is there a relative maximum, minimum, or neither at this point on *f*? Justify your response.

(a) You have two options here: 1) Using the first derivative test: Since *x* = –3 is a critical point, a local minimum will exist when *f*′(*x*) < 0 for *x* < –3 and *f*′(*x*) > 0 for *x* > –3. In this case, those two criteria are satisfied, so *x* = –3 is a relative minimum. 2) You can use the second derivative test: Take the derivative of *f*′(*x*). *f*″(*x*) = (−2*x* − 4)*e ^{−x}*. Plug

*x*= –3 into

*f*″(

*x*). From the second derivative test, if

*f*″(

*c*) < 0, then

*c*is a relative maximum, but if

*f*″(

*c*) > 0, then

*c*is a relative minimum. In this case,

*f*″(−3) = 2

*e*

^{−3}which is greater than 0, so

*x*= –3 is a relative minimum of

*f.*

(b) On what interval, if any, is the graph of *f* both increasing and concave down? Explain your reasoning.

(b) First, use the second derivative from part (a). Find the points of inflection by setting *f*″(*x*) = 0. There is a point of inflection at *x* = –2. Now, test the behavior of *f*″(*x*) around *x* = –2. When *x* < –2, *f*″(*x*) > 0, which means the curve is “concave up.” When *x* > –2, (*x*) < 0, which means the curve is “concave down.” To determine when the curve is decreasing, the first derivative must be tested. We already know from the question stem to part (a) that there is a critical point at *x* = –3. So, test the behavior of *f*′(*x*) around *x* = –3. For *x* < –3, (*x*) < 0, and, thus, falling or decreasing. For *x* > –3, *f*′(*x*) > 0 and, thus, rising or increasing. When the behaviors of the first and second derivatives of *f* are combined, *f*(*x*) is increasing and concave down over the interval *x* > –2.

(c) Find the value of a *f*(5).

(c) In order to find *f*(5), you can integrate to find *f* and then use the initial condition *f*(2) = 15 to find the exact solution. Once the exact solution (the exact equation) is found, then *f* can be evaluated at *x* = 5. A more direct path would be to integrate (*x*) over the range from *x* = 2 to *x* = 5. The value found can then be added to *f*(2). So the solution would be *f*(5) = *f*(2) + (2*x* + 6)*e ^{−x} dx* = 15 − 18

*e*

^{−5}+ 12

*e*

^{−2}.

4. Consider the equation *x*^{3} – 2*x*^{2}*y* + 3*xy*^{2} – 4*y*^{3} = 10.

(a) Write an equation for the slope of the curve at any point.

(a) Find the first derivative of this equation using implicit differentiation:

(b) Find the equation of the normal line to the curve at the point *x* = 1.

(b) At *x* = 1, *y* = –1. Plug these values into the equation for from part (a). Then, the slope of the tangent line is 1. The slope of the normal line is the opposite reciprocal, which is –1. Using point-slope formula, the equation of the normal is *y* + 1 = *x* – 1 or *y* = *x* – 2.

(c) Find at *x* = 1.

(c) The second derivative follows directly from the first in part (a):

Rather than trying to simplify the equation, plug in the value for from part (b) and *x* = 1 and *y* = –1. When simplified using those values, = .

5. Given that *f*(*x*) = sin *x*:

(a) Find the 6^{th} degree Maclaurin series.

(a) In order to determine the 6^{th} degree Maclaurin series, use the general formula: . So, find *f*(*x*) and the first 5 derivatives of *f*(*x*):

*f*(*x*) = sin *x*, *f*′(*x*) = cos *x*, *f*″(*x*) = −sin *x*, *f*″′(*x*) = −cos *x*, *f*^{(4)}(*x*) = sin *x*, and *f*^{(5)}(*x*) = cos *x*

Then, evaluate each of those equations at *x* = 0:

*f*(0) = 0, *f*′(0) = 1, *f*″(0) = 0, *f*″′(0) = −1, *f*^{(4)}(0) = 0, and *f*^{(5)}(0) = 1. Finally, insert the values into the formula for a Maclaurin series and simplify: sin *x* = .

(b) Use the polynomial to estimate sin 0.2.

(b) From part (a), sin *x* = *x* − . To approximate sin 0.2, plug 0.2 in for *x* in the equation and simplify: sin 0.2 ≈ 0.2 − ≈ 0.201336.

(c) Estimate the remainder of the approximation.

(c) The formula for the Lagrange remainder is *R _{n}*(

*x*,

*a*) ≤

*f*

^{(n+1)}(

*c*). Use the series you found in part (a) to generalize the series sin

*x*= . From here,

*R*

_{2n}(

*x*, 0) ≤ . (in this case, it is

*R*

_{2n}not

*R*because every other term is 0.) Recall, |sin

_{n}*x*| ≤ 1, so

*R*

_{2n+1}(

*x*,0) ≤ . Therefore,

*R*

_{7}(

*x*, 0) ≤ . (In general, a good approximation of the remainder/error bound of an

*n*th degree Taylor polynomial is the next nonzero term in a decreasing series.) For the approximation in part (b), plug 0.2 for

*x*into . Therefore, the remainder is ≈ 2.540 × 10

^{−9}.

6. Two particles travel in the *xy*-plane for time *t* ≥ 0. The position of particle *A* is given by *x* = 2*t* – 3 and *y* = (2*t* + 1)^{2} and the position of particle *B* is given by *x* = *t* – 1 and *y* = *t* + 23.

(a) Find the velocity vector for each particle at *t* = 3.

(a) The velocity vector of each particle is found by taking the derivative of each particle’s motion, so the vector is . For particle *A*, = 2 and = 8*t* − 4. For particle *B*, = 1 and = 1. Evaluate each of these derivatives at *t* = 3, so the final velocity vectors are *A* = (2,20) and *B* = (1,1).

(b) Set up, but do not evaluate, an integral expression for the distance traveled by particle A from *t* = 3 to *t* = 5.

(b) The formula for the distance travelled parametrically, or the length of the curve, is . Since the velocity vector does not equal zero at any time over the interval from *t* = 3 to *t* = 5, the particle does not change direction and we can plug in the formulas for the *x* and *y* components of the velocity from part (a): *L* = , which is simplified to *L* = .

(c) At what time do the two particles collide? Justify your answer.

(c) The particles will collide when the *x* and *y* coordinates of their position functions are equal to each other. Therefore, set the position functions equal and solve for *t*. Beginning with the *x*-components: 2*t* – 3 = *t* – 1, so *t* = 2. Plug *t* = 2 into the *y*-components and confirm they are equal. For particle *A*: *y* = (2 • 2 + 1)^{2} = 25. For particle B: *y* = 2 + 23 = 25. Both the *x* and *y* components are equal at *t* = 2, so that is when they collide.

# Part II

# About the AP Calculus Exams

• AB Calculus vs BC Calculus

• The Structure of the AP Calculus Exams

• Overview of Content Topics

• How AP Exams Are Used

• Other Resources

• Designing Your Study Plan

## AB CALCULUS VS BC CALCULUS

AP Calculus is divided into two types: AB and BC. The former is supposed to be the equivalent of a semester of college calculus; the latter, a year. In truth, AB calculus covers closer to three quarters of a year of college calculus. In fact, the main difference between the two is that BC calculus tests some more theoretical aspects of calculus and it covers a few additional topics. In addition, BC calculus is harder than AB calculus. The AB exam usually tests straightforward problems in each topic. They’re not too tricky and they don’t vary very much. The BC exam asks harder questions. But neither exam is tricky in the sense that the SAT is. Nor do they test esoteric aspects of calculus. Rather, both tests tend to focus on testing whether you’ve learned the basics of differential and integral calculus. The tests are difficult because of the breadth of topics that they cover, not the depth. You will probably find that many of the problems in this book seem easier than the problems you’ve had in school. This is because your teacher is giving you problems that are harder than those on the AP.

## THE STRUCTURE OF THE AP CALCULUS EXAMS

Now, some words about the test itself. The AP exam comes in two parts. First, there is a section of multiple-choice questions covering a variety of calculus topics. The multiple choice section has two parts. Part A consists of 28 questions; you are not permitted to use a calculator on this section. Part B consists of 17 questions; you are permitted to use a calculator on this part. These two parts comprise a total of 45 questions.

After this, there is a free response section consisting of six questions, each of which requires you to write out the solutions and the steps by which you solved it. You are permitted to use a calculator for the first two problems but not for the four other problems. Partial credit is given for various steps in the solution of each problem. You’ll usually be required to sketch a graph in one of the questions. The College Board does you a big favor here: You may use a graphing calculator. In fact, The College Board recommends it! And they allow you to use programs as well. But here’s the truth about calculus: Most of the time, you don’t need the calculator anyway. Remember: These are the people who bring you the SAT. Any gift from them should be regarded skeptically!

## OVERVIEW OF CONTENT TOPICS

*Topics in italics are BC Topics*. This list is drawn from the topical outline for AP calculus furnished by the College Board. You might find that your teacher covers some additional topics, or omits some, in your course. Some of the topics are very broad, so we cannot guarantee that this book covers these topics exhaustively.

**I. Functions, Graphs, and Limits**

- Analysis of Graphs
- You should be able to analyze a graph based on “the interplay between geometric and analytic information.” The preceding phrase comes directly from the College Board. Don’t let it scare you. What the College Board really means is that you should have covered graphing in precalculus, and you should know (a) how to graph and (b) how to read a graph.

- Limits
- You should be able to calculate limits algebraically, or to estimate them from a graph or from a table of data.
- You do
**not**need to find limits using the Delta-Epsilon definition of a limit.

- Asymptotes
- You should understand asymptotes graphically and be able to compare the growth rates of different types of functions (namely polynomial functions, logarithmic functions, and exponential functions).
- You should understand asymptotes in terms of limits involving infinity.

- Continuity
- You should be able to test the continuity of a function in terms of limits and you should understand continuous functions graphically.
- You should understand the intermediate value theorem and the extreme value theorem.

*Parametric, Polar, and Vector Functions**You should be able to analyze plane curves given in any of these three forms. Usually, you will be asked to convert from one of these three forms back to Rectangular Form (also known as Cartesian Form).*

**II. Differential Calculus**

- The Definition of the Derivative
- You should be able to find a derivative by finding the limit of the difference quotient.
- You should also know the relationship between differentiability and continuity. That is, if a function is differentiable at a point, it’s continuous there. But if a function is continuous at a point, it’s not necessarily differentiable there.

- Derivative at a Point
- You should know the Power Rule, the Product Rule, the Quotient Rule, and the Chain Rule.
- You should be able to find the slope of a curve at a point, and the tangent and normal lines to a curve at a point.
- You should also be able to use local linear approximation and differentials to estimate the tangent line to a curve at a point.
- You should be able to find the instantaneous rate of change of a function using the derivative or the limit of the average rate of change of a function.
- You should be able to approximate the rate of change of a function from a graph or from a table of values.
- You should be able to find Higher-Order Derivatives and to use Implicit Differentiation.

- Derivative of a Function
- You should be able to relate the graph of a function to the graph of its derivative, and vice-versa.
- You should know the relationship between the sign of a derivative and whether the function is increasing or decreasing (positive derivative means increasing; negative means decreasing).
- You should know how to find relative and absolute maxima and minima.
- You should know the mean value theorem for derivatives and Rolle’s theorem.

- Second Derivative
- You should be able to relate the graph of a function to the graph of its derivative and its second derivative, and vice-versa. This is tricky.
- You should know the relationship between concavity and the sign of the second derivative (positive means concave up; negative means concave down).
- You should know how to find points of inflection.

- Applications of Derivatives
- You should be able to sketch a curve using first and second derivatives and be able to analyze the critical points.
- You should be able to solve Optimization problems (Max/Min problems), and Related Rates problems.
- You should be able to find the derivative of the inverse of a function.
- You should be able to solve Rectilinear Motion problems.

*More Applications of Derivatives**You should be able to analyze planar curves in parametric, polar, and vector form, including velocity and acceleration vectors.**You should be able to use Euler’s Method to find numerical solutions of differential equations.**You should know L’Hôpital’s Rule.*

- Computation of Derivatives
- You should be able to find the derivatives of Trig functions, Logarithmic functions, Exponential functions, and Inverse Trig functions.
*BC students should be able to find the derivatives of parametric, polar, and vector functions.*

**III. Integral Calculus**

- Riemann Sums
- You should be able to find the area under a curve using left, right, and midpoint evaluations and the Trapezoidal Rule.
- You should know the fundamental theorem of calculus:
*f*(*x*)*dx*=*F*(*b*) –*F*(*a*)

- Applications of Integrals
- You should be able to find the area of a region, the volume of a solid of known cross-section, the volume of a solid of revolution, and the average value of a function.
- You should be able to solve acceleration, velocity, and position problems.
*BC students should also be able to find the length of a curve (including a curve in parametric form) and the area of a region bounded by polar curves.*

- Fundamental Theorem of Calculus
- You should know the first and second fundamental theorems of calculus and be able to use them to find the derivative of an integral, and in analytical and graphical analysis of functions.

- Techniques of Antidifferentiation
- You should be able to integrate using the power rule and
*U*-Substitution. *You should be able to do Integration by Parts and simple Partial Fractions.**You should also be able to evaluate Improper Integrals as limits of definite integrals.*

- You should be able to integrate using the power rule and
- Applications of Antidifferentiation
- You should be able to find specific antiderivatives using initial conditions.
- You should be able to solve separable differential equations and logistic differential equations.
- You should be able to interpret differential equations via slope fields. Don’t be intimidated. These look harder than they are.

**IV. Polynomial Approximations and Series**

*The Concept of a Series**You should know that a series is a sequence of partial sums and that convergence is defined as the limit of the sequence of partial sums.*

*Series Concepts**You should understand and be able to solve problems involving Geometric Series, the Harmonic Series, Alternating Series, and P-Series.**You should know the Integral Test, the Ratio Test, and the Comparison Test, and how to use them to determine whether a series converges or diverges.*

*Taylor Series**You should know Taylor Polynomial Approximation; the general Taylor Series centered at x = a; and the Maclaurin Series for e*sin^{x},*x*, cos*x, and*.*You should know how to differentiate and antidifferentiate Taylor Series and how to form new series from known series.**You should know functions defined by power series and radius of convergence.**You should know the Lagrange error bound for Taylor Series.*

## HOW AP EXAMS ARE USED

Different colleges use AP Exams in different ways, so it is important that you go to a particular college’s web site to determine how it uses AP Exams. The three items below represent the main ways in which AP Exam scores can be used:

**College Credit**. Some colleges will give you college credit if you score well on an AP Exam. These credits count towards your graduation requirements, meaning that you can take fewer courses while in college. Given the cost of college, this could be quite a benefit, indeed.**Satisfy Requirements.**Some colleges will allow you to “place out” of certain requirements if you do well on an AP Exam, even if they do not give you actual college credits. For example, you might not need to take an introductory-level course, or perhaps you might not need to take a class in a certain discipline at all.**Admissions Plus**. Even if your AP Exam will not result in college credit or even allow you to place out of certain courses, most colleges will respect your decision to push yourself by taking an AP Course or even an AP Exam outside of a course. A high score on an AP Exam shows mastery of more difficult content than is taught in many high school courses, and colleges may take that into account during the admissions process.

## OTHER RESOURCES

There are many resources available to help you improve your score on the AP Calculus Exam, not the least of which are your **teachers**. If you are taking an AP class, you may be able to get extra attention from your teacher, such as obtaining feedback on your essays. If you are not in an AP course, reach out to a teacher who teaches calculus, and ask if the teacher will review your free response questions or otherwise help you with content.

Another wonderful resource is **AP Central**, the official site of the AP Exams. The scope of the information at this site is quite broad and includes:

- Course Description, which includes details on what content is covered and sample questions
- Sample test questions
- Essay prompts from previous years

The AP Central home page address is: **http://apcentral.collegeboard.com/apc/Controller.jpf**.

The AP Calculus AB Exam Course home page address is: **http://apcentral.collegeboard.com/apc/public/courses/teachers_corner/2178.html**

The AP Calculus BC Exam Course home page address is: **http://apcentral.collegeboard.com/apc/public/courses/teachers_corner/2118.html**

Finally, **The Princeton Review** offers tutoring and small group instruction. Our expert instructors can help you refine your strategic approach and add to your content knowledge. For more information, call 1-800-2REVIEW.

## DESIGNING YOUR STUDY PLAN

As part of the Introduction, you identified some areas of potential improvement. Let’s now delve further into your performance on the diagnostic exam, with the goal of developing a study plan appropriate to your needs and time commitment.

Read the answers and explanations associated with the Multiple Choice questions and respond to the following questions:

- Review the Overview of Content Topics on this page. Next to each topic, indicate your rank of the topic as follows: “1” means “I need a lot of work on this,” “2” means “I need to beef up my knowledge,” and “3” means “I know this topic well.”
- How many days/weeks/months away is your exam?
- What time of day is your best, most focused study time?
- How much time per day/week/month will you devote to preparing for your exam?
- When will you do this preparation? (Be as specific as possible: Mondays & Wednesdays from 3 to 4 pm, for example.)
- What are your overall goals in using this book?

# Part III

# Test-Taking Strategies for the AP Calculus Exams

• How to Approach Multiple Choice Questions

• How to Approach Free Response Questions

• Derivatives and Integrals That You Should Know

• Prerequisite Mathematics

## HOW TO APPROACH MULTIPLE CHOICE QUESTIONS

### Cracking the Multiple Choice Questions

Section I of the AP Calculus Exam consists of 45 multiple choice questions, which you’re given 105 minutes to complete. This section is worth 50 percent of your grade. This section is broken up into two parts. Part A consists of 28 questions that you will have to answer without the use of calculator. Part B consists of 17 multiple choice questions where you are allowed to use a calculator.

All the multiple choice questions will have a similar format: Each will be followed by five answer choices. At times, it may seem that there could be more than one possible correct answer. There is only one! Remember that the committee members who write these questions are calculus teachers. So, when it comes to calculus, they know how students think and what kind of mistakes they make. Answers resulting from common mistakes are often included in the five answer choices to trap you.

### Use the Answer Sheet

For the multiple choice section, you write the answers not in the test booklet but on a separate answer sheet (very similar to the ones we’ve supplied at the very end of this book). Five oval-shaped bubbles follow the question number, one for each possible answer. *Don’t* forget to fill in all your answers on the answer sheet. Don’t just mark them in the test booklet. Marks in the test booklet will not be graded. Also, make sure that your filled-in answers correspond to the correct question numbers! Check your answer sheet after every five answers to make sure you haven’t skipped any bubbles by mistake.

### Should You Guess?

Use process of elimination (POE) to rule out answer choices you know are wrong and increase your chances of guessing the right answer. Read all the answer choices carefully. Eliminate the ones that you know are wrong. If you only have one answer choice left, *choose it,* even if you’re not completely sure why it’s correct. Remember: Questions in the multiple choice section are graded by a computer, so it doesn’t care *how* you arrived at the correct answer.

Even if you can’t eliminate answer choices, go ahead and guess. AP exams no longer include a guessing penalty of a quarter of a point for each incorrect answer. You will be assessed only on the total number of correct answers, so be sure to fill in all the bubbles even if you have no idea what the correct answers are. When you get to questions that are too time-consuming, or that you don’t know the answer to (and can’t eliminate any options), don’t just fill in any answer. Use what we call “your letter of the day” (LOTD). Selecting the same answer choice each time you guess will increase your odds of getting a few of those skipped questions right.

### Use the Two-Pass System

Do not waste time by lingering too long over any single question. If you’re having trouble, move on to the next question. After you finish all the questions, you can come back to the ones you skipped.

The best strategy is to go through the multiple choice section twice. The first time, do all the questions that you can answer fairly quickly—the ones where you feel confident about the correct answer. On this first pass, skip the questions that seem to require more thinking or the ones you need to read two or three times before you understand them. Circle the questions that you’ve skipped in the question booklet so that you can find them easily in the second pass. You must *be very careful* with the answer sheet by making sure the filled-in answers correspond correctly to the questions.

Once you have gone through all the questions, go back to the ones that you skipped in the first pass. But don’t linger too long on any one question even in the second pass. Spending too much time wrestling over a hard question can cause two things to happen: One, you may run out of time and miss out on answering easier questions in the later part of the exam. Two, your anxiety might start building up, and this could prevent you from thinking clearly, which would make answering other questions even more difficult. If you simply don’t know the answer, or can’t eliminate any of them, just use your LOTD and move on.

## HOW TO APPROACH FREE RESPONSE QUESTIONS

### Cracking Free Response Questions

Section II is worth 50 percent of your grade on the AP Calculus Exam. This section is composed of two parts. Part A contains two free response questions (you may use a calculator on this part); Part B contains four free response questions where there are no calculators allowed. You’re given a total of 90 minutes for this section.

### Clearly Explain and Justify Your Answers

Remember that your answers to the free response questions are graded by *readers* and not by computers. Communication is a very important part of AP Calculus. Compose your answers in precise sentences. Just getting the correct numerical answer is not enough. You should be able to *explain* your reasoning behind the technique that you selected and *communicate* your answer in the context of the problem. Even if the question does not explicitly say so, always explain and *justify* every step of your answer, including the final answer. Do not expect the graders to read between the lines. Explain everything as though somebody with no knowledge of calculus is going to read it. Be sure to present your solution in a systematic manner using solid logic and appropriate language. And remember: Although you won’t earn points for neatness, the graders can’t give you a grade if they can’t read and understand your solution!

### Use Only the Space You Need

Do not try to fill up the space provided for each question. The space given is usually more than enough. The people who design the tests realize that some students write in big letters and some students make mistakes and need extra space for corrections. So if you have a complete solution, don’t worry about the extra space. Writing more will not earn you extra credit. In fact, many students tend to go overboard and shoot themselves in the foot by making a mistake after they’ve already written the right answer.

### Read the Whole Question!

Some questions might have several subparts. Try to answer them all, and don’t give up on the question if one part is giving you trouble. For example, if the answer to part (b) depends on the answer to part (a), but you think you got the answer to part (a) wrong, you should still go ahead and do part (b) using your answer to part (a) as required. Chances are that the grader will not mark you wrong twice, unless it is obvious from your answer that you should have discovered your mistake.

### Use Common Sense

Always use your common sense in answering questions. For example, on one free response question that asked students to compute the mean weight of newborn babies from given data, some students answered 70 pounds. It should have been immediately obvious that the answer was probably off by a decimal point. A 70-pound baby would be a giant! This is an important mistake that should be easy to fix. Some mistakes may not be so obvious from the answer. However, the grader will consider simple, *easily recognizable errors* to be *very important.*

## REFLECT

Respond to the following questions:

- How long will you spend on multiple choice questions?
- How will you change your approach to multiple choice questions?
- What is your multiple choice guessing strategy?
- How much time will you spend on each free response question?
- How will you change your approach to the free response questions?
- Will you seek further help, outside of this book (such as a teacher, tutor, or AP Central), on how to approach the calculus exam?

## DERIVATIVES AND INTEGRALS THAT YOU SHOULD KNOW

1. [*ku*] = *k*

2. [*k*] = 0

3. [*uv*] = *u* + *v*

4.

5. [*e ^{u}*] =

*e*

^{u}6. [ln *u*] =

7. [sin *u*] = cos *u*

8. [cos *u*] = − sin *u*

9. [tan *u*] = sec^{2} *u*

10. [cot *u*] = − csc^{2} *u*

11. [sec *u*] = sec *u* tan *u*

12. [csc *u*] = − csc *u* cot *u*

13. [sin^{−1} *u*] =

14. [tan^{−1}] =

15. [sec^{−1} *u*] =

16. ∫*k du* = *ku* + *C*

17. ∫*u ^{n} du* = +

*C*;

*n*≠ −1

18. = ln |*u*| + *C*

19. ∫*e ^{u} du* =

*e*+

^{u}*C*

20. ∫sin *u du* = − cos *u* + *C*

21. ∫cos *u du* = sin *u* + *C*

22. ∫tan *u du* = − ln |cos *u*| + *C*

23. ∫cot *u du* = ln |sin *u*| + *C*

24. ∫sec *u du* = ln |sec *u* + tan *u*| + *C*

25. ∫csc *u du* = − ln |csc *u* + cot *u*| + *C*

26. ∫sec^{2} *u du* = tan *u* + *C*

27. ∫csc^{2} *u du* = − cot *u* + *C*

28. ∫sec *u* tan *u u du* = sec *u* + *C*

29. ∫csc *u* cot *u u du* = − csc *u* + *C*

30. + *C*; |*u*| < *a*

31. + *C*

32. + *C*; |*u*| > *a*

## PREREQUISITE MATHEMATICS

One of the biggest problems that students have with calculus is that their algebra, geometry, and trigonometry are not solid enough. In calculus, you’ll be expected to do a lot of graphing. This requires more than just graphing equations with your calculator. You’ll be expected to look at an equation and have a “feel” for what the graph looks like. You’ll be expected to factor, combine, simplify, and otherwise rearrange algebraic expressions. You’ll be expected to know your formulas for the volume and area of various shapes. You’ll be expected to remember trigonometric ratios, their values at special angles, and various identities. You’ll also be expected to be comfortable with logarithms and exponents.

### Powers

When you multiply exponential expressions with like bases, you add the powers.

*x ^{a}* •

*x*=

^{b}*x*

^{a+b}When you divide exponentiated expressions with like bases, you subtract the powers.

= *x ^{a−b}*

When you raise an exponentiated expression to a power, you multiply the powers.

(*x ^{a}*)

*=*

^{b}*x*

^{ab}When you raise an expression to a fractional power, the denominator of the fraction is the root of the expression, and the numerator is the power.

When you raise an expression to the power of zero, you get one.

*x*^{0} = 1

When you raise an expression to the power of one, you get the expression.

*x*^{1} = *x*

When you raise an expression to a negative power, you get the reciprocal of the expression to the absolute value of the power.

*x ^{−a}* =

### Logarithms

A logarithm is the power to which you raise a base, in order to get a value. In other words, log* _{b}x* =

*a*means that

*b*=

^{a}*x*. There are several rules of logarithms that you should be familiar with.

When you take the logarithm of the product of two expressions, you add the logarithms.

log(*ab*) = log *a* + log *b*

When you take the logarithm of the quotient of two expressions, you subtract the logarithms.

= log *a* – log *b*

When you take the logarithm of an expression to a power, you multiply the logarithm by the power.

log(*a ^{b}*) =

*b*log

*a*

The logarithm of 1 is zero.

log 1 = 0

The logarithm of its base is 1.

log* _{b} b* = 1

You cannot take the logarithm of zero or of a negative number.

In calculus, and virtually all mathematics beyond calculus, you will work with natural logarithms. These are logs with base *e* and are denoted by *ln*. Thus, you should know the following:

ln 1 = 0

ln *e* = 1

ln *e ^{x}* =

*x*

*e*^{ln}* ^{x}* =

*x*

The change of base rule is: log* _{b} x* =

### Geometry

The area of a triangle is (*base*)(*height*).

The area of a rectangle is (*base*)(*height*).

The area of a trapezoid is (*base*_{1} + *base*_{2})(*height*).

The area of a circle is *πr*^{2}.

The circumference of a circle is 2*πr*.

The Pythagorean theorem states that the sum of the squares of the legs of a right triangle equals the square of the hypotenuse. This is more commonly stated as *a*^{2} + *b*^{2} = *c*^{2} where *c* equals the length of the hypotenuse.

The volume of a right circular cylinder is *πr*^{2}*h*.

The surface area of a right circular cylinder is 2*πrh*.

The volume of a right circular cone is *πr*^{2}*h*.

The volume of a sphere is *πr*^{3}.

The surface area of a sphere is 4*πr*^{2}.

### Trigonometry

Given a right triangle with sides *x, y,* and *r* and angle *q* below:

sin(*A* + *B*) = sin *A* cos *B* + cos *A* sin *B*

sin(*A* – *B*) = sin *A* cos *B* – cos *A* sin *B*

cos(*A* + *B*) = cos *A* cos *B* – sin *A* sin *B*

cos(*A* – *B*) = cos *A* cos *B* + sin *A* sin *B*

You must be able to work in radians and know that 2*π* = 360°.

You should know the following:

# Part IV

# Drills

1 Limits Drill 1

2 Limits Drill 1 Answers and Explanations

3 Limits Drill 2

4 Limits Drill 2 Answers and Explanations

5 Functions and Domains Drill

6 Functions and Domains Drill Answers and Explanations

7 Derivatives Drill 1

8 Derivatives Drill 1 Answers and Explanations

9 Derivatives Drill 2

10 Derivatives Drill 2 Answers and Explanations

11 Derivatives Drill 3

12 Derivatives Drill 3 Answers and Explanations

13 Applications of Derivatives Drill 1

14 Applications of Derivatives Drill 1 Answers and Explanations

15 Applications of Derivatives Drill 2

16 Applications of Derivatives Drill 2 Answers and Explanations

17 General and Partial Fraction Integration Drill

18 General and Partial Fraction Integration Drill Answers and Explanations

19 Trigonometric Integration Drill

20 Trigonometric Integration Drill Answers and Explanations

21 Exponential and Logarithmic Integration Drill

22 Exponential and Logarithmic Integration Drill Answers and Explanations

23 Areas, Volumes, and Average Values Drill

24 Areas, Volumes, and Average Values Drill Answers and Explanations

25 Free Response Drill

26 Free Response Drill Answers and Explanations

# Chapter 1

# Limits Drill 1

## LIMITS DRILL 1

1.

(A) –∞

(B) –2

(C) 0

(D) 2

(E) ∞

2.

(A) ∞

(B)

(C) 1

(D) –∞

(E) Does not exist

3.

(A) –∞

(B) 0

(C) 1

(D) ∞

(E) Does not exist

4.

(A) 0

(B)

(C) 2

(D) 5

(E) Does not exist

5.

(A) –

(B) –1

(C) 1

(D)

(E) Does not exist

6.

(A) –∞

(B) –3

(C) 0

(D) 3

(E) ∞

7.

(A) –∞

(B) –

(C) 0

(D)

(E) Does not exist

8.

(A) –∞

(B) –1

(C) 0

(D) ∞

(E) Does not exist

9.

(A) 0

(B)

(C)

(D) 4

(E) Does not exist

10.

(A) 0

(B) 2

(C) 8

(D) 10

(E) 14

11.

(A) 7

(B) 1

(C) 0

(D) –7

(E) Does not exist

12.

(A)

(B) 1

(C)

(D) 2

(E) Does not exist

13.

(A) ∞

(B)

(C)

(D) 0

(E) Does not exist

14.

(A) 1

(B)

(C)

(D) ∞

(E) Does not exist

15.

(A) Does not exist

(B) 0

(C) 1

(D) –1

(E) –

16.

(A) 0

(B) –1

(C) –

(D)

(E) Does not exist

17.

(A) –25

(B) 25

(C) 5

(D) ∞

(E) Does not exist

18.

(A) –49

(B) –14

(C) 0

(D) 14

(E) 49

19.

(A) Does not exist

(B) –1

(C) 0

(D) 1

(E) ∞

20.

(A) Does not exist

(B) 1

(C)

(D) –

(E) –1

21.

(A) 0

(B) 1

(C) 3

(D) 5

(E) The limit does not exist.

# Chapter 2

# Limits Drill 1 Answers and Explanations

## ANSWER KEY

1. D

2. A

3. B

4. B

5. D

6. D

7. B

8. C

9. B

10. E

11. A

12. C

13. D

14. B

15. E

16. C

17. A

18. B

19. E

20. D

21. B

## EXPLANATIONS

1. **D** To do this question, we need to remember how to attack limits. These are generally good candidates for the first pass. When the limit approaches infinity, simply divide both top and bottom by the highest power of *x* in the fraction. Foiling out the bottom produces a 2*x*^{4} as the leading term, so both top and bottom will be divided by an *x*^{4} term.

Simplify.

As *x* gets larger, the fractional terms approach 0 so all that is left is which equals 2.

Also, when limits approach infinity, it is good to keep in mind that when the degrees are the same in the numerator and denominator, then you only need to look to the leading coefficients to find the limits.

2. **A** When calculating limits approaching ∞, you should pay attention to the highest degree of the numerator and denominator. If the highest degree is in the numerator, then the limit will become infinitely larger without bound, thus making it ∞. The long way is to divide each term by the highest term, in this case, *x*^{5}.

Simplify

As *x* approaches ∞, the fractional terms will get smaller and smaller, going to 0. The only value left is in the numerator and as *x* would get larger and larger, the limit increases without bound, making it ∞.

3. **B** When we look to the highest degree, it is in the denominator, which, when limits approach infinity, automatically means the entire expression will approach 0.

Divide by the highest term throughout the expression.

Simplify.

4. **B** Know the rule: When a limit approaches infinity and the degrees in the numerator and denominator are the same, simply look to the leading coefficients to calculate the limit. In this case, the leading coefficients are the 3 in the numerator and the 6 in the denominator . reduces to , making it your answer.

5. **D** This question tests your knowledge of discontinuity. If you plug in –3 directly, the denominator becomes 0, and thus is undefined. If we factor and cancel the numerator and denominator, the problem becomes solvable.

Factor and cancel out like terms.

Plug in –3 to evaluate the limit.

6. **D** This question tests your knowledge of discontinuity. If you plug in 0 directly, the denominator becomes 0, and thus is undefined. If we expand out the numerator and reduce, the problem becomes solvable.

Expand the numerator.

= = =

Plug in 0 to evaluate the limit.

0^{2} − 3(0) + 3 = 0^{2} − 3(0) + 3 = 3

7. **B** This question tests your knowledge of discontinuity. If you plug in 16 directly, the denominator becomes 0, and thus is undefined. If we multiply both the numerator and denominator by 4 + , then the question becomes solvable.

Multiply top and bottom by 4 + .

Plug in 16 to evaluate the limit.

8. **C** This question tests your knowledge of discontinuity. If you plug in 0 directly, the denominator becomes 0, and thus is undefined. If we multiply both the numerator and denominator by 1 + , then the question becomes solvable.

Multiply the top and bottom by 1 + .

Plug in 0 to evaluate the limit.

9. **B** This question tests your knowledge of discontinuity. If you plug in 1 directly, the denominator becomes 0, and thus is undefined. If you factor and cancel like terms, then the question becomes solvable.

Factor and cancel like terms.

Plug in 1 and evaluate the limit.

10. **E** This question tests your knowledge of discontinuity. If you plug in 0 directly, the denominator becomes 0, and thus is undefined. If you expand out the numerator and cancel like terms, then the question becomes solvable.

Expand out numerator and cancel like terms.

Plug in 0 to evaluate the limit.

*x* + 14 = 0 + 14 = 14

11. **A** To calculate a limit as it approaches infinity, look to the terms in the numerator and denominator to find the highest degrees. In this case, they are both 10, so the limit is simply the coefficients divided by each other, , or 7.

12. **C** To calculate a limit as it approaches infinity, look to the terms in the numerator and denominator to find the highest degrees. In this case, they are both 2, so the limit is simply the coefficients divided by each other, .

13. **D** To calculate a limit as it approaches infinity, look to the terms in the numerator and denominator to find the highest degrees. In the numerator, the degree is 2, where in the denominator, the degree is 3. When the degree in the denominator is greater than the degree in the numerator, the limit will approach 0.

14. **B** The first instinct in limit questions is to try to plug in the value *x* is approaching. If you plug 3 in, the denominator becomes 0, thus undefined. However, using a little algebra, we are able to remove a term from the numerator and denominator.

Now that the numerator and denominator are simplified, you can simply plug 3 in for *x* to evaluate the limit.

15. **E** If you plug in 0 directly, the value becomes and can’t be evaluated. However, an *h* is able to be cancelled out of the denominator by factoring.

Now that the numerator and denominator are simplified, you can simply plug 0 in for *h* to evaluate the limit.

16. **C** If you plug in –2, the limit becomes , which is undefined. By factoring the denominator, the discontinuity can be removed which will allow calculation.

Now the limit is able to be evaluated by plugging in –2 for *x.*

17. **A** If you plug in 5 immediately, the limit becomes , which is undefined. By factoring the numerator, the discontinuity can be removed which will allow calculation.

Now the limit is able to be evaluated by plugging in 5 for *x.*

−(5)(5) = −25

18. **B** If you plug in 7 directly to the limit, then it becomes which is undefined. First, the numerator must be factored to remove the discontinuity.

Plug in 7 to evaluate the limit.

−(7) + (7) = −(14) = −14

19. **E** When evaluating limits at infinity, you must pay attention to the degrees in both the numerator and denominator. In this case, the degree in the numerator is 100, while in the denominator, it’s 99. Since the degree is higher in the numerator, the limit will go to infinity as *x* approaches infinity.

20. **D** If you plug in 3 directly, the limit becomes and is undefined. First, the numerator and denominator must be factored.

Now, plug 3 in for *x* to evaluate the limit.

21. **B** .

# Chapter 3

# Limits Drill 2

## LIMITS DRILL 2

1.

(A) Does not exist

(B) 2

(C) 0

(D) –2

(E) ∞

2.

(A) 6

(B) 3

(C) 0

(D) –6

(E) Does not exist

3.

(A) Does not exist

(B) ∞

(C) 6

(D) 0

(E)

4.

(A) ∞

(B) 2

(C) 5

(D) 0

(E) Does not exist

5.

(A) ∞

(B)

(C) 1

(D)

(E) Does not exist

6.

(A) ∞

(B)

(C) 1

(D)

(E) Does not exist

7.

(A) ∞

(B) 7

(C) 6

(D) 0

(E)

8.

(A) 0

(B) –1

(C) 1

(D) ∞

(E) Does not exist

9.

(A) 0

(B) 12

(C)

(D) 1

(E) ∞

10.

(A) ∞

(B) 3

(C) 0

(D) –3

(E) Does not exist

11.

(A) ∞∞

(B) 6

(C) –4

(D) –6

(E) Does not exist

12.

(A) Does not exist

(B) –81

(C) –18

(D) –9

(E) 0

13.

(A) Does not exist

(B) 0

(C)

(D) 1

(E) ∞

14.

(A) –

(B) –1

(C) 0

(D) 1

(E)

15.

(A) ∞

(B) 1

(C)

(D)

(E) 0

16.

(A) 0

(B) 1

(C) 2

(D) –1

(E) Does not exist

17.

(A) –1

(B) 0

(C)

(D) 1

(E) ∞

18.

(A) –

(B)

(C) 1

(D) 7

(E) ∞

19.

(A) Does not exist

(B) 1

(C) 0

(D) –1

(E) –∞

20.

(A) 0

(B) 1

(C) 1.5

(D) 2

(E) Does not exist

# Chapter 4

# Limits Drill 2 Answers and Explanations

## ANSWER KEY

1. B

2. A

3. E

4. C

5. D

6. D

7. E

8. A

9. B

10. B

11. D

12. E

13. C

14. A

15. E

16. A

17. C

18. B

19. D

20. C

## EXPLANATIONS

1. **B** If you plug *x* = 1 into the limit, it becomes , which is undefined. First, a little factoring must be done in the numerator to remove the discontinuity.

= *x* + 1

Now, you can evaluate the limit in its reduced form.

(*x* + 1) = 2

2. **A** If you plug *x* = 0 into the limit, it becomes , which is undefined. First, the numerator must be expanded so an *x* term can cancel out.

= 6 + *x*

Now, you can evaluate the limit in its reduced form.

(6 + 0) = 6

3. **E** If you plug *x* = 0 into the limit, it becomes , which is undefined. First, you must rationalize the expression so the *x*^{2} term can be cancelled out.

= =

Now, you can evaluate the limit in its reduced form.

4. **C** If you plug *x* = 2 into the limit, it becomes , which is undefined. First, you must factor the numerator to cancel out the like terms.

= *x* + 3

Now, you can evaluate the limit in its reduced form.

2 + 3 = 5

5. **D** If you plug *x* = –4 into the limit, it becomes , which is undefined. First, you must factor both the numerator and denominator to cancel like terms.

Now, you can evaluate the limit in its reduced form.

6. **D** If you plug *x* = 4 into the limit, it becomes , which is undefined. First, you must factor both the numerator and denominator to cancel like terms.

Now, you can evaluate the limit in its reduced form.

7. **E** If you plug *x* = 7 into the limit, it becomes , which is undefined. First, you must rationalize the expression to eliminate common terms.

Now, you can evaluate the limit in its reduced form.

8. **A** If you plug *x* = –1 into the limit, it becomes , which is undefined. First, you must factor the numerator and denominator to cancel out like terms.

Now, you can evaluate the limit in its reduced form.

= 0

9. **B** If you plug *x* = 0 into the limit, it becomes , which is undefined. First, you must expand out the numerator to cancel out like terms.

= 12 + 6*x* + *x*^{2}

Now, you can evaluate the limit in its reduced form.

12 + 6(0) + (0)^{2} = 12

10. **B** If you plug 0 in directly, the limit becomes , which is undefined. First, the numerator and denominator must be factored to cancel out like terms.

Now, you can plug 0 in for *x* to evaluate the limit.

0 + 3 = 3

11. **D** If you plug –4 directly into the limit, it becomes , which is undefined. First, the numerator must be factored to cancel out the like term.

= *x* − 2

Now –4 can be plugged in for *x* to evaluate the limit.

(–4) – 2 = –6

12. **E** If you plug –9 directly into the limit, it becomes which is undefined. First, you must factor the numerator and cancel the like term.

= *x* + 9

Plug –9 in for *x* to evaluate the limit.

(–9) + 9 = 0

13. **C** If you plug 2 in for *x*, the limit becomes which is undefined. First, you must factor the denominator to cancel out the like term.

Now, 2 can be plugged in to *x* to evaluate the limit.

14. **A** When evaluating limits at infinity, you must look to the degrees in the numerator and the denominator, which in this case, are the same at 2. Since you would simply divide each term by the highest degree, you must only look to the coefficients of the highest degree terms. Since in the numerator the leading coefficient is 9 and in the denominator the leading coefficient is –5, you only need to divide them to find the limit. The limit is .

15. **E** If you plug zero directly into the limit, the result is , which is undefined. You must first factor the numerator and denominator to cancel out like terms.

Now zero can be plugged in to calculate the limit.

= 0

16. **A** If you plug –1 directly into the limit, the value becomes which is undefined. First, you must factor both numerator and denominator and look for terms to cancel out.

Now –1 can be plugged in to evaluate the limit.

= 0

17. **C** To evaluate this limit, first, you must rationalize the numerator.

= =

Plug 0 into the reduced function to evaluate the limit.

18. **B** If you plug –7 directly into the limit, it becomes which is undefined. First, factor the expression on top and bottom and cancel out any like terms.

Now plug –7 in for *x* to evaluate the limit.

19. **D** If you plug zero directly into the limit, it becomes which is undefined. You must first factor the denominator to cancel any like terms.

Now you can plug in 0 for *x* to evaluate the limit.

20. **C** Begin by separating the functions into sines and cosines to evaluate:

.

# Chapter 5

# Functions and Domains Drill

## FUNCTIONS AND DOMAINS DRILL

1. If *f*(*x*) = , *g*(*x*) = *x*^{2} – 36, and *h*(*x*) = *x*^{3} + 2, then *f*(*g*(*h*(2))) =

(A) –10

(B) –8

(C) 0

(D) 8

(E) 10

2. What is the domain of *f*(*x*) = ?

(A) 0 < *x* < 5

(B) All real numbers; *x* ≠ 0

(C) All real numbers; *x* ≠ 5

(D) All real numbers; *x* ≠ 0, 5

(E) *x* < 0 and *x* > 5

3. Find the domain of *f*(*x*) = .

(A) (–∞,–1)

(B) (–1,1)

(C) (1,∞)

(D) (–∞,1)

(E) (–1,∞)

4. If *f*(*x*) = 3*x* + 2 and *g*(*x*) = (*x* – 2)^{2}, then *g* (*f* (0)) =

(A) –14

(B) –4

(C) 0

(D) 4

(E) 14

5. What is the domain of ?

(A) All real numbers

(B) All real numbers ; *x* ≠ 0

(C) All real numbers ; *x* ≠ 2

(D) All real numbers ; *x* ≠ –2

(E) All real numbers ; *x* ≠ ± 2

6. Consider the function, *f*(*x*) = *x*^{3} + *x*^{2} − 2*x* + 6. Which of the following must be false?

(A) There is a relative maximum at *x* = –2.

(B) There is an absolute maximum on the interval [2,4] at *x* = 4.

(C) There is a relative minimum at *x* = .

(D) There is an absolute minimum at *x* = , on the interval [–4, 1].

(E) There is an absolute maximum at *x* = –2, on the interval [–4, 1].

7. Consider the function *f*(*x*) = . What type of discontinuity occurs at *x* = 2?

(A) point

(B) essential

(C) removable

(D) jump

(E) There is no discontinuity at *x* = 2.

8. For what value of *a* is the function *f*(*x*) = continuous at *x* = 1?

(A) –3

(B) –2

(C) –1

(D) 0

(E) 1

9. Given the function *f*(*x*) = *f*(*x*) = , at what value of *a* will the function be continuous?

(A) –10

(B) –7

(C) –5

(D) 1

(E) 5

# Chapter 6

# Functions and Domains Drill Answers and Explanations

## ANSWER KEY

1. D

2. E

3. B

4. B

5. E

6. D

7. D

8. B

9. C

## EXPLANATIONS

1. **D** We must combine the functions starting from the innermost operation and work outwards.

Calculate *h*(2) first.

*h*(2) = 2^{3} + 2 = 10

Calculate *g*(10) next.

*g*(10) = (10)^{2} – 36 = 64

Calculate *f*(64) next.

*f*(64) = = 8.

2. **E** To find the domain of this function, we must find where the denominator is equal to zero.

Set the denominator equal to zero and solve for *x*.

= 0*x*^{2} − 5*x* = 0*x*(*x* − 5) = 0

*x* = 0; *x* = 5

Since the radical is in the denominator, and requires an even root, the values of *x* must produce a positive radicand, so we must test values to see what yields a result greater than zero. Test values around the *x*-values.

*f*(–1) = + *f*(2) = – *f*(6) = +

Therefore, the domain must be *x* < 0 and *x* > 5.

3. **B** Find the domain of *f*(*x*) = .

Since the function has an even root, it must be greater than or equal to zero. But first, we must set it equal to zero to find what *x* is equal to.

= 0

1 − *x*^{2} = 0

1 = *x*^{2}

±1 = *x*

Test values around the *x*-values found in step 1.

*f*(–2) = – *f*(0) = + *f*(2) = –

Therefore, the domain must be (–1, 1).

4. **B** We must calculate the value of the composite function starting with the innermost value. Find *f*(0) first.

*f*(0) = 3(0) + 2 = 2

Find *g*(2) next.

*g*(2) = (2 – 2)^{2} = 0

5. **E** Set the denominator equal to zero and solve for *x*.

*x*^{2} – 4 = 0

*x*^{2} = 4

*x* = ± 2

6. **D** Set the first derivative equal to zero to determine the critical points: *f*′(*x*) = 3*x*^{2} + 5*x* − 2 = 0. Thus, *x* = and *x* = –2. In order to determine whether these points are maximums or minimums, take the second derivative and evaluate it at each of the critical points: *f*″(*x*) = 6*x* + 5, and *f*″(–2) = –7. Since, the second derivative is positive at *x* = , that *x* = is at a minimum. Similarly, because the second derivative is negative at *x* = –2, *x* = –2 is at a maximum. These points are relative minima and maxima over the entire function, because there are no bounds on the function. These points are absolute minima and maxima on closed intervals if there are no other points that are less or greater than those.

7. **D** Graph the function to see the discontinuity. If you cannot, notice that the function shifts position at *x* = 2 and it is not continuous at that point. The limits on either side of *x* = 2 are not equal, so the discontinuity cannot be a point or removable. Furthermore, there is no asymptote, so the discontinuity is not essential. Therefore, it is a jump discontinuity.

8. **B** There are three conditions that must be satisfied for a function to be continuous: 1. *f*(*c*) exists. 2. *f*(*x*) exists. 3. *f*(*x*) = *f*(*c*). For this function, condition 1 is met because *f*(1) = 0. For the function to be continuous, the left and right hand limits must be equal when approaching *x* = 1. Therefore, we must set *a* + 2 = 0 and *a* = –2. When *a* = –2, the third condition is also met: the limit will equal 0 and *f*(1) = 0.

9. **C** For a function to be continuous, three conditions need to be met: *f*(*c*) exists, *f*(*x*) exists, and *f*(*x*) = *f*(*c*). For a piecewise function, the *f*(*c*) and the limit as *x* approaches *c* must be equal for both pieces, especially at the point where the pieces meet, in this case, *x* = 1. So, first determine *f*(1) by plugging 1 into the top piece: *f*(1) = –2. Then, set the second piece equal to that value and solve for *a*. When this is done, *a* = –5. To verify that the function is truly continuous, take the left and right hand limits of *f*(*x*) as *x* approaches 1. Both the left and right hand limits equal –2, so the limit exists and it is equal to *f*(1), so the function is continuous when *a* = –5.

# Chapter 7

# Derivatives Drill 1

## DERIVATIVES DRILL 1

1. If *y* = (*x*^{2} + 1)^{5}, then *y*′ =

(A) 5 (*x*^{2} + 1)^{4}

(B) 10*x* (*x*^{2} + 1)^{4}

(C) 5*x* (*x*^{2} + 1)^{4}

(D) 5 (2*x*)^{4}

(E) (2*x*)^{5}

2. If *y* = 4*π*^{2}*x*^{2}, then what is *y*′?

(A) 8*π*^{2}*x*

(B) 8*πx*^{2} + 8*π*^{2}*x*

(C) 16*πx*

(D) 2*π*^{2}*x*^{2}

(E) 0

3. If *f*(*x*) = sin *x* cos *x*, then =

(A) –2

(B) –1

(C) 0

(D) 1

(E) 2

4. If *f*(*x*) = 3*x*^{3} – 2*x*^{2} + *x*, then (2) *f*′(2) =

(A) 39

(B) 29

(C) 19

(D) 9

(E) –9

5. If *y* = *x* cos *x*, then find *y*′.

(A) cos *x* + *x* sin *x*

(B) –sin *x*

(C) cos *x* – *x* sin *x*

(D) –*x* sin *x*

(E) *x* sin *x* – cos *x*

6. Find *y*′ if *y* = *x* csc *x* – cot *x*.

(A) csc *x* (csc *x* – cot *x*)

(B) csc *x* (csc *x* + 1 – *x* cot *x*)

(C) csc *x* (csc *x* + cot *x*)

(D) –csc *x* (*x* cot *x* – csc *x*)

(E) *x* csc^{2} *x* – cot *x*

7. What is the derivative of *y* = ?

(A)

(B)

(C)

(D)

(E)

8. If *f*(*x*) = sin *x* ln *x*, then what is *f*′(*x*)?

(A)

(B)

(C)

(D)

(E)

9. If *y* = sin (ln *x*), then *y*′ =

(A) cos (ln *x*)

(B) sin (ln *x*) +

(C)

(D)

(E)

10. If *f*(*x*) = *x*^{3} – *x*^{2}, then what is *f*′(*x*)?

(A) 3*x*^{2} – 3*x* ln 3

(B) 3*x*^{2} – 3*x* – 1

(C) 3*x*^{2} +3*x* ln 3

(D) 3*x*^{2} – *x* ln 3*x*

(E)

11. If *y* = cos *x*, find *y*′.

(A)

(B)

(C) − sin *x*

(D) − *x* sin *x* − cos *x*

(E) *x*^{2} sin 2 +

12. If *y* = *x*^{2} ln (2*x*), then what is the value of at *x* = ?

(A) – ln 4

(B) –

(C) 0

(D)

(E) ln 4

13. Find *f*′(2) if *f*(*x*) = .

(A) –4

(B) –

(C) 0

(D)

(E) 4

14. If *y* = , then find *y*′.

(A)

(B)

(C)

(D)

(E)

15. What is *f*′ if *f*(*x*) = *x*^{3} cos *x*?

(A) 3*x*^{2} cos *x* – *x*^{3} sin *x*

(B) –3*x*^{2} sin *x*

(C) 3*x*^{2} cos *x* + *x*^{3} sin *x*

(D) *x*^{3} sin *x* – 3*x*^{2} cos *x*

(E) 3*x*^{2} sin *x*

16. If *y* = , find .

(A)

(B)

(C)

(D)

(E)

17. If *f*(*θ*) = *θ* sin *θ*, then what is *f*″(*θ*)?

(A) *θ* sin *θ* – 2 cos *θ*

(B) cos *θ* – *θ* sin *θ*

(C) – sin *θ*

(D) *θ* cos *θ* – 2 sin *θ*

(E) 2 cos *θ* – *θ* sin *θ*

18. If *y* = *x*^{2} tan *x*, then *y*′ =

(A) *x*^{2} sec^{2} *x*

(B) 2*x* sec^{2} *x*

(C) 2*x* tan *x* + *x*^{2} sec^{2}*x*

(D) 2*x* tan *x* – *x*^{2} sec^{2}*x*

(E) –*x*^{2} sec^{2}*x*

19. If *f*(*x*) = 3*x*^{2} – 3*x* ln 6, then *f*′(*x*) =

(A) 6*x* + 3 ln 6

(B) 66 ln (*x* + 3)

(C) (*x* + 3) 6*x* + 2

(D) 63 ln (*x* + 3)

(E) (*x* + 3) ln 6

20. If *f*(*x*) = , then *f*′(*x*) =

(A)

(B)

(C)

(D)

(E)

21. If *y* = ln (4*x*^{2}), then *y*′ =

(A) 8*x*

(B) 8*x* ln (4*x*^{2})

(C)

(D)

(E)

22. If *y* = *x*^{3} sin^{2}*x*, then *y*′ =

(A) 3*x*^{2} sin^{2}*x* + 2*x*^{3} sin 2*x*

(B) 3*x*^{2} sin *x* cos *x*

(C) 3*x*^{2} sin^{2}*x* + *x*^{3} sin 2*x*

(D) 6*x*^{2} sin *x* cos *x*

(E) 6*x*^{2} sin *x* + 2*x*^{3} sin *x* cos *x*

23. If *f*(*x*) = (*x* – 2)(2*x* + 3), then *f*′(*x*) =

(A) 2

(B) 2*x*

(C) 4*x*

(D) 4*x* + 1

(E) 4*x* – 1

24. If y = *x*^{6} – 3*x*^{4} + *x*, then *y*′ =

(A) 3*x*^{5} – 12*x*^{3} + 1

(B) – 3*x*^{5} + 12*x*^{3} – 1

(C) *x*^{7} – *x*^{5} + *x*^{2}

(D) *x* (*x*^{5} – 6*x* + 2)

(E) 15*x*^{4} – 36*x*^{2}

25. Find *f*″(*x*) if *f*(*x*) = sin 3*x* + 2cos *x* – sin 2*x*.

(A) 9 sin 3*x* – 4 sin 2*x* + 2 cos *x*

(B) –9 sin 3*x* + 4 sin 2*x* – 2 cos *x*

(C) 4 sin 2*x* – 9 sin 3*x* – 2 cos *x*

(D) 4 sin 3*x* – 9 sin 2*x* – 2 cos *x*

(E) –4 sin 2*x* + 9 sin 3*x* + 2 cos *x*

26. Find *y*′ if 2 + = 10

(A)

(B) –

(C) –

(D)

(E)

27. What is *y*′ if 2*x*^{3} + *x*^{2}*y* – *xy*^{3} = 9?

(A)

(B)

(C)

(D)

(E)

28. What is the derivative of *y*^{5} + *x*^{2}*y*^{3} = 1 + *x*^{4}*y*?

(A)

(B)

(C)

(D)

(E)

29. Find *y*′ if 1 + *x* = sin (*xy*^{2}).

(A)

(B)

(C)

(D) cos (*xy*^{2}) – 1

(E) 2*xy* cos (*xy*^{2}) – 1

30. (2(*x*^{2} + 3)^{3}) =

(A) 6(*x*^{2} + 3)^{2}

(B) 4*x*(*x*^{2} + 3)^{2}

(C) 12*x*(*x*^{2} + 3)^{2}

(D) 6*x*(*x*^{2} + 3)^{2}

(E) 4(*x*^{2} + 3)^{2}

# Chapter 8

# Derivatives Drill 1 Answers and Explanations

## ANSWER KEY

1. B

2. A

3. B

4. B

5. C

6. B

7. A

8. D

9. D

10. A

11. A

12. D

13. A

14. C

15. A

16. E

17. E

18. C

19. A

20. D

21. E

22. C

23. E

24. A

25. C

26. A

27. A

28. D

29. A

30. C

## EXPLANATIONS

1. **B** Take the derivative using the chain rule.

*y*′ = 5(*x*^{2} + 1)^{4} × (2*x*)

= 10*x* (*x*^{2} + 1)^{4}

2. **A** Find the derivative.

*y*′ = 4*π*^{2} × 2*x*

= 8*π*^{2}*x*

3. **B** Take the derivative using the product rule and trigonometric derivatives.

*f*′(*x*) = cos *x* cos *x* + sin *x* (–sin *x*)

Don’t simplify! Plug in .

= (0)(0) – (1)(1)

= –1

4. **B** Take the derivative and plug in 2.

*f*′(*x*) = 9*x*^{2} – 4*x* + 1

*f*′(2) = 9(2)^{2} – 4(2) + 1

= 36 – 8 + 1

= 29

5. **C** This question is testing your knowledge of the product rule and derivatives of trigonometric functions.

Find *y*′.

*y*′ = (1) cos *x* + *x*(–sin *x*)

= cos *x* – *x*sin *x*.

6. **B** This question is testing your knowledge of the product rule and derivatives of trigonometric functions.

The derivative of *x*csc *x* is (1)csc *x* + *x* (–csc *x* cot *x*) and the derivative of cot *x* is –csc^{2}*x*. That makes the derivative all together csc*x* – *x*csc *x* cot *x* + csc^{2}*x*.

Factor out a term of csc *x* to match the answer choices.

*y*′ = csc *x* (csc *x* + 1 – *x*cot *x*)

7. **A** This question is testing you on the quotient rule of derivatives.

Take the derivative of the function.

8. **D** This question is testing you on the product rule, trigonometric differentiation, and the derivative of the natural logarithm.

Take the derivative

*f*′(*x*) = cos *x* × ln *x* + sin *x* ×

= cos *x* ln *x* +

9. **D** This question is testing you on the chain rule, trigonometric differentiation, and the derivative of the natural logarithm.

Take the derivative of the function, starting from the outside and working your way in.

10. **A** Take the derivative.

*f*′(*x*) = 3*x*^{2} – 3*x* ln 3

11. **A** Take the derivative.

12. **D** Take the derivative.

*y*′ = 2*x* ln (2*x*) + *x*^{2} × 2

Don’t simplify! Just plug in for *x*.

13. **A** Take the derivative using the quotient rule.

Don’t simplify! Plug in 2 for *x*.

14. **C** Find the derivative using the quotient rule in combination with the chain rule.

15. **A** Find the first derivative using the product rule.

*f*′(*x*) = (3*x*^{2})(cos *x*) + (*x*^{3})(–sin *x*)

= 3*x*^{2} cos *x* – *x*^{3} sin *x*

16. **E** Find the derivative using the quotient rule and trigonometric derivatives.

*y*′ =

17. **E** Find the second derivative.

*f*′(*θ*) = sin *θ* + *θ* cos *θ*

*f*″(*θ*) = cos *θ* + cos *θ* – *θ* sin *θ*

= 2 cos *θ* – *θ* sin *θ*

18. **C** If *y* = *x*^{2} tan *x*, then *y*′ =

Take the derivative using trigonometric derivatives and the product rule.

*y*′ = (2*x*) tan *x* + (*x*^{2}) sec^{2}*x*.

19. **A** Take the derivative using the rule of raising a number to a function.

*f*′(*x*) = 6*x* + 3 ln 6.

20. **D** Take the derivative using the quotient rule.

21. **E** Take the derivative using the derivative of the natural log function and the chain rule.

22. **C** Take the derivative using the product rule, chain rule, and trigonometric derivatives.

*y*′ = 3*x*^{2} sin^{2}*x* + *x*^{3} (2 sin *x* cos *x*)

Use the trigonometric identity sin 2*x* = 2 sin *x* cos *x* to rewrite the derivative.

*y*′ = 3*x*^{2} sin^{2}*x* + *x*^{3} sin 2*x*.

23. **E** Although the product rule could be used to evaluate the derivative, this question is quickly solved if you FOIL out the function first.

(*x* – 2)(2*x* + 3) = 2*x*^{2} – *x* – 6

Find the derivative.

*f*(*x*) = 2*x*^{2} – *x* – 6

*f*′(*x*) = 4*x* – 1

24. **A** Calculate the derivative using the power rule for each term.

*y* = *x*^{6} – 3*x*^{4} + *x*

*y*′ = 3*x*^{5} – 12*x*^{3} + 1

25. **C** Calculate the first derivative using the trigonometric derivatives and the chain rule.

*f*(*x*) = sin 3*x* + 2 cos *x* – sin 2*x*

*f*′(*x*) = 3 cos 3*x* – 2 sin *x* – 2 cos 2*x*

Now take the second derivative paying attention to the sign changes when applying the trigonometric derivatives.

*f*′(*x*) = 3 cos 3*x* – 2 sin *x* – 2 cos 2*x*

*f*″(*x*) = –9 sin 3*x* – 2 cos *x* + 4 sin 2*x*

*f*″(*x*) = 4 sin 2*x* – 9 sin 3*x* – 2 cos *x*

26. **A** Take the derivative with respect to *x.*

27. **A** Take the derivative with respect to *x* using the power rule and product rule.

28. **D** Take the derivative with respect to *x* using the power rule and product rule.

29. **A** Take the derivative with respect to *x* using the product rule, the chain rule, and the differentiation formula for the sine function.

30. **C** (2(*x*^{2} + 3)^{3}) = 2((*x*^{2} + 3)^{2}(2*x*)) = 12*x*(*x*^{2} + 3)^{2}

# Chapter 9

# Derivatives Drill 2

## DERIVATIVES DRILL 2

1. If *f*(*x*) = sin (csc *x*), then *f*′(*x*) =

(A) csc *x* cot *x* cos (csc *x*)

(B) –csc *x* cot *x* cos (csc *x*)

(C) cot^{2} ^{x} cos (csc *x*)

(D) –cot^{2} ^{x} cos (csc *x*)

(E) cos (–csc *x* cot *x*)

2. If *y* = 6*x*^{5} – 5*x*^{4}, then *y*^{(4)} =

(A) 30*x*^{4} – 20*x*^{3}

(B) 120*x*^{3} – 60*x*^{2}

(C) 360*x*^{2} – 120*x*

(D) 720*x* – 120

(E) 720

3. Find the derivative of *f*(*x*) = *e*^{2}* ^{x}* sin

^{2}3

*x*.

(A) 2*e*^{2}* ^{x}* sin 3

*x*(sin 3

*x*+ 3cos 3

*x*)

(B) 6*e*^{2}* ^{x}* sin

^{2}3

*x*+ 2

*e*

^{2}

*sin 3*

^{x}*x*cos

*x*

(C) 2*e*^{2}* ^{x}* sin 3

*x*(sin 3

*x*– 3cos 3

*x*)

(D) –2*e*^{2}* ^{x}* sin 3

*x*(sin 3

*x*+ 3cos 3

*x*)

(E) 2*e*^{2}* ^{x}* sin 3

*x*cos

*x*– 6

*e*

^{2}

*sin*

^{x}^{2}3

*x*

4. If *y* = *e ^{x}* (ln

*x*)

^{2}, then

*y*′ =

(A) 2*e ^{x}* ln

*x*

(B) *e ^{x}* ln

*x*

(C) *e ^{x}* ln

*x*

(D) 2*x*e* ^{x}* ln

*x*

(E)

5. If *y* = tan^{2} (3*θ*), then *y*′ =

(A) sec^{2} (3*θ*) – 1

(B) 2 tan (3*θ*)

(C) 6 sec^{2} (3*θ*)

(D) 6 tan (3*θ*) sec^{2} (3*θ*)

(E) 3 tan (3*θ*) sec^{2} (3*θ*)

6. If *f*(*x*) = sin (sin (sin *x*)), then what is *f*′(*x*)?

(A) cos (cos (cos *x*))

(B) cos *x* cos (sin *x*) cos (sin (sin *x*))

(C) cos (sin (cos *x*))

(D) cos *x* + sin *x* cos *x* + sin^{2} *x*

(E) cos (cos *x*) cos *x*

7. If *y* = cos^{4} (sin^{3} *x*), then *y*′ =

(A) –12 sin^{2} *x* cos *x* sin (sin^{3} *x*) cos^{3} (sin^{3} *x*)

(B) 4 cos^{3} (sin^{3} *x*)

(C) 4 cos^{3} (3 sin^{2} *x*)

(D) cos^{4} (3 sin^{2} *x*)

(E) 12 sin^{2} *x* cos *x* sin (sin^{3} *x*) cos^{3} (sin^{3} *x*)

8. What is the derivative of *f*(*x*) = sin^{2} *πx*?

(A) 2*π* sin 2*πx*

(B) 2 cos *πx*

(C) sin 2*πx*

(D) *π* sin 2*πx*

(E) sin *πx* cos *πx*

9. If *y* = cos (tan *x*), then *y*′ =

(A) sec^{2} *x* sin (tan *x*)

(B) –sin (sec^{2} *x*)

(C) –sec^{2} *x* sin (tan *x*)

(D) sec *x* tan *x* sin (tan *x*)

(E) –sec *x* tan *x* sin (tan *x*)

10. If *f*(*x*) = (*x*^{–2} + *x*^{–3})(*x*^{5} – 2*x*^{2}), then *f*′(*x*) =

(A) –3*x*^{2} + 2*x* + 2*x*^{–2}

(B) 3*x*^{2} + 2*x* – 2*x*^{–2}

(C) *x*^{3} + *x*^{2} – 2 – 2*x*^{–1}

(D) *x*^{4} + *x*^{3} – 2*x* + 2 ln *x*

(E) –3*x*^{2} – 2*x* – 2*x*^{2}

11. What is the derivative of *y* = ?

(A)

(B)

(C)

(D)

(E)

12. Find the derivative of *y* =

(A)

(B)

(C)

(D)

(E)

13. If *y* = sin *x*, then *y*′ =

(A) [sin *x* + 2*x* cos *x*]

(B) [sin *x* + 2*x* cos *x*]

(C) cos *x*

(D) cos *x*

(E) [sin *x* + 2*x* cos *x*]

14. If *f*(*x*) = *x*^{2} sin *x* tan *x*, then *f*′(*x*) =

(A) 2*x* cos *x* sec^{2}*x*

(B) *x*^{2} (cos *x* tan *x* + sin *x* sec^{2} *x*) + 2*x* (sin *x* tan *x*)

(C) –*x*^{2} (cos *x* tan *x* + sin *x* sec^{2} *x*) + 2*x* (sin *x* tan *x*)

(D) –2*x* cos *x* sec^{2} *x*

(E) *x*^{2} (cos *x* + sec^{2} *x*)

15. What is the derivative of *y* = tan (sin *x*)?

(A) sec^{2} (cos *x*)

(B) cos *x* sec (sin *x*) tan (sin *x*)

(C) cos *x* sec^{2} (sin *x*)

(D) – cos *x* sec^{2} (sin *x*)

(E) – sec^{2} (cos *x*)

16. If *y* = 2 csc (4*x*), then *y*′ =

(A) –2 cot^{2} (4*x*)

(B) 2 cot^{2} (4*x*)

(C) –8 csc (*x*) cot (*x*)

(D) 2*θ* csc (4*x*) cot (4*x*)

(E) –8 csc (4*x*) cot (4*x*)

17. What is the derivative of *y* = (*x*^{2} + 2*x*)*e ^{x}*?

(A) (*x* + 1)*e ^{x}*

(B) 2(*x* + 1)*e ^{x}*

(C) 2*x*(*x* + 1)*e ^{x}*

^{–1}

(D) (*x*^{2} + 4*x* + 2)*e ^{x}*

(E) (*x*^{2} – 4*x* – 2)*e ^{x}*

18. If *y* = *e*^{2} ln *x*, then *y*′ =

(A) 0

(B)

(C)

(D) 2*e* ln *x* +

(E) 2*e* ln *x* −

19. Find *f*′(*x*) if *f*(*x*) = tan *x* + sec^{2} *x*.

(A) sec^{2} *x* (1 + 2 tan *x*)

(B) 2 sec^{2} *x* + tan *x*

(C) sec^{2} *x* (1 – 2 tan *x*)

(D) tan *x* – sec^{2} *x*

(E) 2 sec^{2} *x* – tan *x*

20. If *y* = *e*^{sin} * ^{x}*, then

*y*′(

*π*) =

(A) –2

(B) –1

(C) 0

(D) 1

(E) 2

21. Find *y*′ if *y* = ln (4*x*^{2} – 3*x* + 3)

(A)

(B)

(C)

(D)

(E)

22. What is the derivative of *f*(*x*) = *x*^{4} tan 5*x*?

(A) *x*^{5} sec^{2} 5*x*

(B) 20 *x*^{3} sec^{2} 5*x*

(C) *x*^{3} (4 tan 5*x* – 5*x* sec^{2} 5*x*)

(D) –*x*^{3} (4 tan 5*x* – 5*x* sec^{2} 5*x*)

(E) *x*^{3} (4 tan 5*x* + 5*x* sec^{2} 5*x*)

23. If *y* = tan^{2} (sin *θ*), then *y*′ =

(A) –2 cos *θ* tan (sin *θ*) sec^{2} (sin *θ*)

(B) 2 sec^{2} (cos *θ*)

(C) 2 tan (– cos *θ*)

(D) cos *θ* tan (sin *θ*) sec^{2} (sin *θ*)

(E) 2 cos *θ* tan (sin *θ*) sec^{2} (sin *θ*)

24. If *y* = sec (1 + *x*^{2}), then *y*′ =

(A) sec (2*x*) tan (2*x*)

(B) 2*x* sec (2*x*) tan (2*x*)

(C) 2*x* sec^{2} (1 + *x*^{2})

(D) 2*x* sec (1 + *x*^{2}) tan (1 + *x*^{2})

(E) 2*x* tan^{2} (1+ *x*^{2}) sec (1 + *x*^{2})

25. If *y* sin (*x*^{2}) = *x* sin (*y*^{2}), then *y*′ =

(A) 2 sec (*y*^{2}) – 2 csc (*x*^{2})

(B)

(C) 2 cos *y* – 2 cos *x*

(D)

(E)

26. If *x*^{2} cos *y* + sin 2*y* = *xy*, then find *y*′.

(A)

(B) *x* + *x*^{2} sin *y* – 2 cos 2*y*

(C) 2*x* cos *y* – *y*

(D)

(E)

27. If sec *y* = sec^{2} *x*, then *y*′ =

(A)

(B)

(C) 2 sec^{2} *x* tan *x*

(D) tan *y* – 2 tan *x*

(E) sec *y* tan *y*

# Chapter 10

# Derivatives Drill 2 Answers and Explanations

## ANSWER KEY

1. B

2. D

3. A

4. C

5. D

6. B

7. A

8. D

9. C

10. A

11. B

12. E

13. A

14. B

15. C

16. E

17. D

18. B

19. A

20. B

21. C

22. E

23. E

24. D

25. B

26. A

27. A

## EXPLANATIONS

1. **B** Take the derivative using trigonometric derivatives and the chain rule.

*f*(*x*) = sin (csc *x*)

*f*′(*x*) = cos (csc *x*)(– csc *x* cot *x*)

*f*′(*x*) = –csc *x* cot *x* cos (csc *x*)

2. **D** Start calculating the derivatives to get to the 4th derivative by using the power rule.

*y* = 6*x*^{5} – 5*x*^{4}

*y*′ = 30*x*^{4} – 20*x*^{3}

*y*″ = 120*x*^{3} – 60*x*^{2}

*y*′^{(3)} = 360*x*^{2} – 120*x*

*y*′^{(4)} = 720*x* – 120

3. **A** Take the derivative using the derivative of the exponential function, trigonometric derivatives, the product rule, and the chain rule.

*f*(*x*) = *e*^{2}* ^{x}* sin

^{2}3

*x*

*f*′(*x*) = (2*e*^{2}* ^{x}*)(sin

^{2}3

*x*) + (

*e*

^{2}

*)(2 sin 3*

^{x}*x*cos 3

*x*× 3)

*f*′(*x*) = 2*e*^{2}* ^{x}* sin

^{2}3

*x*+ 6

*e*

^{2}

*sin 3*

^{x}*x*cos 3

*x*

*f*′(*x*) = 2*e*^{2}* ^{x}* sin 3

*x*(sin 3

*x*+ 3 cos 3

*x*)

4. **C** Take the derivative using the differentiation rules of exponential functions, natural logarithms, the product rule and the chain rule.

*y* = *e ^{x}* (ln

*x*)

^{2}

*y*′ = *e ^{x}* (ln

*x*)

^{2}+

*e*

^{x}*y*′ = *e ^{x}* (ln

*x*)

5. **D** Take the derivative using the trigonometric derivative formula for tangent and the chain rule.

*y* = tan^{2} (3*θ*)

*y*′ = 2 tan (3*θ*) sec^{2} (3*θ*) 3

*y*′ = 6 tan (3*θ*) sec^{2} (3*θ*)

6. **B** Take the derivative using the trigonometric formula for the sine function, as well as the chain rule, starting from the outside and working your way in the parentheses.

*f*(*x*) = sin (sin (sin *x*))

*f*′(*x*) = cos (sin (sin *x*)) cos (sin *x*) cos *x*

*f*′(*x*) = cos *x* cos (sin *x*) cos (sin (sin *x*))

7. **A** Take the derivative using the trigonometric formula for the sine and cosine functions and the chain rule.

*y* = cos^{4} (sin^{3} *x*)

*y*′ = 4 cos^{3} (sin^{3} *x*)(– sin (sin^{3} *x*)) 3 sin^{2} *x* cos *x*

*y*′ = –12 sin^{2} *x* cos *x* sin (sin^{3} *x*) cos^{3} (sin^{3} *x*)

8. **D** Take the derivative using the trigonometric formula for the sine function and the chain rule.

*f*(*x*) = sin^{2} *πx*

*f*′(*x*) = 2 sin *πx* cos *πx π*

*f*′(*x*) = 2*π* sin *πx* cos *πx*

*f*′(*x*) = *π* sin 2*πx* (* Note: sin 2*x* = 2 sin *x* cos *x*)

9. **C** Take the derivative using the trigonometric formula for the cosine and tangent functions, as well as the chain rule.

*y* = cos (tan *x*)

*y*′ = –sin (tan *x*) sec^{2}*x*

*y*′ = –sec^{2}*x* sin (tan *x*)

10. **A** Although the product rule can be used to get this derivative, first apply FOIL and then use the power rule of exponents to make it easier.

(*x*^{–2} + *x*^{–3})(*x*^{5} – 2*x*^{2}) = *x*^{3} – 2 + *x*^{2} – 2*x*^{–1}

*f*(*x*) = *x*^{3} + *x*^{2} – 2 – 2*x*^{–1}

*f*′(*x*) = 3*x*^{2} + 2*x* + 2*x*^{–2}

11. **B** Take the derivative using the quotient rule.

12. **E** Take the derivative using the quotient rule.

13. **A** Take the derivative using the trigonometric formula for the sine function and the product rule.

14. **B** Take the derivative by carefully using the product rule, as well as the trigonometric formula for both the sine and tangent functions.

*f*(*x*) = *x*^{2} sin *x* tan *x*

*f*′(*x*) = (2*x*)(sin *x* tan *x*) + (*x*^{2})(cos *x* tan *x* + sin *x* sec^{2} *x*)

15. **C** Take the derivative of the function using the differentiation rules for the tangent and sine functions, as well as the chain rule.

*y* = tan (sin *x*)

*y*′ = sec^{2}(sin *x*) cos *x*

*y*′ = cos *x* sec^{2}(sin *x*)

16. **E** Take the derivative of the function using the differentiation rule for the cosecant function, as well as the chain rule.

*y* = 2 csc (4*x*)

*y*′ = –8 csc (4*x*) cot (4*x*)

17. **D** Take the derivative of the function using the product rule, the power rule, and the exponential function rule.

*y* = (*x*^{2} + 2*x*)*e ^{x}*

*y*′ = (2*x* + 2)(*e ^{x}*) + (

*x*

^{2}+ 2

*x*)(

*e*)

^{x}*y*′ = (*x*^{2} + 4*x* + 2)*e ^{x}*

18. **B** Take the derivative of the function using the constant rule, and the derivative of the natural logarithm function. (Remember, *e*^{2} is a constant!)

*y* = *e*^{2} ln *x*

*y*′ = *e*^{2}

*y*′ =

19. **A** Take the derivative using the rules of differentiation for both the tangent and secant functions, as well as the power rule.

*f*(*x*) = tan *x* + sec^{2} *x*

*f*′(*x*) = sec^{2} *x* + 2 sec *x* sec *x* tan *x*

*f*′(*x*) = sec^{2} *x* (1 + 2 tan *x*)

20. **B** Take the first derivative using the rule of exponential functions and of the sine function.

*y* = *e* ^{sin} ^{x}

*y*′ = *e* ^{sin} * ^{x}* cos

*x*

Do not simplify! Plug in *x* = *π* immediately.

*y*′ = *e* ^{sin} * ^{π}* cos

*π*

= *e*^{0} (–1)

= –1

21. **C** Take the derivative of the function by using the differentiation formula for the natural logarithm function and the chain rule.

*y* = ln (4*x*^{2} − 3*x* + 3)

22. **E** Take the derivative using the product rule, the differentiation formula for the tangent function and the chain rule.

*f*(*x*) = *x*^{4} tan 5*x*

*f*′(*x*) = (4*x*^{3})(tan 5*x*) + (*x*^{4})(sec^{2} 5*x*)(5)

*f*′(*x*) = *x*^{3} (4 tan 5*x* + 5*x* sec^{2} 5*x*)

23. **E** To find this derivative, you must utilize the chain rule and the rules of trigonometric differentiation for the tangent and sine functions.

*y* = tan^{2} (sin *θ*)

*y*′ = 2 tan (sin *θ*) × sec^{2} (sin *θ*) × cos *θ*

So, *y*′ = 2 cos *θ* tan (sin *θ*) sec^{2} (sin *θ*)

24. **D** Take the derivative using the trigonometric rule for the secant function and the chain rule.

*y* = sec (1 + *x*^{2})

*y*′ = sec (1 + *x*^{2}) tan (1 + *x*^{2}) × 2*x*

*y*′ = 2*x* sec (1 + *x*^{2}) tan (1 + *x*^{2})

25. **B** Take the derivative with respect to *x* using the product rule, the chain rule, and the differentiation formula for the sine function.

26. **A** Take the derivative with respect to *x* using the product rule, chain rule, and the differentiation rules for the sine and cosine functions.

27. **A** Take the derivative with respect to *x* using the chain rule and the differentiation rule for the secant function.

# Chapter 11

# Derivatives Drill 3

## DERIVATIVES DRILL 3

1. Find *y*′ if *x*^{6} + *y*^{6} = 6.

(A)

(B)

(C)

(D) −

(E) −

2. Given *h*(*x*) = *f (g(sin x*)), what is *h*′ in terms of *f*′ and *g*′?

(A) cos *x* × *f*′(*g*(sin *x*)) × *g*′(sin *x*)

(B) *f*′(*g*′(cos *x*))

(C) *f*(*g*′(sin *x*)) × *f*′(*g*(sin *x*))

(D) cos *x* × *f*(*g*′(sin *x*)) × *f*′(*g*(sin *x*))

(E) –cos *x* × *f*′(*g*(sin *x*)) × *g*′(sin *x*)

3. If *f*(*x*) = *x*^{2} *g*(*x*), *g*(3) = 2, and *g*′(3) = 1, then what is *f*′(3)?

(A) 30

(B) 21

(C) 3

(D) –21

(E) –30

4. If *y* = 4*x* – tan *x*, then *y*′ =

(A) 4 + sec^{2} *x*

(B) 4 – sec *x* tan *x*

(C) 4 – sec^{2} *x*

(D) 4 + sec *x* tan *x*

(E) 4 sec^{2} *x*

5. If *y* = cot (3*x*^{2} + 5), then *y*′ =

(A) –csc^{2} (6*x*)

(B) –6*x* csc^{2} (3*x*^{2} + 5)

(C) 6*x* csc^{2} (3*x*^{2} + 5)

(D) –6*x* csc (3*x*^{2} + 5) cot (3*x*^{2} + 5)

(E) 6*x* csc (3*x*^{2} + 5) cot (3*x*^{2} + 5)

6. If *y* = csc (cot *x*), then *y*′ =

(A) –csc (csc^{2} *x*) cot (csc^{2} *x*)

(B) –csc^{2} *x* csc (cot *x*) cot (cot *x*)

(C) –csc^{2} *x* cot^{2} (cot *x*)

(D) csc^{2} *x* cot^{2} (cot *x*)

(E) csc^{2} *x* csc (cot *x*) cot (cot *x*)

7. If *y* = tan (sec *x*), then *y*′ =

(A) sec^{2} (sec *x* tan *x*)

(B) sec *x* tan *x* sec^{2} (sec *x*)

(C) sec (sec *x*) tan (sec *x*)

(D) –sec *x* tan *x* sec^{2} (sec *x*)

(E) –sec (sec *x*) tan (sec *x*)

8. Let *r*(*x*) = *f*(*g*(*h*(*x*))), where *h*(1) = 2, *g*(2) = 3, *h*′(1) = 4, *g*′(2) = 5, and *f*′(3) = 6. What is the value of *r*′(1)?

(A) 1

(B) 3

(C) 20

(D) 120

(E) 720

9. If *y* = cot (csc *x*), then *y*′ =

(A) csc^{2} (csc *x* cot *x*)

(B) –csc^{2} (–csc *x* cot *x*)

(C) csc *x* cot *x* csc^{2} (csc *x*)

(D) –csc *x* cot *x* csc^{2} (csc *x*)

(E) csc^{2} (csc *x*) cot (csc *x*)

10. If *y* = sec (tan *x*), then *y*′ =

(A) sec (sec^{2} *x*) tan (sec^{2} *x*)

(B) tan^{2} (sec^{2} *x*) sec (tan *x*)

(C) –sec^{2} *x* sec (tan *x*) tan (tan *x*)

(D) sec^{2} *x* sec (tan *x*) tan (tan *x*)

(E) –sec (sec^{2} *x*) tan (sec^{2} *x*)

11. If *F*(*x*) = *f* (3 *f* (4 *f* (*x*))), where *f*(0) = 0 and *f*′(0) = 2, then what is the value o*f F*′(0)?

(A) 96

(B) 72

(C) 24

(D) 2

(E) 0

12. If *x*^{3} + *y*^{3} = 1, then *y*′ =

(A)

(B)

(C) –

(D)

(E)

13. If *x*^{2} + *xy* – *y*^{2} = 4, then *y*′ =

(A)

(B)

(C)

(D)

(E)

14. Find *y*′ if *x*^{3} + *y*^{3} = 6*xy*.

(A)

(B)

(C)

(D)

(E)

15. Find *y*′ if 4cos *x* sin *y* = 1.

(A) cot *x* cot *y*

(B) tan *x* cot *y*

(C) cot *x* tan *y*

(D) 1 – cos *x* sin *y*

(E) tan *x* tan *y*

16. If *x*^{3} + *y*^{3} = *xy*, then what is *y*′?

(A)

(B)

(C)

(D)

(E)

17. Find *y*′ if *x* = ln (*x*^{2} + *y*^{2}).

(A)

(B)

(C)

(D)

(E)

18. *x*^{4} + *y*^{4} = *π*^{4}. Find .

(A)

(B)

(C)

(D)

(E)

19. If 3*x*^{2} sin *y* = tan *x*, then *y*′ =

(A)

(B)

(C)

(D)

(E) 6*x* cos *y* – sec^{2} *x*

20. Find the derivative of *y*, when *y* = ?

(A)

(B)

(C)

(D)

(E)

21. of *y* = =

(A)

(B)

(C)

(D)

(E)

22. Find if *y*^{3} + 2*y*^{2} = 4*x* − 12.

(A)

(B)

(C)

(D)

(E)

23. Find if *y*^{3} + 2*y*^{2} = 4*x* − 12 and *y* = 1 at *x* = 7.

(A)

(B)

(C)

(D)

(E)

24. What is if *y* = log_{3}(4*x*^{3} − 2*x*)?

(A)

(B)

(C)

(D)

(E)

25. Find the derivative of the inverse of *y* = *x*^{4} − 3 when *y* = –2.

(A)

(B)

(C)

(D) 1

(E)

26. Find for 4*x*^{2} − 2*x*^{2}*y* + 2*xy*^{2} − 3*y*^{2} = *x* at *x* = 1.

(A) –4

(B) 0

(C)

(D) 1

(E)

27. If *f*(*x*) = 2*x*^{2} − 3*x* + 6, find a derivative of *f*^{−1}(*x*) at *y* = 15.

(A)

(B) −

(C)

(D) 3

(E) −

28. Find if *y* = .

(A)

(B)

(C)

(D)

(E)

# Chapter 12

# Derivatives Drill 3 Answers and Explanations

## ANSWER KEY

1. D

2. A

3. B

4. C

5. B

6. E

7. B

8. D

9. C

10. D

11. A

12. C

13. D

14. B

15. E

16. C

17. C

18. A

19. B

20. E

21. A

22. B

23. A

24. B

25. B

26. E

27. A

28. D

## EXPLANATIONS

1. **D** Take the derivative implicitly with respect to *x* and isolate *y*′.

2. **A** You must find the derivative by utilizing the chain rule twice, as well as the trigonometric differentiation formula for the sine function.

*h*(*x*) = *f*(*g*(sin *x*))

*h*′(*x*) = *f*′(*g*(sin *x*)) × *g*′(sin *x*) × cos *x*

*h*′(*x*) = cos *x* × *f*′(*g*(sin *x*)) × *g*′(sin *x*)

3. **B** You must take the derivative using the power and product rules first.

*f*(*x*) = *x*^{2} *g*(*x*)

*f*′(*x*) = 2*x g*(*x*) + *x*^{2}*g*′(*x*)

To find what *f*′(3) is equal to, you must use the other pieces of information and plug in accordingly.

*f*′(3) = 2(3) *g*(3) + (3)^{2}*g*′(3) = (6)(2) + (9)(1) = 21.

4. **C** Take the derivative using the power rule and the trigonometric rule for the tangent function.

*y* = 4*x* – tan *x*

*y*′ = 4 – sec^{2}*x*

5. **B** Take the derivative using the trigonometric rule for the cotangent function and the chain rule.

*y* = cot (3*x*^{2} + 5)

*y*′ = –csc^{2} (3*x*^{2} + 5) × 6*x*

*y*′ = –6*x* csc^{2} (3*x*^{2} + 5)

6. **E** Take the derivative using the trigonometric rules for the cosecant and cotangent functions, as well as the chain rule.

*y* = csc (cot *x*)

*y*′ = –csc (cot *x*) cot (cot *x*) × –csc^{2}*x*

*y*′ = csc^{2}*x* csc (cot *x*) cot (cot *x*)

7. **B** Take the derivative using the trigonometric differentiation rules for the tangent and secant functions, as well as the chain rule.

*y* = tan (sec *x*)

*y*′ = sec^{2} (sec *x*) × sec *x* tan *x*

*y*′ = sec *x* tan *x* sec^{2} (sec *x*)

8. **D** First, you must take the derivative using the chain rule twice.

*r*(*x*) = *f*(*g*(*h*(*x*)))

*r*′(*x*) = *f*′(*g*(*h*(*x*))) × *g*′(*h*(*x*)) × *h*′(*x*)

Evaluate the derivative at *x* = 1 using the information provided in the question.

*r*′(*x*) = *f*′(*g*(*h*(*x*))) × *g*′(*h*(*x*)) × *h*′(*x*)

*r*′(1) = *f*′(*g*(*h*(1))) × *g*′(*h*(1)) × *h*′(1)

*r*′(1) = *f*′(*g*(2)) × *g*′(2) × (4)

*r*′(1) = *f*′(3) × (5) × (4)

*r*′(1) = (6) × (5) × (4) = 120

9. **C** Take the derivative using the trigonometric differentiation rules for the cotangent and cosecant functions, as well as the chain rule.

*y* = cot (csc *x*)

*y*′ = –csc^{2} (csc *x*)(–csc *x* cot *x*)

*y*′ = csc *x* cot *x* csc^{2} (csc *x*)

10. **D** Take the derivative using the trigonometric differentiation rules for the secant and tangent functions, as well as the chain rule.

*y* = sec (tan *x*)

*y*′= sec (tan *x*) tan (tan *x*) × sec^{2} *x*

*y*′ = sec^{2} *x* sec (tan *x*) tan (tan *x*)

11. **A** Take the derivative of *F*(*x*) using the chain rule.

*F*(*x*) = *f* (3 *f* (4 *f* (*x*)))

*F*′(*x*) = *f*′(3 *f* (4 *f* (*x*))) × 3 *f*′(4*f*(*x*)) × 4 *f*′(*x*)

Evaluate the derivative at *x* = 0 using the information provided in the question.

*F*′(*x*) = *f*′(3 *f* (4 *f* (*x*))) × 3 *f*′(4*f* (*x*)) × 4 *f*′(*x*)

*F*′(0) = *f*′(3 *f* (4 *f* (0))) × 3 *f*′(4*f* (0)) × 4 *f*′(0)

*F*′(0) = *f*′(3 *f* (4 (0))) × 3 *f*′(4(0)) × 4(2)

*F*′(0) = *f*′(0) × 3(2) × 4(2)

*F*′(0) = (2) × 3(2) × 4(2) = 96

12. **C** Calculate the derivative with respect to *x*.

13. **D** Calculate the derivative with respect to *x*.

14. **B** Calculate the derivative with respect to *x*.

15. **E** Calculate the derivative with respect to *x*.

16. **C** Calculate the derivative with respect to *x*.

17. **C** Calculate the derivative with respect to *x*.

18. **A** Calculate the derivative with respect to *x*.

19. **B** Calculate the derivative with respect to *x*.

20. **E** This problem can be solved using the Chain rule, Product rule, and Quotient rule, but that can get especially messy. We are going to use logarithmic differentiation to solve it. First, take the natural log of both sides: ln *y* = ln . Use logarithmic rules to simplify the equation: ln *y* = ln(*x*^{3} − 2*x*^{2}) + 2 ln(sin *x*) − 3 ln(*x*^{2} + 1). Now, differentiate both sides with respect to *x*: . Finally, isolate : .

21. **A** Use the chain and quotient rules: .

22. **B** Use implicit differentiation: . Thus,

23. **A** Via implicit differentiation, and . Evaluate at (7, 1), to get, . Plug in these points to solve

24. **B** Recall, . In this problem, *u* = 4*x*^{2} − 2*x* and *du* = (12*x*^{2} − 2)*dx*. Thus,

.

25. **B** . Here, *y* = –2 when *x* = 1 and . Evaluate this derivative at *x* = 1 to get the solution: .

26. **E** First, plug *x* = 1 into the equation and solve for *y.* The result will be two *y* values, *y* = –3 and *y* = 1. Next, use implicit differentiation to find the first derivative: 8*x* − 2*x*^{2} − 4*xy* + 4*xy* + 2*y*^{2} − 6*y* = 1.

Do not simplify because the questions asks for the derivative at *x* = 1. Thus, the final step is to plug (1,1) and (1, –3) into the equation for the first derivative and solve for the derivatives at these points. At (1,1), . At (1, –3), . Since the derivative at (1, –3) is an option, the answer is E.

27. **A** The derivative of the inverse is: . To begin, determine the *x*-value that corresponds to *y* = 15. There are two possible *x*-values: *x* = − and *x* = 3. Next, find the derivative of the function *f*(*x*): *f*(*x*) = 4*x* − 3. Plug this into the left hand side of the formula by first taking the inverse. Then, evaluate the function at both *x*-values: and .

28. **D** Because the quotient rule and chain rule would get messy here, use logarithmic differentiation. First, take the natural log of both sides of the equation and simplify: ln *y* = ln = 2 ln(3*x*^{3} + 2) − ln(*x* − 2). Now, differentiate both sides using implicit differentiation: . Solve for : .

# Chapter 13

# Applications of Derivatives Drill 1

## APPLICATIONS OF DERIVATIVES DRILL 1

1. Find the equation of the line tangent to *y* = at (1,1).

(A) *x* – 2*y* = –1

(B) 2*x* – *y* = 1

(C) *x* – *y* = 2

(D) *x* – 2*y* = 1

(E) 2*x* + *y* = 1

2. If *f*(*x*) = , then find *f*″(2).

(A) –8

(B) –

(C) 0

(D)

(E) 8

3. Find the critical numbers of *y* = 3*x*^{4} + 4*x*^{3} – 12*x*^{2}.

(A) 0

(B) –2, 1

(C) 0, 1

(D) –2, 0, 1

(E) –1, 0, 2

4. What is the maximum value of *f*(*x*) = 2*x*^{3} – 3*x*^{2} – 12*x* + 1 on the interval [–2,3]?

(A) –3

(B) 0

(C) 2

(D) 6

(E) 8

5. Find the interval(s) on which *f* is decreasing for *f*(*x*) = 2*x*^{3} + 3*x*^{2} – 36*x*.

(A) (–∞,–3)

(B) (–2,3)

(C) (–3,2)

(D) (2,3)

(E) (2,∞)

6. Find all critical numbers of *y* = 2*x*^{3} – 3*x*^{2} – 12*x*.

(A) –2

(B) –1

(C) –2, –1

(D) –1, 2

(E) 1, 2

7. Find any points of inflection of *y* = *x*^{4} + 4*x*^{3}.

(A) (–2,–16)

(B) (0,0)

(C) (2,16) and (0,0)

(D) (0,0) and (–2,–16)

(E) (2,16)

8. Find the equation of the line tangent to *y* = sin (sin *x*) at (*π*, 0).

(A) x – *y* = *π*

(B) *x* + *y* = *π*

(C) 2*x* – *y* = *π*

(D) *x* – 2*y* = *π*

(E) *x* + *y* = 2*π*

9. Find the minimum value of *f*(*x*) = 2*x*^{3} + 3*x*^{2} – 36x.

(A) –44

(B) –9

(C) 3

(D) 9

(E) 44

10. What is the point of inflection of *f*(*x*) = (*x* + 1)^{5} – 5*x* – 2?

(A) (–3,1)

(B) (–1,3)

(C) (0,0)

(D) (1,3)

(E) (3,1)

11. On what interval(s) is *f* decreasing if *f*(*x*) = 2 + 2*x*^{2} – *x*^{4}?

(A) (–1,0) only

(B) (1,∞) only

(C) (–∞,–1) and (0,1)

(D) (–1,0) and (1,∞)

(E) (0,1) only

12. A particle is traveling according to *f*(*x*) = *x*^{3} – 12*x*^{2} + 36*x*. What is the velocity at *x* = 3 seconds?

(A) –18

(B) –9

(C) 0

(D) 9

(E) 18

13. If a ball is thrown upward with a velocity of 80 ft/s, then its height after *t* seconds is *s* = 80*t* – 16*t*^{2}. What is the maximum height of the ball?

(A) 2.5

(B) 80

(C) 100

(D) 180

(E) 270

14. Find the equation of the normal line to the curve *y* = at (1, ).

(A) 8*x* + 2*y* = 7

(B) 2*x* + 8*y* = 7

(C) 8*x* – 2*y* = 7

(D) 2*x* – 8*y* = 7

(E) 8*x* – 2*y* = –7

15. For what values of *x* does the graph of *f*(*x*) = *x* + 2sin *x* have a horizontal tangent on [0,2*π*]?

(A) and

(B) and

(C) and

(D) only

(E) no values

16. Find an equation of the tangent line to the curve *y* = 2*x* sin *x* at the point (, *π*).

(A) *y* = 2*x* + *π*

(B) *y* = 2*x* – *π*

(C) *y* = 2*x*

(D) *y* = 2*x* +

(E) *y* = 2*x* −

17. A particle travels in a position governed by the equation *s*(*t*) = 4*t*^{3} – 16*t*^{2}. What is its acceleration at *t* = 2 seconds?

(A) 0

(B) 2

(C) 10

(D) 12

(E) 16

18. If a particle travels along a path according to the equation *s*(*t*) = 6*t*^{2} – 4*t* + 3, then what is the velocity at *t* = 2 seconds?

(A) –20

(B) –10

(C) 0

(D) 10

(E) 20

19. Find the absolute maximum value of *f*(*x*) = (*x*^{2} + 2*x*)^{3} on the interval [–2,1].

(A) –2

(B) –1

(C) 0

(D) 1

(E) 27

20. What is the *x*-coordinate of the point of inflection of *f*(*x*) = 4*x*^{3} + 3*x*^{2} – 6*x*?

(A) –4

(B) –

(C) 0

(D)

(E) 4

21. On what interval(s) is *f* decreasing for *f*(*x*) = ?

(A) (–∞,∞)

(B) (–∞,0)

(C) (0,∞)

(D) (–∞,–3)

(E) (–3,∞)

Questions 22–23 rel*y* on the following information:

Suppose that ** h**(

*x*) =

*f*(

*x*)

*g*(

*x*) and

*F*(

*x*) =

*f*(

*g*(

*x*)), where

*f*(2) = 3,

*g*(2) = 5,

*g*′(2) = 4,

*f*′(2) = –2, and

*f*′(5) = 11.

22. What is the value of *F*′(2)?

(A) 44

(B) 22

(C) 2

(D) –22

(E) –44

23. What is the value of *h*′(2)?

(A) 10

(B) 5

(C) 2

(D) –5

(E) –10

24. A particle moves on a vertical line so that its coordinate at time *t* is given b*y y* = *t*^{3} – 12*t* + 3, where *t* ≥ 0. What is its acceleration at time *t*?

(A) *t*^{4} – 6*t*^{2} + 3*t* + 1

(B) 3*t*^{2} – 12

(C) 6t

(D) 6

(E) 0

25. Find the equation of the tangent line to the equation *x*^{2} + *xy* + *y*^{2} = 3 at the point (1,1).

(A) *x* – *y* = –2

(B) *x* + *y* = –2

(C) 2*x* + *y* = 2

(D) *x* – *y* = 2

(E) *x* + *y* = 2

26. Find the equation of the tangent line to the equation *x*^{2} + 2*xy* – *y*^{2} + *x* = 2 at (1,2).

(A) *x* – 3*y* = 1

(B) 7*x* + 2*y* = 3

(C) 7*x* – 2*y* = 3

(D) 3*x* + *y* = 1

(E) 3*x* – *y* = 3

27. What is the slope of the equation sin (*xy*) = 0 at (2, )?

(A) –*π*

(B) –

(C) –

(D) –

(E)

# Chapter 14

# Applications of Derivatives Drill 1 Answers and Explanations

## ANSWER KEY

1. A

2. B

3. E

4. E

5. C

6. D

7. D

8. B

9. A

10. B

11. D

12. B

13. C

14. C

15. B

16. C

17. E

18. E

19. E

20. D

21. B

22. A

23. C

24. C

25. E

26. C

27. C

## EXPLANATIONS

1. **A** Utilize the quotient rule to evaluate this derivative.

*y*′ =

Find the slope at (1,1) by plugging in 1 for *x.*

*y*′ =

Use the point-slope equation of a line to answer the question.

*y* – *y*_{1} = *m*(*x* – *x*_{1})*y* – 1 = (*x* – 1)

2*y* – 2 = *x* – 1*x* – 2*y* = –1

2. **B** This question is testing your knowledge of higher order derivatives and the chain rule. Start by calculating the first derivative.

Don’t simplify! Plug in 2 for *x.*

3. **E** This question is testing your knowledge of the first derivative and setting it equal to 0.

Find the derivative.

*y*′ = 12*x*^{3} + 12*x*^{2} – 24*x*

Set it equal to 0 and solve for *x.*

*y*′ = 12*x* (*x*^{2} – *x* – 2) = 0

12*x* = 0 and (*x*^{2} – *x* – 2) = 0

*x* = 0 and *x* = 2, –1.

4. **E** To find the maximum, you must take the derivative and test values around the critical numbers.

Take the derivative and set equal to 0 to find the critical numbers.

*f*′(*x*) = 6*x*^{2} – 6*x* – 12 = 0

*x*^{2} – *x* – 2 = 0

(*x* – 2)(*x* + 1) = 0

*x* = 2 ; *x* = –1

Test values around the critical numbers to determine behavior

*f*′(–2) = + *f*′(0) = – *f*′(3) = +

This tells us that before –1, *f* is increasing and after –1, *f* switches to decreasing, making *x* = –1 a relative maximum. To calculate the value, simply plug –1 into the original *f*(*x*).

*f*(–1) = 2(–1)^{3} – 3(–1)^{2} – 12(–1) + 1

= –2 – 3 + 12 + 1 = 8.

Since we are asked to find the maximum over the closed interval [–2,3], we must test the endpoints as being possible maximums before deciding on *f*(–1). So,

*f*(–2) = 2(–2)^{3} – 3(–2)^{2} – 12(–2) + 1 = –3

*f*(3) = 2(3)^{3} – 3(3)^{2} – 12(3) + 1 = –8

Since neither of these are greater than *f*(–1) = 8, 8 is the maximum value.

5. **C** Take the derivative and set it equal to 0 to find the critical numbers.

*f*′ = 6*x*^{2} + 6*x* – 36 = 0

*x*^{2} + *x* – 6 = 0

(*x* + 3)(*x* – 2) = 0

*x* = –3 ; *x* = 2

Test values around the critical numbers to find where *f* is decreasing.

*f*′(–4) = + *f*′(0) = – *f*′(3) = +

This tells us the values in between –3 and 2 are where *f* is decreasing because the first derivative is negative over that interval.

6. **D** Take the derivative and set it equal to 0. Remember that critical numbers are the *x*-values corresponding to when the first derivative is equal to zero.

*y*′ = 6*x*^{2} – 6*x* – 12 = 0

*x*^{2} – *x* – 2 = 0

(*x* – 2)(*x* + 1) = 0

*x* = 2 ; *x* = –1

7. **D** Find the second derivative of the function and set it equal to 0. Remember that points of inflection are found when the second derivative is equal to zero and solved; these are the points when the curve is changing concavity.

*y*′ = 4*x*^{3} + 12*x*^{2}

*y*″ = 12*x*^{2} + 24*x* = 0

12*x*(*x* + 2) = 0

*x* = 0 ; *x* = –2

Find the points by plugging the *x*-values into the original function.

*y* = (0)^{4} + 4(0)^{3} = 0 ; *y* = (–2)^{4} + 4(–2)^{3} = –16

(0,0) (–2,–16)

8. **B** Find the derivative using trigonometric formulas and the chain rule.

*y*′ = cos (sin *x*) × cos *x*

Plug in *x* = *π* to find the slope.

*m* = cos (sin *π*) × cos *π*

= cos ( 0 ) × (–1)

= (1) × (–1)

= –1

Use the point-slope equation of a line to answer the question.

*y* – *y*_{1} = *m*(*x* – *x*_{1})

*y* – 0 = –1(*x* – *π*)

*y* = –*x* + *π*

*x* + *y* = *π*

9. **A** Find the derivative and set it equal to 0 to find the critical points.

*f*′(*x*) = 6*x*^{2} + 6*x* – 36 = 0

*x*^{2} + *x* – 6 = 0

(*x* – 2)(*x* + 3) = 0

*x* = 2 ; *x* = –3

Test values around the critical points to see the behavior of *f.*

*f*′(−4) = + *f*′(0) = – *f*′(3) = +

Since *f* is decreasing before 2 and increasing after it, *x* = 2 is a relative minimum.

Plug 2 into the original function to find the minimum value.

*f*(2) = 2(2)^{3} + 3(2)^{2} – 36(2)

= 16 + 12 – 72

= –44

10. **B** Get the 2nd derivative and set it equal to 0.

*f*′(*x*) = 5(*x* + 1)^{4} – 5

*f*″(*x*) = 20(*x* + 1)^{3} = 0

(*x* + 1)^{3} = 0

*x* + 1 = 0

*x* = –1

Plug –1 into the original function to find the *y*-coordinate.

*f*(–1) = (–1 + 1)^{5} – 5(–1) – 2

= 3

11. **D** Find the derivative and set it equal to 0.

*f*′(*x*) = 4*x* – 4*x*^{3} = 0

4*x* (1 – *x*^{2}) = 0

*x* = 0 ; *x* = 1 ; *x* = –1

Test values around the critical numbers to see the behavior of *f.*

*f*′(–2) = + *f*′ = – *f*′ = + *f*′(2) = –

Since *f*′ is negative on two of the four test points, the interval over which *f* is decreasing is (–1,0) and (1, ∞).

12. **B** Since the derivative of a position function is the velocity, we must find the first derivative.

*f*′(*x*) = 3*x*^{2} – 24*x* + 36

Plug in 3 to find the instantaneous velocity.

*f*′ (3) = 3(3)^{2} – 24(3) + 36 = –9

13. **C** To determine where the ball is at a relative maximum, take the first derivative and set it equal to 0.

*s*′ = 80 – 32*t* = 0

80 = 32*t*

= *t*

Test values around to know the behavior of *s.*

*s*′(2) = + *s*′(3) = –

Since *s* is increasing before and decreasing after it, is a relative max.

Plug in to the position function to find the maximum height.

= 200 − 100 = 100

14. **C** Find the derivative.

*y*′ =

Don’t simplify! Plug in *x* = 1 to find the slope.

Find the normal slope by taking the negative reciprocal and use that in the point-slope formula for a line.

*m*_{normal} = 4

*y* – *y*_{1} = *m*(*x* – *x*_{1})

*y* – = 4 (*x* – 1)

2y – 1 = 8(*x* – 1)

2y – 1 = 8*x* – 8

7 = 8*x* – 2*y*

15. **B** Take the derivative and set it equal to 0 because a horizontal tangent is a line with the slope of 0.

*f*′(*x*) = 1 + 2 cos *x* = 0

1 = –2 cos *x*

– = cos *x*

= *x*

16. **C** Find an equation of the tangent line to the curve *y* = 2*x* sin *x* at the point (, *π*).

Find the derivative.

*y*′ = 2 sin *x* + 2*x* cos *x*

Don’t simplify! Plug in to find the slope.

*m* = 2 sin + 2 cos

= 2 (1) + 0

= 2

Use the point-slope formula for a line to find the equation.

*y* – *y*_{1} = *m*(*x* – *x*_{1})

*y* – *π* = 2 (*x* − )

*y* – *π* = 2*x* – *π*

*y* = 2*x*

17. **E** When given the position function, taking the first derivative will yield the velocity and taking the second derivative will yield the acceleration.

Take the second derivative and plug in *t* = 2.

*s*′(*t*) = *v*(*t*) = 12*t*^{2} – 32*t*

*s*″(*t*) = *v*′(*t*) = *a*(*t*) = 24*t* – 32

*a*(2) = 24(2) – 32 = 16.

18. **E** When given the position function, simply take the first derivative to find the velocity function.

*s*(*t*) = 6*t*^{2} – 4*t* + 3

*v*(*t*) = *s*′(*t*) = 12*t* – 4

Plug in 2 for *t* to find the velocity.

*v*(2) = 12(2) – 4 = 20

19. **E** Take the derivative using the chain rule and then set it equal to zero to find any critical points of the function.

*f*(*x*) = (*x*^{2} + 2*x*)^{3}

*f*′(*x*) = 3(*x*^{2} + 2*x*)^{2}(2*x* + 2) = 0

3(*x*^{2} + 2*x*) = 0 (2*x* + 2) = 0

*x*^{2} + 2*x* = 0 2(*x* + 1) = 0

*x*(*x* + 2) = 0 *x* + 1 = 0

*x* = 0 ; *x* = –2 *x* = –1

To find the absolute maximum value of a function, you must plug critical points as well as interval endpoints into the original function to find the greatest output. Since –2 is already an endpoint on the interval, you only need to check it once.

*f*(–2) = (4 – 4)^{3} = 0 *f*(0) = (0 + 0)^{3} = 0

*f*(–1) = (1 – 2)^{3} = –1 *f*(1) = (1 + 2)^{3} = 27

20. **D** To find the *x*-coordinate of the point of inflection, you must take the second derivative of the function and set it equal to zero. To do this, you must differentiate according to the power rule of derivatives.

*f*(*x*) = 4*x*^{3} + 3*x*^{2} – 6*x*

*f*′(*x*) = 12*x*^{2} + 6*x* – 6

*f*″(*x*) = 24*x* + 6 = 0

24*x* = –6

*x* = –

21. **B** Take the first derivative using the quotient rule and set it equal to zero.

Test values of *x* = –1 and *x* = 1 to determine the behavior of *f.*

*f*′(–1) = –

*f*′(1) = +

*f* is decreasing on (–∞, 0)

22. **A** You must first take the derivative of *F*(*x*) using the chain rule.

*F*(*x*) = *f*(*g*(*x*))

*F*′(*x*) = *f*′(*g*(*x*)) × *g*′(*x*)

To figure out the value of *F*′(2), you must use the other information provided in the initial given paragraph.

*F*′(2) = *f*′(*g*(2)) × *g*′(2) = *f*′(5) × (4) = (11)(4) = 44

23. **C** You must first take the derivative of *h*(*x*) using the product rule.

*h*(*x*) = *f*(*x*)*g*(*x*)

*h*′(*x*) = *f*′(*x*)*g*(*x*) + *f*(*x*)*g*′(*x*)

To figure out the value of *h*′(2), you must use the other information provided in the initial given paragraph.

*h*′(2) = *f*′(2) *g*(2) + *f*(2) *g*′(2) = (–2)(5) + (3)(4) = 2

24. **C** To find the acceleration at time *t*, you must take the derivative of the given position function twice.

*y* = *t*^{3} – 12*t* + 3

*y*′ = 3*t*^{2} – 12

*y*″ = 6*t*

25. **E** Take the derivative with respect to *x* using the power rule and product rule.

*x*^{2} + *xy* + *y*^{2} = 3

2*x* + *y* + *xy*′ + 2*y y*′ = 0

Don’t simplify! Plug in the coordinate immediately to solve for the slope.

Use the point-slope formula to find the equation of the tangent line at (1,1).

*y* – *y*_{1} = *m*(*x* – *x*_{1})

*y* – 1 = –1(*x* – 1)

*y* – 1 = –*x* + 1

*x* + *y* = 2

26. **C** Take the derivative with respect to *x* using the power rule and the product rule.

*x*^{2} + 2*xy* – *y*^{2} + *x* = 2

2*x* + 2(*y* + *xy*′) – 2*y y*′ + 1 = 0

Don’t simplify! Plug in the coordinate immediately to solve for the slope.

Use the point-slope formula to find the equation of the tangent line at (1,2).

*y* – *y*_{1} = *m*(*x* – *x*_{1})

*y* – 2 = (*x* – 1)

*y* – 2 = *x* –

2*y* – 4 = 7*x* – 7

7*x* – 2*y* = 3

27. **C** Take the derivative with respect to *x* using the differentiation formula for the sine function, the chain rule, and the power rule.

sin (*xy*) = 0

cos (*xy*) × (*y* + *xy*′) = 0

Don’t simplify! Plug in the coordinate point immediately

# Chapter 15

# Applications of Derivatives Drill 2

## APPLICATIONS OF DERIVATIVES DRILL 2

1. What is the equation of a parabola *y* = *ax*^{2} + *bx* + *c* that passes throu*g*h (1, 4) and whose tan*g*ent lines at *x* = –1 and *x* = 5 have slopes 6 and –2, respectivel*y*?

(A) *y* = *x*^{2} + *x*

(B) *y* = *x*^{2} + *x*

(C) *y* = *x*^{2} − *x*

(D) *y* = −*x*^{2} − *x*

(E) *y* = −*x*^{2} + *x*

2. At what point(s) on the curve *y* = sin *x* + cos *x*, 0 ≤ *x* ≤ 2*π*, is the tangent line horizontal?

(A)

(B) and

(C)

(D)

(E) and

3. The volume of a cube is increasing at a rate of 10 cm^{3}/min. How fast is the surface area increasing when the length of an edge is 30 cm?

(A) cm^{2}/min

(B) cm^{2}/min

(C) cm^{2}/min

(D) cm^{2}/min

(E) cm^{2}/min

4. How long does it take for a ball to reach 35 m/s if it is pushed down a hill and its position at time *t*, in seconds, is given by *s* = 5*t* + 3*t*^{2}, in meters?

(A) 2 seconds

(B) 3 seconds

(C) 4 seconds

(D) 5 seconds

(E) 6 seconds

5. What is the maximum height reached b*y* a ball if it travels according to the function *s* = 80*t* – 16*t*^{2}, in meters?

(A) 100

(B) 80

(C) 60

(D) 50

(E) 40

6. What is an equation of the line tangent to *y*^{2} = *x*^{3} + 3*x*^{2} at the point (1,–2)?

(A) 4*x* + 9*y* = 1

(B) 9*x* – 4*y* = 1

(C) 4*x* – 9*y* = 1

(D) 4*y* – 9*x* = 1

(E) 9*x* + 4*y* = 1

7. Find the point on the curve *y* = *x*^{} that is a minimum distance from the point (16,0).

(A)

(B) (16, 4)

(C)

(D)

(E) (2, )

8. A cone-shaped funnel has a diameter of 10 m and a height of 12 m. Find the error in the volume if the height is exact, but the diameter is 10.2 m.

(A) 4*π m*^{3}

(B) 12*π m*^{3}

(C) *π m*^{3}

(D) 20*π m*^{3}

(E) *π m*^{3}

9. Find the length of the curve *x* = *t*^{2} + 3 and *y* = 2*t*^{2} − 7 from *t* = 2 to *t* = 5.

(A) 156

(B) 78

(C) 75

(D) 39

(E) 21

10. Find a point on the curve *y* = *x*^{3} − 4*x*^{2} − 3*x* + 13 where the normal is parallel to the *y*-axis.

(A) (3, −5)

(B)

(C) (0, 0)

(D) (3, 5)

(E)

11. Find the slope of the tangent line to the curve *r* = 2 + 3sin *θ*

(A)

(B)

(C)

(D)

(E)

12. Find the equation of the line tangent to the curve 3*x*^{3} − 2*x*^{2} + *x* = *y*^{3} + 2*y*^{2} + 3*y* at *y* = –2.

(A) *y* = 2*x* − 1

(B) *y* = 2*x*

(C) *y* = *x*

(D) *y* = *x*

(E) *y* = 2*x* − 2

13. The curve *y* = *ax*^{3} + *bx*^{2} + *cx* + *d* passes through the point (2, 8) and is normal to *y* = − *x* − 4 at (0, –4). If *b* = 5, what is the value of *a*?

(A) –2

(B) −

(C) −

(D) –1

(E) −

14. At what time does the particle change direction if the position function is given by *x*(*t*) = 2*x*^{4} − 4*x*^{3} + 2*x*^{2} − 8, where *t* > 0?

(A)

(B) 1

(C)

(D) 2

(E)

15. What is the particle’s velocity at *t* = 3 if and *x* = 3*x*^{3} − 2*x*^{2} + 4 and *y* = 2*x*^{2} + 3*x* − 7?

(A) 69

(B) 15

(C)

(D) 4

(E)

16. Use differentials to approximate (5.2)^{3}.

(A) 125

(B) 130

(C) 135

(D) 140

(E) 145

17. The radius of a cylinder is increased from 9 to 9.03 inches. If the height remains constant at 12 inches. Estimate the change in volume.

(A) 0.005*π* in^{3}

(B) 3.24*π* in^{3}

(C) 6.48*π* in^{3}

(D) 9*π* in^{3}

(E) 12*π* in^{3}

18. Use differentials to approximate cos275°.

(A)

(B)

(C)

(D)

(E)

19. Find the length of the curve *y* = from *x* = 0 to *x* = 6.

(A)

(B) 20

(C)

(D)

(E) 21

20. If *x*^{2} + *y*^{2} = 25, then find the slope of the tangent that passes through the point(2,4).

(A) –1

(B) –

(C) 0

(D)

(E) 1

21. Find the equation of the tangent line to the curve *x*^{2} + *xy* + *y*^{2} = 3 at (1,1).

(A) 3*x* – *y* = 2

(B) *x* – *y* = 2

(C) 3*x* + *y* = 2

(D) *x* + *y* = 2

(E) *x* – 3*y* = –2

22. What is the slope of the line normal to the curve *f*(*x*) = *x*^{4} + 3*x*^{2} that passes through the point (2,1)?

(A) 44

(B)

(C) 0

(D) –

(E) –44

23. What dimensions must a rectangle have to maximize the area and have a perimeter of 100 meters?

(A) 40 m b*y* 10 m

(B) 45 m b*y* 5 m

(C) 35 m b*y* 15 m

(D) 30 m b*y* 20 m

(E) 25 m b*y* 25 m

24. What two positive numbers not onl*y y*ield a minimum sum, but also produce a product of 100?

(A) 50 and 2

(B) 25 and 4

(C) 10 and 10

(D) 20 and 5

(E) 100 and 1

25. What two numbers both have a sum of 23 and a product that is maximized?

(A) 12 and 11

(B) 11.5 and 11.5

(C) 10 and 13

(D) 9.5 and 13.5

(E) 8 and 15

26. If *y* = 4*x*^{3} – 9*x*^{2} + 6*x*, then what is the value of the relative minimum, if an*y*?

(A) –1

(B) 0

(C)

(D) 1

(E) 2

# Chapter 16

# Applications of Derivatives Drill 2 Answers and Explanations

## ANSWER KEY

1. E

2. B

3. A

4. D

5. A

6. E

7. C

8. A

9. E

10. A

11. A

12. B

13. B

14. B

15. C

16. D

17. C

18. B

19. D

20. B

21. D

22. D

23. E

24. C

25. B

26. D

## EXPLANATIONS

1. **E** First, you must take the derivative of the given function, keeping in mind that *a*, *b*, and *c* all represent constants, not variables.

*y* = *ax*^{2} + *bx* + *c*

*y*′ = 2*ax* + *b*

Given the 2 coordinates and their respective slopes, you are able to come up with two linear equations relating *a* and *b*, which can be solved to determine the individual values.

*y*′ (–1) = 2*a*(–1) + *b* = 6 and *y*′ (5) = 2*a*(5) + *b* = –2

So, the equations are –2*a* + *b* = 6 and 10*a* + *b* = –2.

Stack the equations on top of one another to start solving for the constants. You can eliminate *b* entirely by subtracting the two equations.

Plugging the value of *a* into either equation yields *b* = .

Now that you know the values of *a* and *b*, you must go back to the original function given and use the given coordinate to determine the value of *c*.

*y* = *ax*^{2} + *bx* + *c*

*y* = −*x*^{2} + *x* + *c*

Now, plug in the coordinate.

4 = −(1)^{2} + (1) + *c*

4 = − + + *c*

0 = *c*

So, the equation of the parabola is *y* = −*x*^{2} + *x*.

2. **B** You must take the derivative of the function using the trigonometric differentiation rules for the sine and cosine functions, and then set it equal to zero to find the critical points.

*y* = sin *x* + cos *x*

*y*′ = cos *x* – sin *x* = 0

cos *x* = sin *x*

1 = tan *x*

= *x*

To find the points, you must plug the values of *x* you found in step one into the original given function.

The points are and .

3. **A** The key to success in any related rates question is to clearly identify what you have and what you need to solve for.

Given: = 10 cm^{3}/min and *s* = 30 cm.

Solve for: .

Recall the formula for volume of a cube (*V* = *s*^{3}) and surface area of a cube (*A* = 6*s*^{2}). You must take the derivatives of each so you can see what they have in common.

You must use the volume equation first because you need to calculate a value for to use in the surface area equation.

= 3*s*^{2}

10 cm^{3}/min = 3(30 cm)^{2}

10 cm^{3}/min = 2700 cm^{2}

cm^{3}/min =

Use the value calculated for in the surface area equation now to solve for .

= 12*s*

= 12(30 cm)( cm/min)

= cm^{2}/min

4. **D** Since the question gives you a velocity, you must take the derivative of the position function to attain the velocity function. When you have that, set the two equal to each other and solve for *t*.

*s* = 5*t* + 3*t*^{2}

*s*′ = 5 + 6*t* = 35

6*t* = 30

*t* = 5 seconds

5. **A** Take the derivative and set it equal to zero to determine the critical numbers of *s*.

*s* = 80*t* – 16*t*^{2}

*s*′ = 80 – 32*t* = 0

–32*t* = – 80

*t* =

Test values around the critical number to determine the behavior of *s* and confirm that *t* = is a maximum.

*s*′(2) = + ; *s*′(3) = –

To find the height, plug in to the original function given.

= 200 – 100 = 100

6. **E** Take the derivative implicitly with respect to *x.*

*y*^{2} = *x*^{3} + 3*x*^{2}

2*y y*′ = 3*x*^{2} + 6*x*

Do not simplify! Plug in the point immediately to find the slope.

2(–2) *y*′ = 3(1)^{2} + 6(1)

–4 *y*′ = 9

*y*′ = –

Use the point-slope equation of a line to find the equation.

*y* – *y*_{1} = *m*(*x* – *x*_{1})

*y* – (–2) = (*x* – 1)

4*y* + 8 = –9*x* + 9

9*x* + 4*y* = 1

7. **C** The distance between the point (16,0) and the curve is found by using the Pythagorean theorem, specifically the equation for the distance can be set up as: *D*^{2} = (*x* − 16)^{2} + (*y* − 0)^{2} = *x*^{2} − 32*x* + 256 + *y*^{2}. When *y* is replaced with the equation of the curve *y* = , the equation becomes *D*^{2} = *x*^{2} − 32*x* + 256 + *x* = *x*^{2} − 31*x* + 256. Rename as *L*, such that *D*^{2} = *L*, then take the derivative of *L*. Set the derivative equal to zero to minimize (or maximize) the distance and solve for *x*. = 2*x* − 31 = 0; *x* = . Find the corresponding *y*-value, . To determine whether this point is a minimum or maximum, find the second derivative of *L*. = 2. Since the second derivative is positive, the point is a relative minimum.

8. **A** In order to approximate the error in the volume of the cone use the approximation formula: *dy* = *f*′(*x*)*dx*. For this problem, *f*(*x*) = *V* = *πr*^{2}*h*, *f*′(*x*) = = *πrh* = 40*π*, and *dx* = *dr* = 0.1. Recall that the diameter is given, so we need to divide that measure in half to get the radius. When these values are input into the equation is *dy* = *dV* = 40*π*(0.1) = 4*π m*^{3}.

9. **E** The length of a curve described parametrically is found by using *L* = . In this problem, *a* = 2 and *b* = 5; = 2*t* and = 4*t*. So the length of the curve is .

10. **A** Since the normal is parallel to the *y*-axis, the tangent to this curve will be perpendicular to the *y*-axis (opposite reciprocals). Thus, = 0. Taking the first derivative of *y*, and setting it equal to zero, we can determine the *x*-coordinate of the point: = 3*x*^{2} − 8*x* − 3 = 0; *x* = − or *x* = 3. When these *x*-values are plugged into *y*, the two possible points are: and (3, −5).

11. **A** . For this problem,

*f*′(*θ*) = 3 cos*θ* and .

With algebra and trig identities, the final solution is A.

12. **B** First, plug *y* = –2 into the equation: 3*x*^{3} − 2*x*^{2} + *x* + 6 = 0. Remember, the AP writers will not give you a very complicated equation that you cannot factor, so always test “easy” values first. For example, test *x* = 1, *x* = –1, *x* = 0, and if necessary, *x* = 2 and *x* = –2. In this case, *x* = –1 works, so the point that the line will be tangent to is (–1,–2). Next, use implicit differentiation to find the first derivative of the equation: 9*x*^{2} − 4*x* + 1 = 3*y*^{2} + 4*y* + 3. Do not simplify; immediately plug in the point (–1, –2) and solve for . Thus, the slope of the tangent will be 2. Finally, plug in the point and slope into the equation of the line: *y* + 2 = 2(*x* + 1). Simplified, the equation of the line is *y* = 2*x*.

13. **B** First, plug in the second point, (0,–4), and solve for *d: d* = –4. Next, plug in the first point, (2,8), into the equation along with *d* = –4: 12 = 8*a* + 4*b* + 2*c*. Now, because the curve is normal to the line *y* = −*x* − 4, the first derivative at (0,–4) is 3. Thus, take the first derivative of the equation for the curve: = 3*ax*^{2} + 2*bx* + *c*. Set the first derivative equal to 3 and plug in *x* = 0: 3 = 3*a*(0)^{2} + 2*b*(0) + *c*. This gives you *c* = 3. You were told *b* = 5 in the question stem, so, to solve for *a*, plug in the values of *b, c,* and *d* into the reduced equation, 12 = 8*a* + 4*b* + 2*c*, and solve for *a*. Therefore, *a* = −.

14. **B** To determine when the particle changes direction, take the first derivative and set it equal to zero; the particle changes direction when the velocity is equal to zero. *x*′(*t*) = 8*x*^{3} − 12*x*^{2} + 4*x* = 0. The particle then changes direction at *x* = 0, *x* = , and *x* = 1.

15. **C** First take the derivative of each parametric function: = 9*x*^{2} − 4*x* and = 4*x* + 3. Now evaluate each of these derivatives at *t* = 3, = 69 and = 15. Finally, the particle’s velocity, .

16. **D** The formula for differentials is *f*(*x* + Δ*x*) = *f*(*X*) + *f*′(*x*)Δ*x*. For this question, *f*(*x*) = *x*^{3}, *f*′(*x*) = 3*x*^{2}, *x* = 5, and Δ*x* = 0.2.

Plug in those values: (5.02)^{3} = 5^{3} + 3(5)^{2}(0.2) = 140.

17. **C** The formula for the volume of a cylinder is *V* = *πr*^{2}*h*. The first derivative is = 2*πrh*. Use differentials to solve for the approximate change in volume: *dV* = 2*πrh dr*. Plug in the given values, where *dr* = 0.03, and solve for *dV*: *dV* = 2π(9)(12)(0.03) = 6.48*π* in^{3}.

18. **B** When dealing with trig functions, the formula for differentials, *f*(*x* + Δ*x*) = *f*(*x*) + *f*′(*x*)Δ*x*, will not work in degrees, only radians. So, first determine the values of *x* and Δ*x* in radians: *x* = 270° = radians and Δ*x* = 5° = radians. Note, *f*(*x*) = cos *x* and *f*′(*x*) = −sin *x*. Plug everything into the formula for differentials: cos 275° = .

19. **D** The length of a curve can be found by the formula *L* = , where *a* and *b* are the endpoints of the curve. *L* = . This integral can be solved by *u*-substitution and the Fundamental Theorem of Calculus, so the length of the curve is .

20. **B** Calculate the derivative with respect to *x*.

2*x* + 2*y y*′ = 0

Do not simplify! Plug in the point (2, 4) to find the slope.

2(2) + 2(4)*y*′ = 0

4 + 8 *y*′ = 0

8 *y*′ = –4

*y*′ = –

21. **D** Calculate the derivative with respect to *x*.

2*x* + *y* + *x y*′ + 2*y y*′ = 0

Do not simplify! Plug in (1, 1) to find the slope.

Use the point-slope form of a line to find the equation.

*y* – *y*_{1} = *m*(*x* – *x*_{1})

*y* – 1 = –1(*x* – 1)

*y* – 1 = –*x* + 1

*x* + *y* = 2

22. **D** Take the derivative and plug in 2 for *x* to get the slope.

*f*′(*x*) = 4*x*^{3} + 6*x*

*f*′(2) = 4(2)^{3} + 6(2)

= 32 + 12

= 44

Take the negative reciprocal of the slope to get the normal slope.

*m* = −

23. **E** For optimization problems, you must find two equations to relate to one another. Establish your equations for perimeter and area of a rectangle, keeping in mind that the total perimeter is provided to you.

100 = 2*x* + 2*y* *A = xy*

Take the perimeter equation and isolate one of the variables. Solve it for *y* and you get

100 – 2*x* = 2*y*

50 – *x* = *y*

Plug the expression from the above into the area formula to gain the new expression in terms of *x*.

*A* = *x*(50 – *x*) = 50*x* – *x*^{2}

Take the derivative and set it equal to zero to determine the critical point.

*A* = 50*x* – *x*^{2}

0 = 50 – 2*x*

2*x* = 50

*x* = 25

Testing the value of *x* = 20 reveals that the function is increasing and testing the value of *x* = 30 reveals that the function is decreasing, confirming that 25 is a maximum.

Find the *y* value by plugging in *x* = 25 to the rewritten perimeter expression.

50 – *x* = *y*

50 – (25) = *y*

25 = *y*

The dimensions are 25 by 25.

24. **C** For optimization problems, you must find two equations to relate to one another. Establish your equations for product and sum, keeping in mind that the product is provided to you.

*S* = *x* + *y* 100 = *xy*

Take the product equation and isolate one of the variables. Solving for *y* yields:

= *y*

Plug the expression from the above into the sum formula to gain the new expression in terms of *x.*

*S* = *x* +

Take the derivative and set it equal to zero to determine the critical point.

*S* = *x* +

0 = 1 –

= 1

100 = *x*^{2}

10 = *x*

Testing the value of *x* = 9 reveals that the function is decreasing and testing the value of *x* = 11 reveals that the function is increasing, confirming that *x* = 10 is a minimum value.

Find the *y* value by plugging in *x* = 10 into the rewritten product function.

= *y*

= *y*

10 = *y*

So, the two numbers are 10 and 10.

25. **B** For optimization problems, you must find two equations to relate to one another. Establish your equations for product and sum, keeping in mind that the sum is provided to you.

23 = *x* + *y P* = *xy*

Take the sum equation and isolate one of the variables. Solving for *y* yields:

23 – *x* = *y*

Plug the expression from the above into the product formula to gain the new expression in terms of *x*.

*P* = *x* (23 – *x*)

*P* = 23*x* – *x*^{2}

Take the derivative and set it equal to zero to determine the critical point.

*P* = 23*x* – *x*^{2}

0 = 23 – 2*x*

2*x* = 23

*x* = 11.5

Testing the value of *x* = 10 reveals that the function is increasing and testing the value of *x* = 12 reveals that the function is decreasing, confirming that *x* = 11.5 is a maximum value.

Find the *y* value by plugging *x* = 11.5 into the rewritten sum function.

23 – *x* = *y*

23 – 11.5 = *y*

11.5 = *y*

So, the numbers are 11.5 and 11.5.

26. **D** Take the derivative and set it equal to zero to determine the critical points.

*y* = 4*x*^{3} – 9*x*^{2} + 6*x*

*y*′ = 12*x*^{2} – 18*x* + 6 = 0

6(2*x*^{2} – 3*x* + 1) = 0

6(2*x* – 1)(*x* – 1) = 0

*x* = ; *x* = 1

*y*′(0) = + ; *y*′ = – ; *y*′(2) = +

By testing values around the determined critical points, it can be concluded that *x* = 1 is a relative minimum. To find the value, simply plug *x* = 1 into the original function.

*y*(1) = 4(1)^{3} – 9(1)^{2} + 6(1)

= 4 – 9 + 6 = 1

# Chapter 17

# General and Partial Fraction Integration Drill

## GENERAL AND PARTIAL FRACTION INTEGRATION DRILL

1. Find *f*(*x*) if *f*(0) = 8 and *f*′(*x*) = 1 – 6*x*.

(A) 8 + *x* + 3*x*^{2}

(B) 8 + *x* – 3*x*^{2}

(C) 6 – *x* + 3*x*^{2}

(D) –8 + *x* + 3*x*^{2}

(E) 1 – *x* – 3*x*^{2}

2. If *f*′(*x*) = 8*x*^{3} + 12*x* + 3, find *f*(*x*) if *f*(1) = 6.

(A) 2*x*^{4} + 6*x*^{2} + 3*x* – 5

(B) 5*x*^{4} + 3*x*^{2} + 2*x* + 5

(C) –4*x*^{4} + 6*x*^{2} + 3*x* – 5

(D) 5*x*^{4} + 6*x*^{2} + 3*x* + 5

(E) 2*x*^{4} – 6*x*^{2} – 3*x* – 5

3. If *f* = 4, and *f*′(*x*) = 2cos *x* + sec^{2} *x*, then find *f*(*x*).

(A) –2cos *x* + tan *x* + 4

(B) 2sin *x* – tan *x* + 4

(C) 2 sin *x* + tan *x* + 4 − 2

(D) −2 sin *x* + 2 sec *x* + 4 + 2

(E) 2 sin *x* + 2 tan *x* + 4 + 2

4. If *f*′′(*x*) = 24*x*^{2} + 2*x* + 10, *f*(1) = 5, and *f*′(1) = –3, then *f*(*x*) =

(A) 2*x*^{4} + *x*^{3} + 5*x*^{2} − 22*x* −

(B) 2*x*^{4} − − 5*x*^{2} − 22*x* +

(C) −2*x*^{4} + + 5*x*^{2} − 22*x* +

(D) 2*x*^{4} − 4*x*^{3} + 5*x*^{2} − 22*x* +

(E) 2*x*^{4} + + 5*x*^{2} − 22*x* +

5. Given *f*(0) = 2, *f*′(0) = 1, *f*″ = 4 – 6*x* – 40*x*^{3}, what is *f*(*x*)?

(A) 2 – *x* – 2*x*^{2} – *x*^{3} + 2*x*^{5}

(B) 2 + 2*x* + 3*x*^{2} – *x*^{3} – 6*x*^{5}

(C) –2 + *x* + 2*x*^{2} – *x*^{3} – 2*x*^{5}

(D) 2 + *x* + 2*x*^{2} – *x*^{3} – 2*x*^{5}

(E) 2 + *x* + 2*x*^{2} + *x*^{3} + 6*x*^{5}

6. Given *f*″(*x*) = sin *x* + cos *x*, *f*(0) = 3, and *f*′(0) = 4, then what is *f*(*x*)?

(A) sin *x* + cos *x* + 5*x* + 4

(B) cos *x* – sin *x* + 5*x* + 4

(C) sin *x* – cos *x* − 5*x* – 4

(D) sin *x* – cos *x* + 5*x* + 4

(E) –sin *x* – cos *x* + 5*x* + 4

7. If *f*″(*x*) = 2 – 12*x*, *f*(0) = 9, and *f*(2) = 15, then find *f*(*x*).

(A) 2*x*^{3} – *x*^{2} + 9*x* + 9

(B) –2*x*^{3} + *x*^{2} – 9*x* + 9

(C) 2*x*^{3} + *x*^{2} + 9*x* + 9

(D) –2*x*^{3} – *x*^{2} + 9*x* – 9

(E) –2*x*^{3} + *x*^{2} + 9*x* + 9

8. Find *f*(*x*) if *f*″(*x*) = 20*x*^{3} + 12*x*^{2} + 4, *f*(0) = 8, and *f*(1) = 5.

(A) –*x*^{5} + *x*^{4} + 2*x*^{2} – 5*x* + 2

(B) *x*^{5} + *x*^{4} + 2*x*^{2} – 7*x* + 8

(C) 6*x*^{5} + 5*x*^{4} + 2*x*^{2} – 7*x* + 8

(D) –*x*^{5} + *x*^{4} + 2*x*^{2} – 7*x* + 8

(E) *x*^{5} + *x*^{4} + 2*x*^{2} – 5*x* + 2

9. Find *f*(*x*) if *f*′(0) = 2, *f*(1) = 1, and *f*″(*x*) = 20*x*^{3} + 12*x*^{2} + 4.

(A) *x*^{5} + *x*^{4} + 2*x*^{2} – 5*x* + 2

(B) 6*x*^{5} + *x*^{4} + 2*x*^{2} – 5*x* + 2

(C) *x*^{5} – *x*^{4} – 2*x*^{2} – 5*x* – 2

(D) –*x*^{5} + *x*^{4} + 2*x*^{2} – 5*x* + 2

(E) *x*^{5} + 5*x*^{4} + 3*x*^{2} – 5*x* + 2

10. Find *f*(*x*) if *f*″(*x*) = 48*x*^{2} – 6*x* + 1, *f*(0) = 1, and *f*′(0) = 2.

(A) *x*^{4} – 4*x*^{3} + *x*^{2} + 2*x* + 1

(B) *x*^{4} – *x*^{3} + *x*^{2} + 2*x* + 1

(C) 4*x*^{4} – *x*^{3} + *x*^{2} + 2*x* + 1

(D) –4*x*^{4} + *x*^{3} + *x*^{2} + 2*x* + 1

(E) 4*x*^{4} + *x*^{3} + 2*x*^{2} + *x* + 1

11. (5 − 2*x* + 3*x*)*dx* =

(A) –63

(B) –14

(C) 2

(D) 14

(E) 63

12. *x*(2 + *x*^{5})*dx* =

(A) –33

(B) –

(C) 0

(D)

(E) 33

13.

(A) *x*^{4} – *x*^{2} + *C*

(B) –*x*^{} − x + *C*

(C) + *C*

(D) + *C*

(E) *x*^{} − 4 + *C*

14. ∫(3*x*^{2} + 2*x*)(*x*^{3} + *x*^{2})^{5} *dx* =

(A) (*x*^{3} + *x*^{2})^{6} + *C*

(B) (3*x*^{2} + 2*x*)^{6} + *C*

(C) (*x*^{3} + *x*^{2})^{6} + *C*

(D) (3*x*^{2} + 2*x*)^{6} + *C*

(E) –(*x*^{3} + *x*^{2})^{6} + *C*

15. − 3∫*x*^{2} *dx* =

(A) *x* – *x*^{3} + *C*

(B) *x*^{2} – *x*^{3} + *C*

(C) – *x*^{3} + *C*

(D) 2 – *x*^{3} + *C*

(E) – *x*^{3} + *C*

16. ∫(*x*^{2} + *x* − *x*^{−1} + 2*x*^{−2})*dx* =

(A) 2*x* + 1 + *x*^{−2} – 4*x*^{−3} + *C*

(B) –*x*^{3} – *x*^{2} + ln *x* + 2*x*^{−1} + *C*

(C) *x*^{3} + *x*^{2} – ln *x* – 2*x*^{−1} + *C*

(D) 2*x* – 1 – *x*^{−2} + 4*x*^{−3} + *C*

(E) 3*x* + *C*

17. ∫(2*x*^{2} + 3)*dx* =

(A) 2*x*^{3} + 3*x* + *C*

(B) *x*^{3} + 3*x* + *C*

(C) *x*^{3} + 3*x* + *C*

(D) 4*x* + *C*

(E) 4*x*^{3} + 3 + *C*

18. ∫*x*(2*x*^{2} + 7)^{5} *dx* =

(A) + *C*

(B) + *C*

(C) (2*x*^{2} + 7)^{6} + *C*

(D) + *C*

(E) (2*x*^{2} + 7)^{4} + *C*

19. (*x*^{2} − *x*)(6*x*^{3} − 9*x*^{2}) *dx* =

(A)

(B)

(C)

(D)

(E)

20. Find (2*t* − 3*t*^{3}) *dt* =

(A) −*x*^{3}

(B) 6*x*^{5} − 9*x*^{11}

(C) 2*x*^{3} − 3*x*^{6}

(D) 2*x*^{3} − 3*x*^{9}

(E) 6*x*^{6} − 3*x*^{18}

21.

(A) 0

(B)

(C)

(D)

(E) The integral diverges.

22.

(A)

(B)

(C)

(D)

(E)

23.

(A)

(B)

(C)

(D)

(E)

24.

(A) – ln |*x* – 1| + ln |*x* + 1| + *C*

(B) 2 |ln *x* – 1| – ln |*x* + 1| + *C*

(C) ln |*x* – 1| – ln |*x* + 1| + *C*

(D) ln |*x* – 1| – 2 ln |*x* + 1| + *C*

(E) 2 ln |*x* – 1| –2 ln |*x* + 1| + *C*

25.

(A)

(B)

(C)

(D)

(E)

26.

(A)

(B)

(C)

(D)

(E)

27.

(A)

(B)

(C)

(D)

(E)

28.

(A)

(B)

(C)

(D)

(E)

29.

(A)

(B)

(C)

(D)

(E)

30.

(A) ln *x* + ln |*x* + 1| + *C*

(B) – ln *x* + ln |*x* + 1| + *C*

(C) ln *x* – ln |*x* + 1| + *C*

(D) – ln *x* – ln |*x* + 1| + *C*

(E) ln *x* – ln |*x* – 1| + *C*

31.

(A) – ln *x* – ln |2*x* + 1| + *C*

(B) ln *x* + ln |2*x* + 1| + *C*

(C) – ln *x* + ln |2*x* + 1| + *C*

(D) ln *x* – ln |2*x* + 1| + *C*

(E) ln *x* – ln |2*x* – 1| + *C*

32. ∫*y*(*y*^{2} + 1)^{5} dy =

(A) (*y*^{2} + 1)^{6} + *C*

(B) (*y*^{2} + 1)^{4} + *C*

(C) (*y*^{2} + 1)^{5} + *C*

(D) (*y*^{2} + 1)^{6} + *C*

(E) (*y*^{2} + 1)^{6} + *C*

33.

(A) − *x*^{2} + *C*

(B) − *x*^{2} + *C*

(C) − *x*^{3} + *C*

(D) 2 − *x*^{2} + *C*

(E) 3 − *x*^{3} + *C*

34.

(A) 19.333

(B) 24.667

(C) 33

(D) 38.333

(E) 46

35. If the series converges, then which of the following series will not converge?

(A)

(B)

(C)

(D)

(E)

36. To what value does the series (*n* + 1)2* ^{n}* converge?

(A) 0

(B)

(C) 1

(D) 2

(E) The series diverges.

37. A particle’s acceleration is given by *a*(*t*) = 6*t*^{2} − 4. Its initial position is 1 and its velocity at *t* = 2 is 9. What is the position of the particle at *t* = 4?

(A) 113

(B) 110

(C) 101

(D) 100

(E) 92

38. Find the average value of *f*(*x*) = *x*^{3} − 2 from *x* = 1 to *x* = 3.

(A) 2

(B) 4

(C) 8

(D) 16

(E) 32

# Chapter 18

# General and Partial Fraction Integration Drill Answers and Explanations

## ANSWER KEY

1. B

2. A

3. C

4. E

5. D

6. D

7. E

8. B

9. A

10. C

11. E

12. D

13. D

14. A

15. C

16. C

17. A

18. B

19. C

20. B

21. E

22. D

23. B

24. C

25. A

26. E

27. A

28. E

29. B

30. C

31. D

32. E

33. D

34 A

35. B

36. E

37. C

38. C

## EXPLANATIONS

1. **B** To find *f*(*x*), you must integrate to find the general antiderivative.

∫(1 − 6*x*)*dx* = *x* − 3*x*^{2} + *C*

To find the value of *C*, you must use the given point *f*(0) = 8.

*f*(0) = (0) – 3(0)^{2} + *C* = 8 ; so, *C* = 8

*f*(*x*) = 8 + *x* – 3*x*^{2}

2. **A** To find *f*(*x*), you must integrate to find the general antiderivative.

∫(8*x*^{3} + 12*x* + 3)*dx* = 2*x*^{4} + 6*x*^{2} + 3*x* + *C*

To find the value of *C*, you must use the given point *f*(1) = 6.

*f*(1) = 2(1)^{4} + 6(1)^{2} + 3(1) + *C* = 6 ; so *C* = –5

*f*(*x*) = 2*x*^{4} + 6*x*^{2} + 3*x* – 5

3. **C** To find *f*(*x*), you must integrate to find the general antiderivative.

∫(2 cos *x* + sec^{2}*x*)*dx* = 2 sin *x* + tan *x* + *C*

To find the value of *C*, you must use the given point = 4.

= 4 so *C* = 4 − 2

*f*(*x*) = 2 sin *x* + tan *x* + 4 − 2

4. **E** To find *f*(*x*), you must integrate twice. Integrate once to find the first general antiderivative.

∫(24*x*^{2} + 2*x* + 10)*dx* = 8*x*^{3} + *x*^{2} + 10*x* + *C*

To find the value of *C*, you must first use the coordinate affecting the first derivative, *f*′(1) = –3.

*f*′(1) = 8(1)^{3} + (1)^{2} + 10(1) + *C* = –3 ; so *C* = –22

Now, you must integrate again to find the general antiderivative *f.*

∫(8*x*^{3} + *x*^{2} + 10*x* − 22)*dx* = 2*x*^{4} + + 5*x*^{2} − 22 + *C*

Use the coordinate *f*(1) = 5 to find the value of the constant.

*f*(1) = 2(1)^{4} + + 5(1)^{2} − 22(1) + *C* = 5; so *C* =

*f*(*x*) = 2*x*^{4} + + 5*x*^{2} − 22*x* +

5. **D** To find *f*(*x*), you must integrate twice. Integrate once to find the first antiderivative.

∫(4 − 6*x* − 40*x*^{3})*dx* = 4*x* − 3*x*^{2} − 10*x*^{4} + *C*

Use the coordinate involving the first derivative to find the value of *C*.

*f*(0) = 4(0) – 3(0)^{2} – 10(0)^{4} + *C* = 1 ; so *C* = 1

Now integrate once again to find *f.*

∫(4*x* − 3*x*^{2} − 10*x*^{4} + 1)*dx* = 2*x*^{2} − *x*^{3} − 2*x*^{5} + *x* + *C*

Use the other coordinate to find the value of the constant of integration, *C.*

*f*(0) = 2(0)^{2} – (0)^{3} – 2(0)^{5} + (0) + *C* = 2 ; so *C* = 2

*f*(*x*) = 2 + *x* + 2*x*^{2} – *x*^{3} – 2*x*^{5}

6. **D** To find *f*(*x*), you must integrate twice. Integrate once to find the first antiderivative.

∫(sin *x* + cos *x*)*dx* = − cos *x* + sin *x* + *C*

To find the value of *C*, you must use the coordinate that affects the first derivative.

*f*′(0) = –cos (0) + sin (0) + *C* = 4 ; so *C* = 5

Now integrate again to find the general function *f.*

∫(−cos *x* + sin *x* + 5)*dx* = − sin *x* − cos *x* + 5*x* + *C*

To find the value of *C*, you must use the other coordinate given affecting the function.

*f*(0) = –sin (0) – cos (0) + 5(0) + *C* = 3 ; so *C* = 4

*f*(*x*) = –sin *x* – cos *x* + 5*x* + 4

7. **E** To find *f*(*x*), you must integrate twice.

∫(2 − 12*x*)*dx* = 2*x* − 6*x*^{2} + *C*

∫(2*x* − 6*x*^{2} + *C*)*dx* = *x*^{2} − 2*x*^{3} + C*x* + *D*

Use the coordinates given to find the values of the constants.

*f*(0) = (0)^{2} – 2(0)^{3} + *C*(0) + *D* = 9 ; so *D* = 9

*f*(2) = (2)^{2} – 2(2)^{3} + *C*(2) + 9 = 15 ; so *C* = 9

*f*(*x*) = –2*x*^{3} + *x*^{2} + 9*x* + 9

8. **B** To find *f*(*x*), you must integrate twice.

∫(20*x*^{3} + 12*x*^{2} + 4)*dx* = 5*x*^{4} + 4*x*^{3} + 4*x* + *C*

∫(5*x*^{4} + 4*x*^{3} + 4*x* + *C*)*dx* = *x*^{5} + *x*^{4} + 2*x*^{2} + C*x* + *D*

Use the coordinates given to find the values of the constants.

*f*(0) = (0)^{5} + (0)^{4} + 2(0)^{2} + *C*(0) + *D* = 8 ; so *D* = 8

*f*(1) = (1)^{5} + (1)^{4} + 2(1)^{2} + *C*(1) + 8 = 5 ; so *C* = –7

*f*(*x*) = *x*^{5} + *x*^{4} + 2*x*^{2} – 7*x* + 8

9. **A** To find *f*(*x*), you must integrate twice.

∫(20*x*^{3} + 12*x*^{2} + 4)*dx* = 5*x*^{4} + 4*x*^{3} + 4*x* + *C*

∫(5*x*^{4} + 4*x*^{3} + 4*x* + *C*)*dx* = *x*^{5} + *x*^{4} + 2*x*^{2} + C*x* + D

Use the coordinates given to find the values of the constants.

*f*(0) = (0)^{5} + (0)^{4} + 2(0)^{2} + *C*(0) + *D* = 2 ; so *D* = 2

*f*(1) = (1)^{5} + (1)^{4} + 2(1)^{2} + *C*(1) + 2 = 1 ; so *C* = –5

*f*(*x*) = *x*^{5} + *x*^{4} + 2*x*^{2} – 5*x* + 2

10. **C** To find *f*(*x*), you must integrate the function.

∫(48*x*^{2} − 6*x* + 1)*dx* = 16*x*^{3} − 3*x*^{2} + *x* + *C*

Use the coordinate given that affects the first derivative.

*f*(0) = 16(0)^{3} – 3(0)^{2} + (0) + *C* = 2 ; so *C* = 2

Integrate once again to find the general antiderivative function.

∫(16*x*^{3} − 3*x*^{2} + *x* + 2)*dx* = 4*x*^{4} − *x*^{3} + *x*^{2} + 2*x* + *C*

Now, plug the other coordinate into the function to determine the value of *C*.

*f*(0) = 4(0)^{4} – (0)^{3} + (0)^{2} + 2(0) + *C* = 1 ; so *C* = 1

*f*(*x*) = 4*x*^{4} – *x*^{3} + *x*^{2} + 2*x* + 1

11. **E** Calculate the anti-derivative.

Evaluate the function at the endpoints.

[5(4) – (4)^{2} + (4)^{3}] – [5(1) – (1)^{2} + (1)^{3}]

[20 – 16 + 64] – [5 – 1 + 1]

[68] – [5] = 63

12. **D** Calculate the anti-derivative.

Evaluate the function at the endpoints.

13. **D** Evaluate the integral first by dividing *x* into each term in the numerator.

*x*^{3} − 4 + *C*

14. **A** Evaluate the integral using the substitution *u* = *x*^{3} + *x*^{2}, then *du* = 3*x*^{2} + 2*x dx*

∫(3*x*^{2} + 2*x*)(*x*^{3} + *x*^{2})*dx* = ∫(*u*^{5} *du* =

*u*^{6} + *C* = (*x*^{3} + *x*^{2})^{6} + *C*

15. **C** This is truly a friendly problem if you identify what is actually given to you. The coefficients should be moved into the integrand to see that the integrands are perfect derivatives. Utilizing this fact, we have:

Which yields:

*x*^{} − *x*^{3} + *C* = − *x*^{3} + *C*

16. **C** Evaluate the integral using the power rule for integration, paying attention to the *x*^{–1} term, which will turn out to be the natural logarithm.

∫(*x*^{2} + *x* − *x*^{−1} + 2*x*^{−2})*dx* = *x*^{3} + *x*^{2} − ln *x* − 2*x*^{−1} + *C*

17. **A** Use the addition and power rules of integration: ∫(2*x*^{2} + 3)*dx* = *x*^{3} + 3*x* + *C*

18. **B** Use *u*-substitution. *u* = (2*x*^{2} + 7) and *du* = 4 *x dx*. Plug this back into the integral:

19. **C** Use *u*-substitution and the Fundamental Theorem of Calculus to evaluate this integral. *u* = 6*x*^{3} – 9*x*^{2} and *du* = 18*x*^{2} – 18*x*. Thus,

20. **B** From the Second Fundamental Theorem of Calculus,

(2*t* − 3*t*^{3})*dt* = (2*x*^{3} − 3(*x*^{3})^{3}) = 6*x*^{5} − 9*x*^{11}.

21. **E** Since is undefined at *x* = 0, this is an improper integral and we must evaluate it in two parts: . Thus, the integral diverges.

22. **D** You must use the method of partial fractions to evaluate this integral. First begin by separating the function to solve for the individual numerators.

*x* – 9 = *A*(*x* – 2) + *B*(*x* + 5)

*x* – 9 = *Ax* – 2*A* + *Bx* + 5*B*

*x* – 9 = (*A* + *B*)*x* – 2*A* + 5*B*

So, it follows that:

(1) *A* + *B* = 1 and (2) –2*A* + 5*B* = –9

Multiply the first equation by 2 and then stack and add them to start solving simultaneously.

If *B* = –1, using equation (1) means that *A* must equal 2. So, the partial fraction decomposition must be:

23. **B** To solve this integral, you must use the method of partial fractions. First, start by breaking down the function into its individual components.

1 = *A*(*x* – 1) + *B*(*x* + 4)

1 = *Ax* – *A* + *Bx* + 4*B*

1 = (*A* + *B*) *x* – *A* + 4*B*

So, it follows that:

*A* + *B* = 0

–*A* + 4*B* = 1

Add the two equations together to start solving for the individual values of the constants. This yields that *B* = and *A* equals −. So, the partial fraction decomposition is:

24. **C** To solve this integral, you must use the method of partial fractions. First, start by breaking down the function into its individual components.

2 = *A*(*x* – 1) + *B*(*x* + 1)

2 = *Ax* – *A* + *Bx* + *B*

2 = (*A* + *B*) *x* – *A* + *B*

So, it follows that:

*A* + *B* = 0

–*A* + *B* = 2

Add the two equations together to start solving for the individual values of the constants. This yields that *B* = 1 and *A* = –1. So, the partial fraction decomposition is:

−ln |*x* + 1| + ln |*x* − 1| + *C* =

ln |*x* − 1| − ln |*x* + 1| + *C*

25. **A** To solve this integral, you must use the method of partial fractions. First, start by breaking down the function into its individual components.

3 = *A*(*x* – 1) + *B*(*x* + 2)

3 = *Ax* – *A* + *Bx* + 2*B*

3 = (*A* + *B*)*x* – *A* + 2*B*

So, it follows that:

*A* + *B* = 0

–*A* + 2*B* = 3

Add the two equations together to start solving for the individual values of the constants. This yields that *B* = 1 and *A* = –1. So, the partial fraction decomposition is:

26. **E** To solve this integral, you must use the method of partial fractions. First, start by breaking down the function into its individual components.

5 – *x* = *A*(*x* + 1) + *B*(2*x* – 1)

5 – *x* = *Ax* + *A* + 2*Bx* – *B*

5 – *x* = (*A* + 2*B*)*x* + *A* – *B*

So, it follows that:

*A* + 2*B* = –1

*A* – *B* = 5

Subtract the two equations to start solving for the individual values of the constants. This yields that *B* = –2 and *A* = 3. So, the partial fraction decomposition is:

27. **A** To solve this integral, you must use the method of partial fractions. First, start by breaking down the function into its individual components.

*x*^{2} + 12*x* + 12 = *A*(*x*^{2} – 4) + *Bx*(*x* – 2) + *C x*(*x* + 2)

*x*^{2} + 12*x* + 12 = *Ax*^{2} – 4*A* + *Bx*^{2} – 2*Bx* + *Cx*^{2} + 2*Cx*

*x*^{2} + 12*x* + 12 = (*A* + *B* + *C*)*x*^{2} + (–2*B* + 2*C*)*x* – 4*A*

So, it follows that:

*A* + *B* + *C* = 1

–2*B* + 2*C* = 12

–4*A* = 12

You must start solving the equations simultaneously. Start with the third one since it only has the *A* in it. *A* must equal –3. Plugging that value into the first equation now means that *B* + *C* = 4. Multiply this equation by 2 and add it to the remaining equation.

2*B* + 2*C* = 8

–2*B* + 2*C* = 12

This yields that *C* = 5 and *B* = –1. So, the partial fraction decomposition is:

28. **E** To solve this integral, you must use the method of partial fractions. First, start by breaking down the function into its individual components.

4*x*^{2} + 2*x* – 1 = *Ax*(*x* + 1) + *B*(*x* + 1) + *Cx*^{2}

4*x*^{2} + 2*x* – 1 = *Ax*^{2} + *Ax* + *Bx* + *B* + *Cx*^{2}

4*x*^{2} + 2*x* – 1 = (*A* + *C*)*x*^{2} + (*A* + *B*)*x* + *B*

So, it follows that:

*A* + *C* = 4

*A* + *B* = 2

*B* = –1

The value of *B* is given immediately, so use that to determine the values of *A* and *C* which are 3 and 1, respectively. So, the partial fraction decomposition is:

29. **B** To solve this integral, you must use the method of partial fractions. First, start by breaking down the function into its individual components.

(* A quadratic denominator requires a linear numerator)

*x*^{2} – 1 = *A*(*x*^{2} + 1) + (*Bx* + *C*)*x*

*x*^{2} – 1 = *Ax*^{2} + *A* + *Bx*^{2} + *Cx*

*x*^{2} – 1 = (*A* + *B*)*x*^{2} + *Cx* + *A*

So, it follows that:

*A* + *B* = 1

*C* = 0

*A* = –1

*A* = –1, *B* = 2, and *C* = 0

Now that you know the values of the constants, the partial fraction decomposition becomes:

30. **C** To solve this integral, you must use the method of partial fractions. First, start by breaking down the function into its individual components.

1 = *A*(*x* + 1) + *Bx*

1 = *Ax* + *A* + *Bx*

1 = (*A* + *B*)*x* + *A*

So, it follows that:

*A* + *B* = 0

*A* = 1

So, *A* = 1 and *B* = –1.

Now that you know the values of the constants, the partial fraction decomposition becomes:

= ln *x* − ln |*x* + 1| + *C*

31. **D** To solve this integral, you must use the method of partial fractions. First, start by breaking down the function into its individual components.

1 = *A*(2*x* + 1) + *Bx*

1 = 2*Ax* + *A* + *Bx*

1 = (2*A* + *B*)*x* + *A*

So, it follows that

2*A* + *B* = 0 and *A* = 1; so *B* must equal –2.

Now that you know the values of the constants, the partial fraction decomposition becomes:

= ln *x* − ln |2*x* + 1| + *C*

32. **E** Evaluate the integral using the substitution *u* = *y*^{2} + 1, then *du* = *y dy.*

∫*y*(*y*^{2} + 1)^{5} *dy* = ∫*u*^{5} *du* =

*u*^{6} + *C* = (*y*^{2} + 1)^{6} + *C*

33. **D** First, divide the numerator by *x* to begin solving the integral.

2 − *x*^{2} + *C*

34. **A** From the First Fundamental Theorem of Calculus, = 19.333

Answer is B.

35. **B** First, recognize that the series is a geometric series with *a* = 6 and *r* = . Since geo-metric series with *r* < 1, always converge, look through the answer choices for any other geometric series. Notice, answer choice C is also a geometric series with *a* = 1 and *r* = . The other four answer choices are not geometric series. Therefore, we need to use other tests to determine if these series converge or not. Although you may choose any test you like, look for indicators that one test may be better suited than the others. For instance, answers A and E are good candidates for the Comparison Test. Recall the Comparison Test says let 0, ≤ *a _{n}* ≤

*b*, if

*b*converges, then

_{n}*a*converges. In this case, the question stem gives

_{n}*b*= , and the answer choices give

_{n}*a*. In both answer choices A and E,

_{n}*a*<

_{n}*b*for all values of

_{n}*n*, so both A and E converge. For answer choice D, the Ratio Test is a good option to use. In the Ratio Test,

*ρ*= , where

*a*. If

_{n}*ρ*< 1, the series converges. As

*ρ*< 1, in D, that series converges. If you use the Ratio Test for answer choice B, you will find that

*ρ*= 1, the test is insufficient and we don’t know whether the series converges or diverges. To determine with certainty if the series in B will converge or not, you will need to use another test, such as the Integral Test or the P-Series Test. If you use the Integral Test, remember

*a*=

_{n}*f*(

*n*) where

*f*is positive, continuous and decreasing for

*x*≥ 1. If

*f*(

*x*)

*dx*converges, then

*a*converges. In B, the integral diverges, so the series diverges and the answer is B. The P-Series Test is even easier; recognize that any series in the form , where

_{n}*p*> 0, converges if

*p*> 1 and diverges if 0 <

*p*≤ 1, so B diverges.

36. **E** In order to determine to what value the series converges, we must first determine if the series converges. You may use whichever test you please. Here, we will use the Ratio Test. In the Ratio Test, *ρ* = , where *a _{n}*. If

*ρ*< 1, the series converges and if

*ρ*< 1, the series diverges. In this case,

*ρ*= 2, so the series diverges and cannot converge to a single value.

37. **C** First, integrate, *a*(*t*) = 6*t*^{2} − 4, to determine *v*(*t*). Thus, *v*(*t*) = 2*t*^{5} − 4*t* + *C*. Determine *C* by plugging in (2,9), so *v*(*t*) = 2*t*^{3} − 4*t* + 1. Next, integrate *v*(*t*) to determine the position function, *x*(*t*): *x*(*t*) = *t*^{4} − 2*t*^{2} + *t* + *C*. At the point, (0,1), *x*(*t*) = *t*^{4} − 2*t*^{2} + *t* + 1. Now, evaluate *x*(4) to determine the position of the particle. *x*(4) = 101.

38. **C** *f*(*c*) = *f*(*x*)*dx*. For this problem *f*(*c*) = (*x*^{3} − 2)*dx* = 8.

# Chapter 19

# Trigonometric Integration Drill

## TRIGONOMETRIC INTEGRATION DRILL

1. ∫(csc 5*x* cot 5*x*)*dx* =

(A) + *C*

(B) + *C*

(C) + *C*

(D) + *C*

(E) + *C*

2. ∫(2sin *x* + 3cos *x*)*dx* =

(A) 3sin *x* + 2cos *x* + *C*

(B) 3sin *x* − 2cos *x* + *C*

(C) 3cos *x* − 2sin *x* + *C*

(D) 2cos *x* − 3sin *x* + *C*

(E) −2cos *x* − 3sin *x* + *C*

3. (sin^{−1}(*x*^{3} + *x*^{2})) =

(A)

(B)

(C)

(D)

(E)

4. ∫sin^{3} *x* cos *x dx* =

(A) sin^{4} *x* + *C*

(B) sin^{4} *x* + *C*

(C) cos^{4} *x* + *C*

(D) cos^{4} *x* + *C*

(E) 4sin^{4} *x* + *C*

5.

(A) cot(3*x*) − tan + *C*

(B) tan(3*x*) − cot + *C*

(C) 3tan(3*x*) − cot + *C*

(D) tan(3*x*) + cot + *C*

(E) cot − tan(3*x*) + *C*

6. ∫sin^{2} *x* cos *x* =

(A) + *C*

(B) cos^{3}*x* + *C*

(C) sin^{3}*x* + *C*

(D) + *C*

(E) + *C*

7. Find (3cos *x* − 2sin *x*)*dx*.

(A) + 2

(B) − 2

(C) 5

(D) + 2

(E) + 2

8. ∫csc(8*x*) cot(8*x*)*dx* =

(A) + *C*

(B) + *C*

(C) + *C*

(D) + *C*

(E) + *C*

9. ∫(cos 3*x* − csc^{2}3*x*)*dx* =

(A) sin 3*x* + cot 3*x* + *C*

(B) (sin 3*x* + cot 3*x*) + *C*

(C) 3(sin 3*x* + cot 3*x*) + *C*

(D) sin 3*x* − cot 3*x* + *C*

(E) (sin 3*x* − cot 3*x*) + *C*

10. ∫(3sin *x* − 2 cos *x*)*dx* =

(A) −3cos *x* − 2sin *x* + *C*

(B) 3cos*x* + 2sin *x* + *C*

(C) 2sin *x* − 3*cos x* + *C*

(D) 3cos *x* − 2sin *x* + *C*

(E) 2cos *x* + 3sin *x* + *C*

11. ∫(8 sin^{2}*x* + 8cos^{2}*x* − 8*x*^{2})*dx* =

(A) 8*x* − *x*^{3} + *C*

(B) 8*x* + *x*^{3} + *C*

(C) 16*x* + sin *x* − cos *x* + *C*

(D) 16*x* + *C*

(E) sin^{2}*x* − cos^{2}*x* + *x*^{3} + *C*

12. ∫3 csc 2*x dx* =

(A) −ln |csc 3*x* − cot 3*x*| + *C*

(B) ln |csc 3*x* + cot 3*x*| + *C*

(C) ln |csc 3*x* − cot 3*x*| + *C*

(D) −ln |csc 3*x* + cot 3*x*| + *C*

(E) ln |csc 3*x* − cot 3*x*| + *C*

13. ∫*x*^{2} cos *x dx* =

(A) *x*^{2} sin *x* + 2*x* cos *x* – 2sin *x* + *C*

(B) –2*x* sin *x* + *C*

(C) *x*^{3} sin *x* + *C*

(D) *x*^{2} sin *x* – 2*x* cos *x* + 2sin *x* + *C*

(E) –*x*^{2} sin *x* – 2x cos *x* + 2sin *x* + *C*

14. ∫*x* sec^{2} *x dx* =

(A) 2 sec^{2} *x* tan *x* + *C*

(B) *x* tan *x* + ln |cos *x*| + *C*

(C) *x*^{2} tan *x* + *C*

(D) *x* tan *x* – ln |cos *x*| + *C*

(E) ln |sec *x*| + *C*

15. ∫*e*^{2}* ^{x}* sin

*x dx*=

(A) –*e*^{2}* ^{x}* cos

*x*+

*C*

(B) −*e*^{2}* ^{x}* + ∫

*e*

^{2}

*sin*

^{x}*x dx*

(C) −2*e*^{2}* ^{x}* cos

*x*− 2 ∫

*e*

^{2}

*cos*

^{x}*x dx*

(D) −*e*^{2}* ^{x}* cos

*x*+ 2 ∫

*e*

^{2}

*cos*

^{x}*x dx*

(E) *e*^{2}* ^{x}* cos

*x*− 2 ∫

*e*

^{2}

*cos*

^{x}*x dx*

16. ∫arctan *x dx* =

(A) *x* arctan *x* +

(B) −*x* arctan *x* +

(C) *x* arctan *x* −

(D) −*x* arctan *x* −

(E) −*x* arctan *x* +

17. ∫*x* cos *x dx*

(A) *x* sin *x* + cos *x* + *C*

(B) *x* cos *x* + sin *x* + *C*

(C) (*x* + 1) cos *x* + *C*

(D) (*x* + 1) sin *x* + *C*

(E) *x* sin *x* – cos *x* + *C*

18. ∫*θ* cos 5*θ dθ* =

(A) *θ*^{2} sin 5*θ* + cos 5*θ* + *C*

(B) *θ*^{2} sin 5*θ* + cos 5*θ* + *C*

(C) *θ* sin 5*θ* + cos 5*θ* + *C*

(D) *θ* cos 5*θ* + sin 5*θ* + *C*

(E) *θ* sin 5*θ* – cos 5*θ* + *C*

19. ∫*x* sin 2*x dx* =

(A) *x* sin 2*x* + cos 2*x* + *C*

(B) *x* cos 2*x* + sin 2*x* + *C*

(C) – *x* sin 2*x* + cos 2*x* + *C*

(D) – *x* cos 2*x* + sin 2*x* + *C*

(E) *x* cos 2*x* – sin 2*x* + *C*

20. *x* sin 3*x dx* =

(A) –

(B) –

(C) 0

(D)

(E)

21. ∫*x* sec *x* tan *x dx* =

(A) *x* sec *x* – ln |sec *x* + tan *x*| + *C*

(B) *x* sec *x* – ln |sec *x* tan *x*| + *C*

(C) *x* sec *x* + *C*

(D) *x*^{2} sec *x* + *C*

(E) *x* sec *x* + ln |sec *x* + tan *x*| + *C*

22. ∫sec^{3} *x* tan *x dx* =

(A) sec^{3} *x* + *C*

(B) sec^{3} *x* + *C*

(C) 3 sec^{3} *x* + *C*

(D) tan^{2} *x* + *C*

(E) sec *x* tan *x* + *C*

23. ∫ csc^{2} *x dx* =

(A) – cot *x* + *C*

(B) – (cot *x*)^{} + *C*

(C) – (cot *x*)^{} + *C*

(D) (csc *x*)^{} + *C*

(E) (cot *x*)^{} + *C*

24. ∫*e*^{tan} * ^{x}* sec

^{2}

*x dx*=

(A) *e*^{sec} * ^{x}* +

*C*

(B) *e*^{tan} ^{x} + *C*

(C) *e*^{sec} ^{x}^{ tan} ^{x} + *C*

(D) *e*^{csc} ^{x}^{ cot} ^{x} + *C*

(E) *e*^{cot} ^{x} + *C*

25.

(A) 1 –

(B) 0

(C) 1 +

(D) *π*

(E)

26.

(A) ln (1 + cos^{2}*x*) + *C*

(B) ln (sin^{2}*x*) + *C*

(C) ln (sec^{2}*x*) + *C*

(D) ln (1 – cos^{2}*x*) + *C*

(E) –ln (1 + cos^{2}*x*) + *C*

27.

(A) (tan^{–1} *x*)^{2} + *C*

(B) (tan^{–1} *x*)^{2} + *C*

(C) (tan^{–1} *x*)^{2} + *C*

(D) (tan^{–1} *x*) + *C*

(E) 2 tan^{–1} *x* + *C*

28. ∫*x*^{2} cos(*x*^{3}) *dx* =

(A) sin (*x*^{3}) + *C*

(B) sin (*x*^{2}) + *C*

(C) sin (*x*^{3}) + *C*

(D) sin (*x*^{3}) + *C*

(E) sin (*x*^{3}) + *C*

29. ∫sin *x* sec^{2} (cos *x*) *dx* =

(A) tan (cos *x*) + *C*

(B) –tan (cos *x*) + *C*

(C) –cos (cos *x*) + *C*

(D) –sin (cos *x*) + *C*

(E) sin (cos *x*) + *C*

30. ∫cot *x dx* =

(A) ln (sin *x*) + *C*

(B) ln (cos *x*) + *C*

(C) ln (tan *x*) + *C*

(D) ln (csc^{2} *x*) + *C*

(E) –ln (sin *x*) + *C*

31. ∫sec 2*θ* tan 2*θ* d*θ* =

(A) sec 2*θ* + *C*

(B) sec 2*θ* + *C*

(C) sec 2*θ* + *C*

(D) sec 2*θ* + *C*

(E) – sec 2*θ* + *C*

32. ∫ tan *x dx* =

(A) ln (cos *x*) + *C*

(B) ln (sin *x*) + *C*

(C) ln (tan *x*) + *C*

(D) ln (sec *x*) + *C*

(E) ln (cot *x*) + *C*

33.

(A) sin^{–1} *x* + ln (1 + *x*^{2}) + *C*

(B) sec^{–1} *x* + ln (1 + *x*^{2}) + *C*

(C) tan *x* + ln (1 + *x*^{2}) + *C*

(D) sin *x* + ln (1 + *x*^{2}) + *C*

(E) tan^{–1} *x* + ln (1 + *x*^{2}) + *C*

34.

(A) ln (sin^{–1} *x*) + *C*

(B) ln (cos^{–1} *x*) + *C*

(C) ln (tan^{–1} *x*) + *C*

(D) ln (csc^{–1} *x*) + *C*

(E) ln (cot^{–1} *x*) + *C*

35. ∫(1 − cos^{2} *x*)cos *x dx* =

(A) cos^{3} *x* + *C*

(B) cos^{3} *x* + *C*

(C) sin^{3} *x* + *C*

(D) sin^{3} *x* + *C*

(E) (1 – cos^{2} *x*)^{2} + *C*

36. ∫(tan^{2} *x* + 1) *dx* =

(A) tan^{3} *x* + *C*

(B) tan^{3} *x* + *x* + *C*

(C) cos *x* + *C*

(D) sec^{2} *x* + *C*

(E) tan *x* + *C*

37. ∫*x*^{2} cos (*x*^{3}) *dx* =

(A) *x*^{3} sin (*x*^{3}) + *C*

(B) *x*^{3} + sin (2*x*^{3}) + *C*

(C) *x*^{3} – sin (2*x*^{3}) + *C*

(D) *x*^{3} + sin (2*x*^{3}) + *C*

(E) 2*x* cos (*x*^{3}) – 3*x*^{5} sin (*x*^{3}) + *C*

38. ∫sin^{2} (3*x*) *dx* =

(A) 2*x* – 12 sin (6*x*) + *C*

(B) *x* – sin (6*x*) + *C*

(C) sin^{3} (3*x*) + *C*

(D) *x* + sin (6*x*) + *C*

(E) 2*x* + 12 sin (6*x*) + *C*

39. ∫(4 sec^{2} *x* + csc *x* cot *x*) *dx* =

(A) 4 tan *x* – csc *x* + *C*

(B) 4 tan *x* + csc *x* + *C*

(C) 8 sec *x* tan *x* – csc *x* + *C*

(D) 8 sec *x* – csc *x* + *C*

(E) 4 tan *x* – csc^{2} *x* + *C*

40. ∫sec *x* (sec *x* + tan *x*)*dx* =

(A) tan *x* – sec *x* + *C*

(B) 2 sec *x* tan *x* + sec *x* + *C*

(C) sec^{2} *x* + sec *x* tan *x* + *C*

(D) 2 sec *x* tan *x* – sec *x* + *C*

(E) tan *x* + sec *x* + *C*

41.

(A) 1 – cos *x* + *C*

(B) tan *x* – sec *x* + *C*

(C) tan *x* + sec *x* + *C*

(D) *x* – csc *x* cot *x* + *C*

(E) sec *x* – tan *x* + *C*

42. ∫tan^{2} *x dx* =

(A) *x* – tan *x* + *C*

(B) sec^{2} *x* – 1 + *C*

(C) tan *x* – *x* + *C*

(D) 2tan *x* sec^{2} *x* + *C*

(E) tan *x* + *x* + *C*

43. ∫sec *x* (tan *x* + cos *x*)*dx* =

(A) sec *x* – *x* + *C*

(B) sec *x* + sin *x* + *C*

(C) *x* – sec *x* + *C*

(D) sec *x* tan *x* – sin *x* + *C*

(E) sec *x* + *x* + *C*

44.

(A) arctan (2*x*) + *C*

(B) 2 arctan (2*x*) + *C*

(C) arcsin (2*x*) + *C*

(D) 2 arcsin (2*x*) + *C*

(E) arccos (2*x*) + *C*

45.

(A) arctan (2*x*) + *C*

(B) + *C*

(C) + *C*

(D) + *C*

(E) arcsin (2*x*) + *C*

46. ∫ sin *x e*^{cos} * ^{x} dx* =

(A) –*e*^{sin} * ^{x}* +

*C*

(B) *e*^{cos} * ^{x}* +

*C*

(C) *e*^{sin} * ^{x}* +

*C*

(D) –*e*^{sec} * ^{x}* +

*C*

(E) –*e*^{cos} * ^{x}* +

*C*

47.

(A) –csc (ln *x*) + *C*

(B) –cos (ln *x*) + *C*

(C) cos (ln *x*) + *C*

(D) sin (ln *x*) + *C*

(E) –sin (ln *x*) + *C*

48. cos^{3}*x* sin 2*x dx* =

(A)

(B)

(C)

(D)

(E) 1

49.

(A) (sin^{–1} *x*)^{2} + *C*

(B) (tan^{–1} *x*) + *C*

(C) (tan^{–1} *x*)^{2} + *C*

(D) (sin^{–1} *x*) + *C*

(E) cos^{−1}*x* + *C*

# Chapter 20

# Trigonometric Integration Drill Answers and Explanations

## ANSWER KEY

1. B

2. B

3. C

4. A

5. B

6. E

7. B

8. A

9. B

10. A

11. A

12. D

13. A

14. B

15. D

16. C

17. A

18. C

19. D

20. E

21. A

22. A

23. B

24. B

25. C

26. E

27. C

28. A

29. B

30. A

31. C

32. D

33. E

34. A

35. D

36. E

37. D

38. B

39. A

40. E

41. B

42. C

43. E

44. A

45. D

46. E

47. B

48. B

49. A

## EXPLANATIONS

1. **B** Recall ∫ csc *u* cot *u du* = − csc *u* + *C*, so you need to use *u*-substitution. Here, *u* = 5*x* and *du* = 5 *dx*. When these are replaced into the integral, we get: ∫ csc 5*x* cot 5*x dx* = ∫ csc *u* cot *u du* = −csc *u* + *C*. Replacing 5*x* back in for *u*, the solution is + *C*.

2. **B** Recall ∫ cos *u du* = sin *u* + *C* and ∫ sin *u du* = − cos *u* + *C*. So the integral becomes: ∫(2 sin *x* + 3 cos *x*)*dx* = −2 cos *x* + 3 sin *x* + *C*.

3. **C** Recall . In this case, *u* = *x*^{3} + *x*^{2} and *du* = (3*x*^{2} + 2*x*)*dx*. Thus, .

4. **A** Use *u*-substitution: *u* = sin *x* and *du* = cos *x dx*. Thus, ∫ sin^{3} *x* cos *x dx* = ∫*u*^{3} *du* = sin^{4} *x* + *C*.

5. **B** Treat this as two separate integrals added together and then use *u*-substitution on each part. For the first integral, ∫3sec^{2}(3*x*)*dx*, *u* = 3*x* and *du* = 3 *dx*. Then, ∫3sec^{2}(3*x*)*dx* = tan(3*x*) + *C*. For the second integral, and *du* = (*x*^{3} − 2)*dx*. So, .

Thus, the final solution is .

6. **E** Use *u*-substitution where *u* = sin *x* and *du* = cos *x dx*. Therefore, ∫sin^{2}*x* cos *x dx* = + *C*.

7. **B** Use the first fundamental theorem of calculus:

(3 cos *x* − 2 sin *x*)*dx* = (3 sin *x* + 2 cos *x*) .

8. **A** Remember the integrals of trig functions and use *u*-substitution:

∫csc(8*x*)cot(8*x*)*dx* = + *C*.

9. **B** Use the addition rule, *u*-substitution, and the integrals of trig functions:

∫(cos 3*x* − csc^{2} 3*x*)*dx* =

10. **A** Remember coefficients do not change a derivative. ∫(3 sin *x* − 2 cos *x*)*dx* = −3 cos *x* − 2 sin *x* + *C*.

11. **A** Remember the trig identities. In this case, notice sin^{2}*x* + cos^{2}*x* = 1. Thus, you can rewrite the integral as ∫(8(1) − 8*x*^{2})*dx*. Now, using the power rule, you can integrate: 8*x* − *x*^{3} + *C*

12. **D** Using *u*-substitution and the integral of cosecant, ∫3 csc 2*x dx* = −ln |csc 3*x* + cot 3*x*| + *C*.

13. **A** Evaluate the integral using Integration by Parts. Make the following substitutions:

*u = x*^{2} *dv* = cos *x dx*

Then,

*du* = 2*x dx v* = sin *x*

Recall the formula for Integration by Parts is *uv* − ∫*v du*. So,

∫*x*^{2} cos *x dx* = *x*^{2} sin *x* − 2 ∫*x* sin *x dx*

Integration by Parts is required again for this new integral, so make the following substitutions:

*u = x* *dv* = sin *x dx*

Then,

*du = dx* *v* = –cos *x*

Proceed with Integration by Parts once again.

∫*x*^{2} cos *x dx* = *x*^{2} sin *x* − 2[−*x* cos *x* + ∫ cos *x dx*]

= *x*^{2} sin *x* + 2*x* cos *x* − 2 sin *x* + *C*

14. **B** Evaluate the integral using Integration by Parts. Make the following substitutions:

*u = x dv* = sec^{2} *x dx*

Then,

*du = dx v* = tan *x*

Recall the formula for Integration by Parts is *uv* − ∫*v du*. So,

∫*x* sec^{2} *x dx* = *x* tan *x* − ∫tan *x dx*

= *x* tan *x* + ln |cos *x*| + *C*

15. **D** Integration by parts is required to evaluate this integral, but looking at the answers, you do not have to solve it out entirely. So, proceed with the following substitutions:

*u* = *e*^{2}^{x}*dv* = sin *x dx*

Then,

*du* = 2*e*^{2}^{x} dx*v* = –cos *x*

Recall the formula for Integration by parts is *uv* − ∫*v du*. So,

∫*e*^{2}* ^{x}* sin

*x dx*= −

*e*

^{2}

*cos*

^{x}*x*+ 2 ∫

*e*

^{2}

*cos*

^{x}*x dx*

16. **C** Evaluate the integral using Integration by Parts. Looking at the answers reveals that you are not required to solve out the integral entirely, so proceed using the following substitutions:

*u* = arctan *x* *dv = dx*

Then,

*du* = *v = x*

Recall the formula for Integration by Parts is *uv* − ∫*v du*. So,

∫arctan *x dx* = *x* arctan *x* −

17. **A** Evaluate the integral by using the Integration by Parts method.

*u = x* *dv* = cos *x dx*

*du = dx* *v* = sin *x*

Recall, the proper integration by parts format is *uv* − ∫*v du*. So,

∫*x* cos *x dx* = *x* sin *x* − ∫sin *x dx*

= *x* sin *x* + cos *x* + *C*

18. **C** Evaluate the integral by using the Integration by Parts method.

*u* = *θ* *dv* = cos 5*θ dx*

*du = dθ* *v* = sin 5*θ*

Recall, the proper integration by parts format is *uv* − ∫*v du*. So,

19. **D** Evaluate the integral by using the Integration by Parts method.

*u = x* *dv* = sin 2*x dx*

*du = dx* *v* = −cos 2*x*

Recall, the proper integration by parts format is *uv* − ∫*v du*. So,

20. **E** Evaluate the integral by using the Integration by Parts method.

*u = x* *dv* = sin 3*x dx*

*du = dx* *v* = −cos 3*x*

Recall, the proper integration by parts format is . So,

21. **A** Evaluate the integral by using the Integration by Parts method.

*u = x* *dv* = sec *x* tan *x dx*

*du = dx* *v* = sec *x*

Recall, the proper integration by parts format is *uv* − ∫*v du*. So,

∫*x* sec *x* tan *x dx* = *x* sec *x* − ∫sec *x dx*

= *x* sec *x* − ln |sec *x* + tan *x*| + *C*

Note: ∫sec *x dx* can be found by multiplying the top and the bottom by sec *x* + tan *x* and performing a *u*-substitution.

22. **A** Rewrite the integral to make integration easier.

∫sec^{3} *x* tan *x dx* = ∫sec^{2} *x* sec *x* tan *x dx* =

Now integrate with u-substitution, allowing *u* to equal sec *x*, since the derivative, *du,* shows up in the integrand.

Let *u* = sec *x*, so *du* = sec *x* tan *x dx.*

∫ sec^{2} *x* sec *x* tan *x dx* = ∫*u*^{2} *du* =

*u*^{3} = sec^{3} *x* + *C*

23. **B** Evaluate the integral substituting *u* = cot *x*, then – *du* = csc^{2} *x dx.*

∫ csc^{2} *x dx* = ∫ *du* =

24. **B** Evaluate the integral using the substitution *u* = tan *x*, then *du* = sec^{2} *x dx*.

∫*e*^{tan} * ^{x}* sec

^{2}

*x dx*= ∫

*e*=

^{u}du*e ^{u}* +

*C*=

*e*

^{tan}

*+*

^{x}*C*

25. **C** Evaluate the integral first by dividing each term in the numerator by cos^{2} *θ*.

Calculate the value at the endpoints.

26. **E** Rewrite the integral using the identity sin 2*x* = 2 sin *x* cos *x*

Evaluate the integral using the substitution *u* = 1 + cos^{2} *x*, because then –*du* = 2sin *x* cos *x dx*.

–ln (*u*) = –ln (1 + cos^{2} *x*) + *C.*

27. **C** Evaluate the integral using the substitution *u* = tan^{–1} *x*, then *du* = .

*u*^{2} + *C* = (tan^{–1} *x*)^{2} + *C*.

28. **A** Evaluate the integral by using the substitution *u* = *x*^{3}, then *du* = *x*^{2} *dx.*

∫*x*^{2} cos(*x*^{3})*dx* = ∫cos(*u*) *du* =

sin *u* + *C* = sin (*x*^{3}) + *C*

29. **B** Evaluate the integral using the substitution *u* = cos *x*, then –*du* = sin *x dx.*

∫ sin *x* sec^{2}(cos *x*)*dx* = − ∫sec^{2} *u du* =

–tan *u* + *C* = –tan (cos *x*) + *C*

30. **A** Rewrite the integral to make integration easier:

∫ cot *x dx* =

Evaluate the integral using the substitution *u* = sin *x*, then *du* = cos *x dx.*

ln (*u*) + *C* = ln (sin *x*) + *C*.

31. **C** Evaluate the integral using the substitution *u* = 2*θ*, then *du* = *dθ*.

∫sec 2*θ* tan 2*θ dθ* = ∫sec *u* tan *u du* =

sec *u* + *C* = sec 2*θ* + *C*

32. **D** Rewrite the integral to make it easier to integrate.

Evaluate the integral using the substitution *u* = cos *x*, then –*du* = sin *x dx.*

–ln *u* + *C* = –ln (cos *x*) + *C* = ln (sec *x*) + *C*

33. **E** Separate the integral into the sum of 2 integrals:

The first integral is the derivative of the tan^{–1} *x*. The second integral requires the substitution of *u* = 1 + *x*^{2}, which allows *du* = *x dx.*

tan^{−1}*x* + ln (1 + *x*^{2}) + *C*

34. **A** Evaluate the integral using the substitution *u* = sin^{–1} *x*, then *du* = .

ln (*u*) + *C* = ln (sin^{–1} *x*) + *C*

35. **D** Evaluate the integral by first rewriting it using the Pythagorean identity sin^{2} *x* + cos^{2} *x* = 1.

∫(1 − cos^{2} *x*) cos *x dx* = ∫sin^{2} *x* cos *x dx*

Evaluate the integral now by using the substitution *u* = sin *x*, then *du* = cos *x dx*

∫sin^{2} *x* cos *x dx* = ∫*u*^{2} *du* =

*u*^{3} + *C* = sin^{3} *x* + *C*

36. **E** Evaluate the integral by first rewriting it using the Pythagorean identity tan^{2} *x* + 1 = sec^{2} *x*.

∫(tan^{2} *x* + 1)*dx* = ∫sec^{2} *x dx* = tan *x* + *C*

37. **D** First, to evaluate the integral, you must make the substitution *u* = *x*^{3}, which makes *du* = *x*^{2} *dx.*

Make the substitution back of *u* = *x*^{3} to get the final answer.

38. **B** First, make a *u*-substitution to evaluate this integral. Let *u* = 3*x,* which makes *du = dx.*

Make the substitution back of *u* = 3*x* to get the final answer.

39. **A** Evaluate the integral using the trigonometric integration rules for the sec^{2} *x* function and the csc *x* cot *x* function.

∫(4 sec^{2} *x* + csc *x* cot *x*)*dx* = 4 tan *x* − csc *x* + *C*

40. **E** Evaluate the integral by first rewriting it applying a distribution of sec *x* throughout the integrand. Then, use the trigonometric integration rules for the sec^{2} *x* function and the sec *x* tan *x* function.

∫sec *x*(sec *x* + tan *x*)*dx* = ∫(sec^{2} *x* + sec *x* tan *x*)*dx* =

tan *x* + sec *x* + *C*

41. **B** To tackle this integral, you must first perform a little algebraic manipulation. Multiply both the top and the bottom by (1 – sin *x*).

The resulting integrand can be split up into two integrals now. will now become sec^{2}*x*, and will now become –sec *x* tan *x*.

= ∫sec^{2} *x dx* = ∫sec *x* tan *x dx* = tan *x* − sec *x* + *C*

42. **C** Evaluate the integral by first using the Pythagorean identity 1 + tan^{2}*x* = sec^{2} *x*.

∫tan^{2} *x dx* = ∫(sec^{2} *x* − 1)*dx* = tan *x* − *x* + *C*

43. **E** Evaluate the integral first by rewriting it after an application of distribution of the sec *x* term. Then, evaluate using the trigonometric integration rule for the sec *x* tan *x* function.

∫sec *x* (tan *x* + cos *x*)*dx* = ∫(sec *x* tan *x* + 1)*dx* = sec *x* + *x* + *C*

44. **A** This integral takes on the form of the inverse tangent function, but the denominator must be rewritten as 1 + (2*x*)^{2} to be able to apply a *u*-substitution.

Now, let *u* = 2*x*, which makes *du = dx.*

45. **D** Evaluate the integral using the substitution *u* = *x*^{2} + 4*x*, which makes *du* = *x* + 2 *dx.*

46. **E** Evaluate the integral making the substitution *u* = cos *x*, then –*du* = sin *x dx.*

∫sin *x e*^{cos} * ^{x} dx* = −∫

*e*=

^{u}du−*e*^{cos}* ^{x}* +

*C*

47. **B** Evaluate the integral using the substitution *u* = ln *x*, then *du* = .

= ∫sin *u du* =

−cos *u* + *C*

−cos (ln *x*) + *C*

48. **B** First, rewrite the integral making the substitution

Evaluate the integral making the substitution *u* = cos *x*, making –*du* = sin *x dx*. The new limits of integration, when *x* = 0 and respectively, become 1 and 0.

49. **A** Evaluate the integral making the substitution *u* = sin^{–1}*x*, which makes *du* = .

# Chapter 21

# Exponential and Logarithmic Integration Drill

## EXPONENTIAL AND LOGARITHMIC INTEGRATION DRILL

1.

(A) (ln *x*)^{3} + *C*

(B) 3*x* (ln *x*)^{3} + *C*

(C) 3 (ln *x*)^{3} + *C*

(D) (ln *x*)^{3} + *C*

(E) (ln *x*)^{3} + *C*

2.

(A) –4

(B) –2

(C) 0

(D) 2

(E) 4

3.

(A) ln (*e* + 1)

(B) ln 2

(C) 0

(D) –ln (*e* + 1)

(E) –ln 2

4. ∫*e*^{5}* ^{x} dx* =

(A) 5*e*^{5}* ^{x}* +

*C*

(B) *e*^{5}* ^{x}* +

*C*

(C) *e*^{5}* ^{x}* +

*C*

(D) *e ^{x}* +

*C*

(E) (5*e*)^{5}* ^{x}* +

*C*

5. ∫*e*^{−3}* ^{x} dx* =

(A) – *e*^{–}* ^{x}* +

*C*

(B) – *e*^{–3}* ^{x}* +

*C*

(C) *e*^{–3}* ^{x}* +

*C*

(D) –3 *e*^{–3}* ^{x}* +

*C*

(E) –3 *e*^{–x} + *C*

6.

(A) (1 + e* ^{x}*)

^{}+

*C*

(B) (1 + e* ^{x}*)

^{}+

*C*

(C) (1 + e* ^{x}*)

^{}+

*C*

(D) (1 + e* ^{x}*)

^{}+

*C*

(E) 3 (1 + e* ^{x}*)

^{}+

*C*

7. ∫(*e ^{x}* +

*e*)

^{−x}*dx*=

(A) *e*^{2}* ^{x}* + 2

*x*–

*e*

^{–2}

*+*

^{x}*C*

(B) 2*e*^{2}* ^{x}* + 2

*x*– 2

*e*

^{–2}

*+*

^{x}*C*

(C) *e*^{2}* ^{x}* + 2x – 2

*e*

^{–2}

*+*

^{x}*C*

(D) *e*^{2}* ^{x}* +

*x*–

*e*

^{–2}

*+*

^{x}*C*

(E) *e*^{2}* ^{x}* + 2

*x*+ 2

*e*

^{–2}

*+*

^{x}*C*

8. ∫*e ^{x}* (4 +

*e*)

^{x}^{4}

*dx*=

(A) (4 + *e ^{x}*)

^{5}+

*C*

(B) (4 + *e ^{x}*)

^{5}+

*C*

(C) (4 + *e ^{x}*)

^{5}+

*C*

(D) (4 + *e ^{x}*)

^{5}+

*C*

(E) *e ^{x}* (4 +

*e*)

^{x}^{5}+

*C*

9. ∫*e ^{x}* sin (

*e*)

^{x}*dx*=

(A) –cos (*e ^{x}*) +

*C*

(B) cos (*e ^{x}*) +

*C*

(C) sin (*e ^{x}*) +

*C*

(D) –sin (*e ^{x}*) +

*C*

(E) –csc (*e ^{x}*) +

*C*

10.

(A) –*e ^{–x}* – 2

*x*+

*e*+

^{x}*C*

(B) 2*x* + *e ^{x}* +

*C*

(C) 2*e ^{x}* + 2

*x*+

*C*

(D) –*e ^{–x}* + 2

*x*+

*C*

(E) –*e ^{–x}* + 2

*x*+

*e*+

^{x}*C*

11.

(A) + ln 2

(B) – + ln 2

(C) + ln 2

(D) + ln 2

(E) – ln 2

12.

(A) ln 3

(B) ln 4

(C) ln 8

(D) ln 16

(E) ln 24

13.

(A)

(B) 3 ln

(C) 3 ln

(D)

(E) ln (5)

14.

(A) –[1 – *e*^{–1}]

(B) [1 + e^{–1}]

(C) –[1 + *e*^{–1}]

(D) [*e*^{–1} – 1]

(E) [1 – *e*^{–1}]

15.

(A) –*e*^{} + *C*

(B) –*e*^{} + *C*

(C) *e*^{} + *C*

(D) *e*^{–} + *C*

(E) –*e*^{–} + *C*

16.

(A) *e*^{} + *C*

(B) 2*e*^{} + *C*

(C) 2*e*^{} + *C*

(D) –2*e*^{} + *C*

(E) 2*e*^{–} + *C*

17. 1* _{x} dx* =

(A) 10* ^{x}* ln 10

(B)

(C)

(D)

(E)

18.

(A) ln (*e ^{x}* – 1) +

*C*

(B) (*e ^{x}* + 1)

^{2}+

*C*

(C) ln (*e*^{2}* ^{x}* + 1) +

*C*

(D) *e ^{x}* ln (

*e*+ 1) +

^{x}*C*

(E) ln (*e ^{x}* + 1) +

*C*

19.

(A) *x* – 6 ln |*x* – 6| + *C*

(B) *x* – 6 ln |*x* + 6| + *C*

(C) *x* + 6 ln |*x* + 6| + *C*

(D) *x* + 6 ln |*x* – 6| + *C*

(E) *x* – 6 ln |*x* + 6| + *C*

20.

(A) (1 + *e*^{2}* ^{x}*) +

*C*

(B) *e*^{2}* ^{x}* +

*C*

(C) tan^{–1}(*e ^{x}*) +

*C*

(D) sin^{–1}(*e ^{x}*) +

*C*

(E)

21. ∫*e*^{2}* ^{x}* cos

*x dx*

(A) *e*^{2}* ^{x}* cos

*x*+

*e*

^{2}

*sin*

^{x}*x*+

*C*

(B) *e*^{2}* ^{x}* cos

*x*+

*e*

^{2}

*sin*

^{x}*x*+

*C*

(C) *e*^{2}* ^{x}* cos

*x*−

*e*

^{2}

*sin*

^{x}*x*+

*C*

(D) *e*^{2}* ^{x}* cos

*x*−

*e*

^{2}

*sin*

^{x}*x*+

*C*

(E) *e*^{2}* ^{x}* cos

*x*+

*e*

^{2}

*sin*

^{x}*x*+

*C*

22. ∫*x*^{2}ln^{2} *x dx* =

(A) (9ln^{2}*x* + 6ln*x* + 2) + *C*

(B) (9ln^{2}*x* − 6ln*x* + 2) + *C*

(C) (9ln^{2}*x* − 6ln*x* − 2) + *C*

(D) (9ln^{2}*x* + 6ln*x* − 2) + *C*

(E) −(9ln^{2}*x* − 6ln*x* + 2) + *C*

23. ∫*x*2^{x2+7} *dx* =

(A)

(B)

(C)

(D)

(E)

24. ∫3(4^{6}* ^{x}*)

*dx*=

(A)

(B)

(C)

(D)

(E)

25.

(A) ln|*x* − 3| + ln|*x* + 2| + *C*

(B) ln|*x* − 3| − ln|*x* + 2| + *C*

(C) ln|*x* − 3| + ln|*x* + 2| + *C*

(D) ln|*x* − 3| − ln|*x* + 2| + *C*

(E) ln|*x* + 2| − ln|*x* − 3| + *C*

26. ∫*xe*^{2}* ^{x} dx* =

(A) *xe*^{2}* ^{x}* –

*e*

^{2}

*+*

^{x}*C*

(B) 2*xe*^{2}* ^{x}* – 4

*e*

^{2}

*+*

^{x}*C*

(C) 2*xe*^{2}* ^{x}* –

*e*

^{2}

*+*

^{x}*C*

(D) *xe*^{2}* ^{x}* – 4

*e*

^{2}

*+*

^{x}*C*

(E) *xe*^{2}* ^{x}* +

*e*

^{2}

*+*

^{x}*C*

27. ∫*x*^{2}*e*^{2}* ^{x} dx* =

(A) 2*x*^{2}*e*^{2}* ^{x}* – 2

*xe*

^{2}

*+ 4*

^{x}*e*

^{2}

*+*

^{x}*C*

(B) *x*^{2}*e*^{2}* ^{x}* +

*xe*

^{2}

*–*

^{x}*e*

^{2}

*+*

^{x}*C*

(C) –*x*^{2}*e*^{2}* ^{x}* +

*xe*

^{2}

*– 4*

^{x}*e*

^{2}

*+*

^{x}*C*

(D) *x*^{2}*e*^{2}* ^{x}* –

*xe*

^{2}

*+*

^{x}*e*

^{2}

*+*

^{x}*C*

(E) 2 *x*^{2}*e*^{2}* ^{x}* –

*xe*

^{2}

*+*

^{x}*e*

^{2}

*+*

^{x}*C*

28. ∫*xe*^{−2}* ^{x} dx* =

(A) *xe*^{–2}* ^{x}* – 4

*e*

^{–2}

*+*

^{x}*C*

(B) –*xe*^{–2}* ^{x}* –

*e*

^{–2}

*+*

^{x}*C*

(C) *xe*^{–2}* ^{x}* +

*e*

^{–2}

*+*

^{x}*C*

(D) – 2*xe*^{–2}* ^{x}* – 4

*e*

^{–2}

*+*

^{x}*C*

(E) *xe*^{–2}* ^{x}* –

*e*

^{–2}

*+*

^{x}*C*

29. ∫*x*^{3}*e ^{x} dx* =

(A) *x*^{4}*e ^{x}* +

*C*

(B) 3*x*^{2}*e ^{x}* +

*x*

^{3}

*e*+

^{x}*C*

(C) *x*^{3}*e ^{x}* + 3

*x*

^{2}

*e*+ 6

^{x}*xe*+ 6

^{x}*e*+

^{x}*C*

(D) – *x*^{3}*e ^{x}* + 3

*x*

^{2}

*e*– 6

^{x}*xe*+ 6

^{x}*e*+

^{x}*C*

(E) *x*^{3}*e ^{x}* – 3

*x*

^{2}

*e*+ 6

^{x}*xe*– 6

^{x}*e*+

^{x}*C*

30. ∫*x*^{3} ln *x dx* =

(A) 3*x* + *C*

(B) *x*^{4} ln *x* + *C*

(C) *x*^{4} ln *x* – *x*^{4} + *C*

(D) – *x*^{4} ln *x* + *x*^{4} + *C*

(E) 4*x*^{4} ln *x* – 16*x*^{4} + *C*

31. ∫ln (3*x*)*dx* =

(A) + *C*

(B) + *C*

(C) 3*x* ln (3*x*) + *C*

(D) *x* ln (3*x*) + *x* + *C*

(E) *x* ln (3*x*) – *x* + *C*

32. ∫*x*^{2} ln *x dx*

(A) *x*^{3} ln *x* – *x*^{3} + *C*

(B) *x*^{3} ln *x* – *x*^{3} + *C*

(C) *x*^{3} (–1 + ln *x*) + *C*

(D) ln *x* (*x*^{3} – 1) + *C*

(E) 3*x*^{3} ln *x* – *x*^{3} + *C*

33. ∫*xe ^{−x} dx* =

(A) *e ^{–x}* (

*x*+ 1) +

*C*

(B) *e ^{x}* (

*x*+ 1) +

*C*

(C) –*e ^{x}* (

*x*– 1) +

*C*

(D) *e ^{–x}* (

*x*– 1) +

*C*

(E) –*e ^{–x}* (

*x*+ 1) +

*C*

34. ∫(ln *x*)^{2} dx =

(A) *x*^{2}(ln *x*)^{2} – 2*x* ln *x* + 2*x* + *C*

(B) *x* (ln *x*)^{2} + 2*x* ln *x* + 2*x* + *C*

(C) *x* (ln *x*)^{2} – 2*x* ln *x* + 2*x* + *C*

(D) *x* (ln *x*)^{2} – 2*x* ln *x* – 2*x* + *C*

(E) –*x* (ln *x*)^{3} – 2*x* ln *x* + 2*x* + *C*

35. ∫ln *x dx* =

(A) *x* ln *x* + *x* + *C*

(B) *x* – *x* ln *x* + *C*

(C) + *C*

(D) *x* ln *x* – *x* + *C*

(E) *x* – ln *x* + *C*

36. ∫*x*^{5} ln *x dx* =

(A) *x*^{6} ln *x* + *x*^{6} + *C*

(B) *x*^{6} ln *x* – *x*^{6} + *C*

(C) *x*^{6} – *x*^{6} ln *x* + *C*

(D) 5*x*^{4} ln *x* + *x*^{4} + *C*

(E) *x*^{6} ln *x* – *x*^{6} + *C*

37.

(A) .

(B) .

(C) .

(D) .

(E) .

# Chapter 22

# Exponential and Logarithmic Integration Drill Answers and Explanations

## ANSWER KEY

1. A

2. B

3. A

4. B

5. B

6. C

7. A

8. D

9. A

10. E

11. D

12. C

13. D

14. E

15. A

16. C

17. D

18. E

19. D

20. C

21. A

22. D

23. A

24. E

25. D

26. A

27. D

28. B

29. E

30. C

31. E

32. B

33. E

34. C

35. D

36. B

37. A

## EXPLANATIONS

1. **A** Evaluate the integral using the substitution *u* = ln *x*, then *du* = .

= ∫*u*^{2} *du* =

*u*^{3} + *C* = (ln *x*)^{3} + *C*

2. **B** Evaluate the integral using the substitution *u* = ln *x*, then *du* = . The new endpoints of the integral would be from 1 to 0.

3. **A** Evaluate the integral using the substitution *u* = *e ^{z}* +

*z*, then

*du*=

*e*+ 1

^{z}*dz.*

− ln *u* = ln(*e ^{z}* +

*z*)

Calculate the function at the endpoints.

ln (*e*^{1} + 1) – ln (*e*^{0} + 0) = ln (*e* + 1).

4. **B** Evaluate the integral using the substitution *u* = 5*x*, then *du* = *dx*.

∫*e*^{5}* ^{x} dx* = ∫

*e*=

^{u}du*e ^{u}* +

*C*=

*e*

^{5}

*+*

^{x}*C*

5. **B** Evaluate the integral making the substitution *u* = –3*x*, then –*du = dx*.

∫*e*^{−3}* ^{x} dx* = −∫

*e*=

^{u}du−*e ^{u}* +

*C*=

−*e*^{−3}* ^{x}* +

*C*

6. **C** Evaluate the integral making the substitution *u* = 1 + *e ^{x}*, making

*du*=

*e*

^{x}dx.

7. **A** Expand out the integrand by squaring the terms in parentheses.

(*e ^{x}* +

*e*)

^{–x}^{2}=

*e*

^{2}

*+ 2 +*

^{x}*e*

^{–2}

^{x}Integrate each piece of the integrand separately to find the antiderivative.

∫(*e ^{x}* +

*e*)

^{−x}^{2}

*dx*= ∫

*e*

^{2}

*+ 2∫*

_{x}dx*dx*+ ∫

*e*

^{−2}

*=*

^{x}dx*e*^{2}* ^{x}* + 2

*x*−

*e*

^{−2}

*+*

^{x}*C*

8. **D** Evaluate the integral making the substitution *u* = 4 + *e ^{x}*, then

*du*=

*e*

^{x}dx.∫*e ^{x}* (4 +

*e*)

^{x}^{4}

*dx*= ∫

*u*

^{4}

*du*=

*u*^{5} + *C* =

(4 + *e ^{x}*)

^{5}+

*C*

9. **A** Evaluate the integral using the substitution *u* = *e ^{x}*, then

*du*=

*e*

^{x}dx.∫*e ^{x}* sin (

*e*)

^{x}*dx*= ∫sin

*u dx*=

−cos *u* + *C* =

−cos (*e ^{x}*) +

*C*

10. **E** Expand out the integrand by squaring the numerator.

(1 + *e ^{x}*)

^{2}= 1 + 2

*e*+

^{x}*e*

^{2}

^{x}Divide the denominator into the numerator.

= *e ^{−x}* + 2 +

*e*

^{x}Perform the integration on each part of the reduced integrand.

∫*e ^{−x} dx* + 2∫

*dx*+ ∫

*e*=

^{x}dx−*e ^{−x}* + 2

*x*+

*e*+

^{x}*C*

11. **D** Simplify the integrand by dividing the denominator into the numerator.

= 4*x*^{−3} + *x*^{−1}

Perform the integration on each part of the reduced integrand.

12. **C** Evaluate the integral and plug in the endpoints immediately.

3[ln 4 − ln 2]

Use the subtraction and exponent properties of natural logarithms to simplify the answer.

3[ln 4 − ln 2] = 3[ln 2] = ln 8

13. **D** Evaluate the integral by using the substitution *u* = 8 – 3*x*, then –*du* = *dx*. The new limits of integration then become, when *x* is 1 and 2 respectively, 5 and 2.

−[ln 2 − ln 5]

Use the subtraction and exponent properties of natural logarithms to simplify the answer.

14. **E** Evaluate the integral making the substitution *u* = –*x*^{2}, the –*du* = *x dx*. The new limits of integration, when *x* = 0 and 1 respectively, become 0 and –1.

−[*e*^{−1} − *e*^{0}] =

−[*e*^{−1} − 1] = [1 − e^{−1}]

15. **A** Evaluate the integral making the substitution *u* = , making –*du* = .

= −∫*e ^{u} du* =

−*e ^{u}* +

*C*=

−*e*^{} + *C*

16. **C** Evaluate the integral making the substitution *u* = , making 2 *du* = .

= 2∫*e ^{u} du* =

2*e ^{u}* +

*C*= 2

*e*

^{}+

*C*

17. **D** Recall that ∫*A ^{x} dx* = , then plug in the limits immediately.

18. **E** Evaluate the integral making the substitution *u* = *e ^{x}* + 1, making

*du*=

*e*.

^{x}

ln *u* + *C* =

ln (*e ^{x}* + 1) +

*C*

19. **D** Evaluate the integral making the substitution *u* = *x* – 6. This makes *du* = *dx*. To tackle the *x* in the numerator, you must add 6 to *u* in the substitution, so *x* = *u* + 6.

*u* + 6 ln *u* =

*x* − 6 + 6 ln |*x* − 6| + *C*

The final simplification is to have the general constant *C* absorb the –6 to look like the answer choice.

*x* − 6 + 6 ln |*x* − 6| + *C* = *x* + 6 ln |*x* − 6| + *C*

20. **C** Evaluate the integral by making the substitution *u* = *e ^{x}*, which makes

*du*=

*e*.

^{x}dx

tan^{−1}*x* + *C* =

tan^{−1}(*e ^{x}*)

*C*

21. **A** Since *u*-substitution will not work, we need to use integration by parts. Set *u* = cos *x* and *dv* = *e*^{2}* ^{x} dx*, then

*du*= −sin

*x dx*and

*v*=

*e*

^{2}

*. The integral then becomes: ∫*

^{x}*e*

^{2}

*cos*

^{x}*x dx*=

*e*

^{2}

*cos*

^{x}*x*+ ∫

*e*

^{2}

*sin*

^{x}*x dx*. We need to use integration by parts again to evaluate the resulting integral. Set

*u*= sin

*x*and

*dv*=

*e*

^{2}

*, then*

^{x}dx*du*= cos

*x dx*and

*v*=

*e*

^{2}

*. So the full solution is: ∫*

^{x}*e*

^{2}

*cos*

^{x}*x dx*=

*e*

^{2}

*cos*

^{x}*x*+ (

*e*

^{2}

*sin*

^{x}*x*− ∫

*e*

^{2}

*cos*

^{x}*x dx*). With some algebra, the final solution is: ∫

*e*

^{2}

*cos*

^{x}*x dx*=

*e*

^{2}

*cos*

^{x}*x*+

*e*

^{2}

*sin*

^{x}*x*+

*C*.

22. **D** Since *u*-substitution will not work, we need to use integration by parts. Set *u* = ln^{2} *x* and *dv* = *x*^{2} *dx*, then *du* = and *v* = . The integral then becomes: ∫*x*^{2}ln^{2}*x dx* = ln^{2}*x* + ∫*x*^{2} ln *x dx*. We need to use integration by parts again to evaluate the resulting integral. Set *u* = ln *x* and *dv* = *x*^{2}*dx*, then *du* = and *v* = . So the full solution is: ∫*x*^{2}ln^{2}*x dx* = . With some calculus and algebra, the final solution is: ∫*x*^{2}ln^{2}*x dx* = (9ln^{2}*x* + 6ln *x* + 2) + *C*.

23. **A** Recall, ∫*a ^{u} du* =

*a*+

^{u}*C*. In this problem,

*u*=

*x*

^{2}+ 7 and

*du*= 2

*x dx*. Thus, ∫

*x*2

^{x2+7}

*dx*= .

24. **E** ∫*a ^{u} du* = +

*C*,

*u*= 6

*x*, and

*du*=

*dx*. ∫3(4

^{6}

*)*

^{x}*dx*= .

25. **D** To solve this integral, we must use the method of partial fractions. Set up the fraction as an equation with two parts: . Solve for A and B: *A* = and *B* = −. Plug in those values as coefficients into the re-written integral and solve: .

26. **A** You must evaluate this integral by parts using the following substitutions:

*u = x* *dv* = *e*^{2}^{x} dx

Then,

*du = dx* *v* = *e*^{2}^{x}

Recall the Integration by Parts formula is *uv* − ∫*v du*, so,

∫*xe*^{2}* ^{x} dx* =

*xe*

^{2}

*− ∫*

^{x}*e*

^{2}

*=*

^{x}dx*xe*

^{2}

*−*

^{x}*e*

^{2}

*+*

^{x}*C*

27. **D** You must evaluate this integral by parts using the following substitutions:

*u = x*^{2} *dv = e*^{2}^{x} dx

Then,

*du* = 2*x dx* *v* = *e*^{2}^{x}

Recall the Integration by Parts formula is *uv* − ∫*v du*, so,

∫*x*^{2}*e*^{2}* ^{x} dx* =

*x*

^{2}

*e*

^{2}

*− ∫*

^{x}*xe*

^{2}

^{x}dxYou must use Integration by parts once again to evaluate the integral.

*u = x* *dv* = *e*^{2}^{x} dx

Then,

*du = dx* *v* = *e*^{2}^{x}

∫*x*^{2}*e*^{2}* ^{x} dx* =

*x*

^{2}

*e*

^{2}

*− ∫*

^{x}*xe*

^{2}

*=*

^{x}dx*x*^{2}*e*^{2}* ^{x}* − [

*xe*

^{2}

*− ∫*

^{x}*e*

^{2}

*] =*

^{x}dx*x*^{2}*e*^{2}* ^{x}* −

*xe*

^{2}

*+*

^{x}*e*

^{2}

*+*

^{x}*C*

28. **B** You must evaluate this integral by parts using the following substitutions:

*u = x* *dv = e*^{–2}^{x} dx

Then,

*du = dx* *v* = −*e*^{−2}^{x}

Recall the Integration by Parts formula is *uv* − ∫*v du*, so,

∫*xe*^{−2}* ^{x} dx* = −

*xe*

^{−2}

*+ ∫*

^{x}*e*

^{−2}

*=*

^{x}dx−*xe*^{−2}* ^{x}* −

*e*

^{−2}

*+*

^{x}*C*

29. **E** You must evaluate the integral using Integration by Parts. Make the following substitutions:

*u = x*^{3} *dv = e ^{x} dx*

Then,

*du* = 3*x*^{2} *dx* *v = e ^{x}*

Recall the Integration by Parts formula is *uv* − ∫*v du*, so,

∫*x*^{3}*e ^{x} dx* =

*x*

^{3}

*e*− 3∫

^{x}*x*

^{2}

*e*

^{x}dxIntegration by parts is required again for this new integral.

*u = x*^{2} *dv = e ^{x} dx*

*du* = 2*x dx* *v = e ^{x}*

Set up the formula again, recalling the first step of integration.

∫*x* *e ^{x} dx* =

*x*

^{3}

*e*− 3[

^{x}*x*

^{2}

*e*− 2∫

^{x}*xe*]

^{x}dxIntegration by parts is required once again for the new integral.

*u = x* *dv = e ^{x} dx*

*du = dx* *v = e ^{x}*

Set up the formula once again, recalling the previous steps.

∫*x*^{3}*e ^{x} dx* =

*x*

^{3}

*e*− 3[

^{x}*x*

^{2}

*e*− 2[

^{x}*xe*− ∫

^{x}*e*]] =

^{x}dx*x*^{3}*e ^{x}* − 3

*x*

^{2}

*e*+ 6

^{x}*xe*− 6

^{x}*e*+

^{x}*C*

30. **C** Evaluate the integral using Integration by Parts. Make the following substitutions:

*u* = ln *x* *dv = x*^{3} *dx*

Then,

*du* = *dx* *v* = *x*^{4}

Recall the Integration by Parts formula is *uv* − ∫*v du*. So,

∫*x*^{3} ln *x dx* = *x*^{4} ln *x* − ∫*x*^{3} *dx*

= *x*^{4} ln *x* − *x*^{4} + *C*

31. **E** You must evaluate this using the Integration by Parts format. Make the following substitutions:

*u* = ln (3*x*) *dv = dx*

Then,

*du* = *dx* *v = x*

Recall the formula for Integration by Parts is *uv* − ∫*v du*. So,

∫ln (3*x*) *dx* = *x* ln (3*x*) − ∫*dx*

= *x* ln (3*x*) − *x* + *C*

32. **B** Evaluate the integral by using the Integration by Parts method.

*u* = ln *x* *dv* = *x*^{2} *dx*

*du* = *v* = *x*^{3}

Recall, the proper integration by parts format is *uv* − ∫*v du*. So,

∫*x*^{2} ln *x dx* = *x*^{3} ln *x* − ∫*x*^{2} *dx*

= *x*^{3} ln *x* − *x*^{3} + *C*

33. **E** Evaluate the integral by using the Integration by Parts method.

*u = x* *dv* = *e ^{–x} dx*

*du = dx* *v = –e ^{–x}*

Recall, the proper integration by parts format is *uv* − ∫*v du*. So,

∫*xe ^{−x} dx* = −

*xe*+ ∫

^{−x}*e*

^{−x}dx = −*xe ^{−x}* +

*e*+

^{−x}*C*

= −*e ^{−x}*(

*x*+ 1) +

*C*

34. **C** Evaluate the integral by using the Integration by Parts method.

*u* = (ln *x*)^{2} *dv = dx*

*du* = 2(ln *x*)*dx* *v = x*

Recall, the proper integration by parts format is *uv* − ∫*v du*. So,

∫(ln *x*)^{2} *dx* = *x*(ln *x*)^{2} − 2∫ln *x dx*

Integration by parts is required again to complete the solution.

*u* = ln *x* *dv = dx*

*du* = *v = x*

∫(ln *x*)^{2} *dx* = *x*(ln *x*)^{2} − 2∫ln *x dx*

= *x*(ln *x*)^{2} − 2*x* ln *x* − ∫*dx*

= *x*(ln *x*)^{2} − 2*x* ln *x* + 2*x* + *C*

35. **D** Evaluate the integral by using the Integration by Parts method.

*u* = ln *x* *dv = dx*

*du* = *v = x*

Recall, the proper integration by parts format is *uv* − ∫*v du*. So,

∫ln *x dx* = *x* ln *x* − ∫*dx*

= *x* ln *x* − *x* + *C*

36. **B** Evaluate the integral by using the Integration by Parts method.

*u* = ln *x* *dv = x*^{5} *dx*

*du* = *v* = *x*^{6}

Recall, the proper integration by parts format is *uv* − ∫*v du*. So,

∫*x*^{5} ln *x dx* = *x*^{6} ln *x* − ∫*x*^{5} *dx*

= *x*^{6} ln *x* − *x*^{6} + *C*

37. **A** First, divide *x* into each term in the numerator.

Evaluate the integral with the power rule, then plug in the endpoints.

# Chapter 23

# Areas, Volumes, and Average Values Drill

## AREAS, VOLUMES, AND AVERAGE VALUES DRILL

1. What is the area between the curve *y* = *x*^{3} − 8 and the *x*-axis from *x* = 0 to *x* = 2.

(A) 0

(B) 4

(C) 8

(D) 12

(E) 16

2. What is the area enclosed by the curve *x* = *y*^{2} − *y* − 2 and the *y*-axis?

(A) 3

(B) 3.5

(C) 4

(D) 4.5

(E) 5

3. What is the volume of the solid formed by revolving the curve *y* = 2*x*^{2} − 8 about the *x*-axis?

(A) 21.333

(B) 67.021

(C) 136.533

(D) 221.867

(E) 428.932

4. What is the volume of the solid formed by the curves *y* = *x*^{2} and *y* = 3*x* revolved around the line *y* = 7?

(A)

(B)

(C)

(D)

(E)

5. Find the area of the region in the plane enclosed by the cardioid *r* = 2 + 2 sin *θ*

(A) *π*

(B) 2*π*

(C) 3*π*

(D) 6*π*

(E) 12*π*

6. Find the volume of the solid formed by revolving *y* = *x*^{2} from *x* = 1 to *x* = 3 over the *x*-axis.

(A) 20*π*

(B)

(C)

(D)

(E) 50*π*

7. Find the area between the curves *y* = *x*^{4} and *y* = *x*^{2} from *x* = 0 to *x* = 1.

(A)

(B)

(C)

(D)

(E) 1

8. Approximate the area under the curve *f*(*x*) = *x*^{3} + 4 from *x* = 0 to *x* = 2 using four inscribed trapezoids.

(A)

(B) 10

(C)

(D) 12

(E)

9. Given the following table of values for *x* and *y:*

Use a left-hand Riemann sum with eight subintervals to approximate *f*(*x*) *dx*.

(A) 110

(B) 121

(C) 123

(D) 126

(E) 137

10. What is the area between the curves *y* = 6*x*^{2} − *x* and *y* = *x*^{2} − 6*x*?

(A) 4

(B)

(C)