By Dr. Steve Warner
Legal Notice This book is copyright 2016 with all rights reserved. It is illegal to copy, distribute, or create derivative works from this book in whole or in part or to contribute to the copying, distribution, or creating of derivative works of this book.
320 AP Calculus BC Problems arranged by Topic and Difficulty Level 320 Level 1, 2, 3, 4, and 5 AP Calculus Problems Dr. Steve Warner
ยฉ 2016, All Rights Reserved Second Edition
iii
BOOKS BY DR. STEVE WARNER FOR COLLEGE BOUND STUDENTS 28 New SAT Math Lessons to Improve Your Score in One Month Beginner Course Intermediate Course Advanced Course New SAT Math Problems arranged by Topic and Difficulty Level New SAT Verbal Prep Book for Reading and Writing Mastery 320 SAT Math Subject Test Problems Level 1 Test Level 2 Test The 32 Most Effective SAT Math Strategies SAT Prep Official Study Guide Math Companion Vocabulary Builder 320 ACT Math Problems arranged by Topic and Difficulty Level 320 GRE Math Problems arranged by Topic and Difficulty Level 320 SAT Math Problems arranged by Topic and Difficulty Level 320 AP Calculus AB Problems 320 AP Calculus BC Problems SHSAT Verbal Prep Book to Improve Your Score in Two Months 555 Math IQ Questions for Middle School Students 555 Advanced Math Problems for Middle School Students 555 Geometry Problems for High School Students Algebra Handbook for Gifted Middle School Students
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iv
Table of Contents Introduction: The Proper Way to Prepare 1. Using this book effectively 2. Overview of the AP Calculus exam 3. Structure of this book 4. Practice in small amounts over a long period of time 5. Redo the problems you get wrong over and over and over until you get them right 6. Check your answers properly 7. Take a guess whenever you cannot solve a problem 8. Pace yourself
7 7 8 8 9 10 10 10 11
Problems by Level and Topic with Fully Explained Solutions Level 1: Differentiation Level 1: Integration Level 1: Limits and Continuity Level 1: Series Level 2: Differentiation Level 2: Integration Level 2: Limits and Continuity Level 2: Series Level 3: Differentiation Level 3: Integration Level 3: Series Level 4: Differentiation Level 4: Integration Level 4: Series Level 5: Free Response Questions
12 12 22 36 44 53 63 75 79 87 102 119 128 141 155 161
Supplemental Problems โ Questions Level 1: Differentiation Level 1: Integration Level 1: Limits and Continuity Level 1: Series Level 2: Differentiation Level 2: Integration Level 2: Limits and Continuity
180 180 180 181 182 184 185 186
v
Level 2: Series Level 3: Differentiation Level 3: Integration Level 3: Series Level 4: Differentiation Level 4: Integration Level 4: Series Level 5: Free Response Questions
187 189 191 192 194 196 198 199
Supplemental Problems - Answers
203
About the Author
212
Books by Dr. Steve Warner
213
vi
PROBLEMS BY LEVEL AND TOPIC WITH FULLY EXPLAINED SOLUTIONS
LEVEL 1: DIFFERENTIATION 1.
If ๐(๐ฅ) = 7๐ฅ 4 + ๐ฅ + 3๐ โ sec ๐ฅ, then ๐ โฒ (๐ฅ) = (A) 28๐ฅ 3 + 1 โ sec ๐ฅ tan ๐ฅ (B) 28๐ฅ 3 + 1 + sec ๐ฅ tan ๐ฅ (C) 28๐ฅ 3 + 3 โ sec ๐ฅ tan ๐ฅ 7
๐ฅ2
5
2
(D) ๐ฅ 5 +
+ 3๐๐ฅ โ ln|sec ๐ฅ + tan ๐ฅ|
Solution: ๐ โฒ (๐ฅ) = 28๐ฅ 3 + 1 โ sec ๐ฅ tan ๐ฅ. This is choice (A). Notes: (1) If ๐ is any real number, then the derivative of ๐ฅ ๐ is ๐๐ฅ ๐โ1 . Symbolically, For example,
๐
๐
๐ ๐ ๐๐ฅ
[๐๐ ] = ๐๐๐โ๐ . [๐ฅ 4 ] = 4๐ฅ 3 .
As another example,
๐ ๐๐ฅ
[๐ฅ ] =
๐ ๐๐ฅ
[๐ฅ 1 ] = 1๐ฅ 0 = 1(1) = 1.
(2) Of course it is worth just remembering that
๐ ๐๐ฅ
[๐ฅ ] = 1.
(3) The derivative of a constant is 0. A constant is just a real number. For example, 3๐ is a constant. So
๐ ๐๐ฅ
[3๐] = 0.
(4) The derivative of a constant times a function is the constant times the derivative of the function. Symbolically, For example,
๐ ๐๐ฅ ๐ ๐๐ฅ
[๐๐(๐ฅ)] = ๐ [7๐ฅ 4 ] = 7
๐ ๐๐ฅ
๐ ๐๐ฅ
[๐(๐ฅ)].
[๐ฅ 4 ] = 7 โ
4๐ฅ 3 = 28๐ฅ 3 .
12
(5) You should know the derivatives of the six basic trig functions: ๐
๐
๐ ๐
๐
๐ ๐
๐
๐
[๐ฌ๐ข๐ง ๐] =
๐
๐๐จ๐ฌ ๐
๐
๐ ๐
[๐๐จ๐ฌ ๐] = โ ๐ฌ๐ข๐ง ๐ [๐ญ๐๐ง ๐] =
๐
๐ ๐
๐ฌ๐๐ ๐ ๐
๐
๐
[๐๐ฌ๐ ๐] = โ ๐๐ฌ๐ ๐ ๐๐จ๐ญ ๐ [๐ฌ๐๐ ๐] =
๐ฌ๐๐ ๐ ๐ญ๐๐ง ๐
[๐๐จ๐ญ ๐] = โ ๐๐ฌ๐ ๐ ๐
(6) If ๐ and โ are functions, then (๐ + โ)โฒ (๐ฅ) = ๐โฒ (๐ฅ) + โโฒ(๐ฅ). In other words, when differentiating a sum, we can simply differentiate term by term. Similarly, (๐ โ โ)โฒ (๐ฅ) = ๐โฒ (๐ฅ) โ โโฒ(๐ฅ). (7) In the given problem we differentiate each of ๐ฅ 4 , ๐ฅ, 3๐ and sec ๐ฅ separately and then use notes (4) and (6) to write the final answer. 2.
If ๐(๐ฅ) =
๐ 4๐ฅโ4 4
5
โ ln(๐ฅ 2 ) + (2๐ฅ โ 1)2 , then ๐โฒ (1) =
(A) 1 (B) 2 (C) 3 (D) 4 3
2
Solution: ๐โฒ (๐ฅ) = ๐ 4๐ฅโ4 โ + 5(2๐ฅ โ 1)2 . ๐ฅ
Therefore ๐โฒ (1) = 1 โ 2 + 5 = 4, choice (D). Notes: (1) The derivative of ๐ ๐ฅ is ๐ ๐ฅ . Symbolically,
๐ ๐๐ฅ
[๐ ๐ฅ ] = ๐ ๐ฅ . 1
(2) The derivative of ln ๐ฅ is . ๐ฅ
Symbolically,
๐ ๐๐ฅ
1
[ln ๐ฅ] = . ๐ฅ
(3) In this problem we need the chain rule which says the following: If f (๐ฅ) = (๐ โ โ)(๐ฅ) = ๐(โ(๐ฅ)) , then ๐ โฒ (๐ฅ) = ๐โฒ (โ(๐ฅ)) โ
โโฒ(๐ฅ)
13
For example, if ๐(๐ฅ) = ln(๐ฅ 2 ), then ๐(๐ฅ) = ๐(โ(๐ฅ)) where ๐(๐ฅ) = ln ๐ฅ 1 2 and โ(๐ฅ) = ๐ฅ 2 . So ๐ โฒ (๐ฅ) = ๐โฒ (โ(๐ฅ)) โ
โโฒ (๐ฅ) = 2 โ
2๐ฅ = . ๐ฅ
Similarly, we have ๐ ๐๐ฅ
5 2
๐ ๐๐ฅ
[
๐ 4๐ฅโ4 4
1
๐
4
๐๐ฅ
]= โ
1
[๐ 4๐ฅโ4 ] = ๐ 4๐ฅโ4 โ
4 = ๐ 4๐ฅโ4 , and 4
3 2
5
๐ฅ
3 2
[(2๐ฅ โ 1) ] = (2๐ฅ โ 1) (2) = 5(2๐ฅ โ 1) . 2
(4) As an alternative to using the chain rule to differentiate ln(๐ฅ 2 ), we ๐ ๐ 1 2 can rewrite ln(๐ฅ 2 ) as 2 ln ๐ฅ. Then [2 ln ๐ฅ] = 2 [ln ๐ฅ] = 2 โ
= . ๐๐ฅ ๐๐ฅ ๐ฅ ๐ฅ See the first table in problem 3 for the rule of logarithms used here. (5) In the given problem we differentiate each of 5
๐ 4๐ฅโ4 4
, ln(๐ฅ 2 ), and
(2๐ฅ โ 1)2 separately and then use note (6) from problem 1 to write the final answer. (6) If we could use a calculator for this problem, we can compute ๐โฒ(๐ฅ) at ๐ฅ = 1 using our TI-84 calculator by first selecting nDeriv( (or pressing 8) under the MATH menu, then typing the following: e^(4X โ 4)/4 โ ln(X^2) + (2X โ 1)^(5/2), X, 1), and pressing ENTER. The display will show approximately 4. ๐
3.
๐๐ฅ
5
๐ฅ ln ๐ ๐ฅ
[
6
]=
(A) 6๐ฅ 5 (B) ๐ฅ 5 (C) 6๐ฅ 5 + ๐ฅ 6 (D) ๐ฅ 5 + ๐ฅ 6 5
5
Solution: ln ๐ ๐ฅ = ๐ฅ 5 , so that ๐ ๐๐ฅ
5
[
๐ฅ ln ๐ ๐ฅ 6
]=
๐
1
[ ๐ฅ
๐๐ฅ 6
6]
1
๐ฅ ln ๐ ๐ฅ
=
6 5
๐ฅโ
๐ฅ 5 6
1
= ๐ฅ 6 . Therefore we have 6
5
= โ
6๐ฅ = ๐ฅ , choice (B). 6
Notes: (1) ๐(๐ฅ) = log ๐ ๐ฅ is called the natural logarithmic function and is usually abbreviated as ๐(๐ฅ) = ln ๐ฅ. 5
(2) Here are two ways to simplify ln ๐ ๐ฅ .
14
5
Method 1: Recall that ln ๐ = 1. We have ln ๐ ๐ฅ = ๐ฅ 5 ln ๐ = ๐ฅ 5 (1) = ๐ฅ 5 . Here we have used the last law in the following table: Laws of Logarithms: Here is a review of the basic laws of logarithms. Law logb1 = 0 logbb = 1 logbx + logby = logb(xy) ๐ logbx โ logby = logb( ) ๐
logbxn = nlogbx
Example log21 = 0 log66 = 1 log57 + log52 = log514 log321 โ log37 = log33 = 1 log8 35 = 5log83
Method 2: Recall that the functions ๐ ๐ฅ and ln ๐ฅ are inverses of each other. This means that ๐ ln ๐ฅ = ๐ฅ and ln ๐ ๐ฅ = ๐ฅ. Replacing ๐ฅ by ๐ฅ 5 in the 5 second equation gives ln ๐ ๐ฅ = ๐ฅ 5 . (3) Geometrically inverse functions have graphs that are mirror images across the line ๐ฆ = ๐ฅ. Here is a picture of the graphs of ๐ฆ = ๐ ๐ฅ and ๐ฆ = ln ๐ฅ together with the line ๐ฆ = ๐ฅ. Notice how the line ๐ฆ = ๐ฅ acts as a mirror for the two functions.
(4) ๐ฅ โ
๐ฅ 5 = ๐ฅ 1 โ
๐ฅ 5 = ๐ฅ 1+5 = ๐ฅ 6 .
15
Here is a complete review of the laws of exponents: Law x0 = 1 x1 = x xaxb = xa+b xa/xb = xa-b (xa)b = xab (xy)a = xaya (x/y)a = xa/ya x-1 = 1/x x-a = 1/xa ๐ x1/n = โ๐ ๐ ๐ ๐ xm/n = โ๐๐ =( โ๐) 4.
Example 30 = 1 91 = 9 x3x5 = x8 x11/x4 = x7 (x5)3 = x15 (xy)4 = x4y4 (x/y)6 = x6/y6 3-1 = 1/3 9-2 = 1/81 3 x1/3 = โ๐ฅ 9
x9/2 =โ๐ฅ 9 =(โ๐ฅ)
The slope of the tangent line to the graph of ๐ฆ = ๐ฅ๐ 2๐ฅ at ๐ฅ = ln 3 is (A) 9 (B) 18 (C) 18 ln 3 (D) 18 ln 3 + 9
Solution: ๐ฆ โฒ = ๐ฅ๐ 2๐ฅ โ
2 + ๐ 2๐ฅ โ
1 = 2๐ฅ๐ 2๐ฅ + ๐ 2๐ฅ = ๐ 2๐ฅ (2๐ฅ + 1). When ๐ฅ = ln 3, we have that the slope of the tangent line is 2
๐ฆ โฒ |๐ฅ=ln 3 = ๐ 2 ln 3 (2 ln 3 + 1) = ๐ ln 3 (2 ln 3 + 1) = 32 (2 ln 3 + 1) = 9(2 ln 3 + 1) = 18 ln 3 + 9. This is choice (D). Notes: (1) To find the slope of a tangent line to the graph of a function, we simply take the derivative of that function. If we want the slope of the tangent line at a specified ๐ฅ-value, we substitute that ๐ฅ-value into the derivative of the function. (2) The derivative of ๐(๐ฅ) = ๐ ๐ฅ is ๐ โฒ (๐ฅ) = ๐ ๐ฅ (3) In this problem we used the product rule which says the following: If f (๐ฅ) = ๐ข(๐ฅ)๐ฃ(๐ฅ) , then ๐ โฒ (๐ฅ) = ๐ข(๐ฅ)๐ฃ โฒ (๐ฅ) + ๐ฃ(๐ฅ)๐ขโฒ (๐ฅ)
16
(4) When differentiating ๐ 2๐ฅ we needed to use the chain rule. See problem 2 for details. (5) See problem 3 for information on logarithms. (6) The functions ๐ ๐ฅ and ln ๐ฅ are inverses of each other. This means that 2 ๐ ln ๐ฅ = ๐ฅ and ln ๐ ๐ฅ = ๐ฅ. In particular, ๐ ln 3 = 32 . (7) ๐ ln ๐ฅ = ln ๐ฅ ๐ . In particular, 2 ln 3 = ln 32 . See the first table in problem 3 for the rule of logarithms used here. 2
(8) Using notes (6) and (7) together we get ๐ 2 ln 3 = ๐ ln 3 = ๐ ln 9 = 9. (9) As an alternative to using the rule of logarithms as was done in note 2
(8), we can use a law of exponents instead to write ๐ 2 ln 3 = (๐ ln 3 ) . 2
Since ๐ ln 3 = 3, we have ๐ 2 ln 3 = (๐ ln 3 ) = 32 = 9. The rule that we used here is (๐ฅ ๐ )๐ = ๐ฅ ๐๐ with ๐ = ln 3 and ๐ = 2. (9) If we could use a calculator for this problem, we can compute ๐ฆโฒ at ๐ฅ = ln 2 using our TI-84 calculator by first selecting nDeriv( (or pressing 8) under the MATH menu, then typing the following: Xe^(2X), X, ln 3), and pressing ENTER. The display will show approximately 28.775. When we put choice (D) in our calculator we also get approximately 28.775. 5.
If ๐ฅ = ln(๐ก 2 + 1) and ๐ฆ = cos 3๐ก, then (A) โ (B) โ (C) โ (D) โ
Solution: ๐๐ฆ ๐๐ฅ
=
๐๐ฆ ๐๐ก ๐๐ฅ ๐๐ก
๐๐ฆ ๐๐ก
๐๐ฆ ๐๐ฅ
=
3 sin 3๐ก ๐ก 2 +1 3 sin 3๐ก 2๐ก(๐ก 2 +1) 3(๐ก 2 +1) sin 3๐ก 2๐ก 3 sin 3๐ก 2๐ก
= โ3 sin 3๐ก and
= (โ3 sin 3๐ก) รท
๐๐ฅ ๐๐ก
2๐ก ๐ก 2 +1
=
2๐ก
. Therefore
๐ก 2 +1
= (โ3 sin 3๐ก) โ
This is choice (C).
17
๐ก 2 +1 2๐ก
=โ
3(๐ก 2 +1) sin 3๐ก 2๐ก
.
Notes: (1) In this problem we are given a parametrically defined curve. The variable ๐ก is called the parameter, and the two given equations are called parametric equations. For example, when ๐ก = 0, we have that ๐ฅ = ln(02 + 1) = ln 1 = 0 and ๐ฆ = cos(3 โ
0) = 1. So the point (0,1) is on the given parametrically defined curve, and this point corresponds to the parameter value ๐ก = 0. Each value for ๐ก corresponds to a point (๐ฅ, ๐ฆ) in the ๐ฅ๐ฆ-plane. (2) The derivative
๐๐ฆ ๐๐ฅ
is equal to
๐๐ฆ ๐๐ก ๐๐ฅ ๐๐ก
.
1
(3) The derivative of ln ๐ฅ is . ๐ฅ
Symbolically,
๐ ๐๐ฅ
1
[ln ๐ฅ] = . ๐ฅ
๐๐ฆ
๐๐ฅ
(4) The derivatives and both required the chain rule. See problem 2 ๐๐ก ๐๐ก for a detailed explanation of this rule. 6.
If ๐(๐ฅ) = ๐๐ 3 + 3
1
โ๐ฅ 2
+(
๐ฅ+2 2 ๐ฅโ2
) โ 11๐ฅ , then ๐โฒ (๐ฅ) = 2
Solution: We first rewrite ๐ as ๐(๐ฅ) = ๐๐ 3 + ๐ฅ โ3 + (
๐ฅ+2
2
) โ 11๐ฅ .
๐ฅโ2
2 5 ๐ฅ + 2 (๐ฅ โ 2)(1) โ (๐ฅ + 2)(1) ) ๐โฒ (๐ฅ) = 0 โ ๐ฅ โ3 + 2 ( โ 11๐ฅ (ln 11) (๐ฅ โ 2)2 3 ๐ฅโ2 =โ
2 3
๐ฅ+2
3 โ๐ฅ 5
โ 8 (๐ฅโ2)3 โ (ln 11)11๐ฅ .
Notes: (1) ๐๐ 3 is a constant. Therefore (2) 3
1
โ๐ฅ 2 2
1
=
2 ๐ฅ3
2
= ๐ฅ โ3 . So
5
(3) โ ๐ฅ โ3 = โ 3
2 5 3๐ฅ 3
๐
[3
1
๐๐ฅ โ๐ฅ
=โ
2 3
3 โ๐ฅ 5
]= 2
๐ ๐๐ฅ
๐ ๐๐ฅ 2
๐(๐ฅ) ๐ท(๐ฅ)
2
2
2
5
[๐ฅ โ3 ] = โ ๐ฅ โ3โ1 = โ ๐ฅ โ3 . 3
.
(4) The quotient rule says the following: If f (๐ฅ) =
[๐๐ 3 ] = 0.
, then
18
3
๐ โฒ (๐ฅ) =
๐ท(๐ฅ)๐ โฒ (๐ฅ) โ ๐(๐ฅ)๐ทโฒ(๐ฅ) [๐ท(๐ฅ)]2
I like to use the letters ๐ for โnumeratorโ and D for โdenominator.โ (5) The derivative of ๐ฅ + 2 is 1 because the derivative of ๐ฅ is 1, and the derivative of any constant is 0. Similarly, the derivative of ๐ฅ โ 2 is also 1. Now using the quotient rule we see that the derivative of (๐ฅโ2)(1)โ(๐ฅ+2)(1) (๐ฅโ2)2
=
๐ฅโ2โ๐ฅโ2 (๐ฅโ2)2
(6) Differentiating (
In particular,
๐ ๐๐ฅ
๐ ๐๐ฅ
๐ฅโ2
is
.
๐ฅ+2 2 ๐ฅโ2
) requires the chain rule. Using note (5) we see
that this derivative is 2 ( (7) If ๐ > 0, then
=
โ4 (๐ฅโ2)2
๐ฅ+2
๐ฅ+2 ๐ฅโ2
โ4
๐ฅ+2
) ((๐ฅโ2)2 ) = โ8 (๐ฅโ2)3 .
[๐ ๐ฅ ] = ๐ ๐ฅ (ln ๐).
[11๐ฅ ] = 11๐ฅ (ln 11).
(8) For ๐ > 0, ๐ ๐ฅ = ๐ ๐ฅ ln ๐ . ๐ฅ
To see this, first observe that ๐ ๐ฅ ln ๐ = ๐ ln ๐ by the power rule for logarithms (see problem 3 for the laws of logarithms). Second, recall that the functions ๐ ๐ฅ and ln ๐ฅ are inverses of each other. This means that ๐ ln ๐ฅ = ๐ฅ and ln ๐ ๐ฅ = ๐ฅ. Replacing ๐ฅ by ๐ ๐ฅ in the first ๐ฅ formula yields ๐ ln ๐ = ๐ ๐ฅ . (9) The formula in note (8) gives an alternate method for differentiating 11๐ฅ . We can rewrite 11๐ฅ as ๐ ๐ฅ ln 11 and use the chain rule. Here are the details: ๐ ๐๐ฅ
[11๐ฅ ] =
๐ ๐๐ฅ
[๐ ๐ฅ ln 11 ] = ๐ ๐ฅ ln 11 (ln 11) = 11๐ฅ (ln 11).
Note that in the last step we rewrote ๐ ๐ฅ ln 11 as 11๐ฅ . (10) There is one more method we can use to differentiate 11๐ฅ . We can use logarithmic differentiation. We start by writing ๐ฆ = 11๐ฅ .
19
We then take the natural log of each side of this equation: ln ๐ฆ = ln 11๐ฅ . We now use the power rule for logarithms to bring the ๐ฅ out of the exponent: ln ๐ฆ = ๐ฅ ln 11. 1
๐๐ฆ
๐ฆ
๐๐ฅ
Now we differentiate implicitly to get โ
Solve for ๐๐ฆ ๐๐ฅ
๐๐ฆ ๐๐ฅ
= ln 11.
by multiplying each side of the last equation by ๐ฆ to get
= ๐ฆ ln 11.
Finally, replacing ๐ฆ by 11๐ฅ gives us
๐๐ฆ ๐๐ฅ
= 11๐ฅ (ln 11).
(11) Logarithmic differentiation is a general method that can often be used to handle expressions that have exponents with variables. (12) See problem 35 for more information on implicit differentiation. 7.
Differentiate ๐(๐ฅ) =
๐ cot 3๐ฅ โ๐ฅ
and express your answer as a simple
fraction. Solution: 1
๐
โฒ (๐ฅ)
=
Notes: (1) ๐ ๐๐ฅ ๐ ๐๐ฅ
โ๐ฅ(๐ cot 3๐ฅ )(โ csc2 3๐ฅ)(3)โ๐ cot 3๐ฅ (2โ๐ฅ) ๐ฅ ๐ ๐๐ฅ
=
โ6๐ฅ(csc2 3๐ฅ)๐ cot 3๐ฅ โ๐ cot 3๐ฅ 2๐ฅ โ๐ฅ
.
[๐ ๐ฅ ] = ๐ ๐ฅ
[cot ๐ฅ] = โ csc 2 ๐ฅ [3๐ฅ ] = 3
๐
๐
1
1
1
1
[โ๐ฅ] = [๐ฅ 2 ] = ๐ฅ โ2 = โ
๐๐ฅ ๐๐ฅ 2 2
1 1 ๐ฅ2
1
= โ
1
2 โ๐ฅ
=
1 2โ๐ฅ
(2) We start off using the quotient rule (see problem 6 for a detailed explanation of the quotient rule). Here we get ๐
๐
โ๐ฅโ
๐๐ฅ[๐ cot 3๐ฅ ]โ๐ cot 3๐ฅ โ
๐๐ฅ[โ๐ฅ] (โ๐ฅ)
2
๐
(3) [๐ cot 3๐ฅ ] requires two applications of the chain rule. See problem 2 ๐๐ฅ for a detailed explanation of the chain rule. Here we get
20
๐ ๐๐ฅ
[๐ cot 3๐ฅ ] = ๐ cot 3๐ฅ (โ csc 2 3๐ฅ)(3).
(4) After differentiating we wind up with a complex fraction: 1
โ๐ฅ(๐ cot 3๐ฅ )(โ csc2 3๐ฅ)(3)โ๐ cot 3๐ฅ (2โ๐ฅ) ๐ฅ
We simplify this complex fraction by multiplying the numerator and denominator by 2โ๐ฅ. Note the following: ๐ฅ(2โ๐ฅ) = 2๐ฅ โ๐ฅ (this is where the final denominator comes from). โ๐ฅ(๐ cot 3๐ฅ )(โ csc 2 3๐ฅ)(3)(2โ๐ฅ) = โ6โ๐ฅ โ๐ฅ (csc 2 3๐ฅ)๐ cot 3๐ฅ = โ6๐ฅ (csc 2 3๐ฅ)๐ cot 3๐ฅ
๐ cot 3๐ฅ (
1 2โ๐ฅ
) (2โ๐ฅ) = ๐ cot 3๐ฅ
The last two results give the final numerator. 8.
If ๐ญ is the vector-valued ln ๐ก ๐ญ(๐ก) = โฉ , cos 2 ๐กโช, then ๐ญโฒโฒ (๐ก) =
function
defined
by
, โ2 cos ๐ก sin ๐กโช,
and
๐ก
Solution: ๐ญ
โฒ (๐ก)
=โฉ
1 ๐ก
๐ก( )โ(ln ๐ก)(1)
so ๐ญโฒโฒ (๐ก) = โฉ =โฉ
, 2 (cos ๐ก)(โ sin ๐ก)โช = โฉ
๐ก2 โ1 2 ๐ก ( )โ(1โln ๐ก)(2๐ก) ๐ก
๐ก4 โ๐กโ2๐ก+2๐ก ln ๐ก ๐ก4
1โln ๐ก ๐ก2
, โ2 cos ๐ก cos ๐ก โ 2(sin ๐ก)(โ sin ๐ก)โช
, โ2 (cos 2 ๐ก โ sin2 ๐ก)โช = โฉ
๐ ๐ฅ๐ง ๐โ๐ ๐๐
, โ๐ ๐๐จ๐ฌ ๐๐โช.
Notes: (1) A 2-dimensional vector-valued function ๐ญ has the form ๐ญ(๐ก) = โฉ๐ฅ(๐ก), ๐ฆ(๐ก)โช where ๐ฅ and ๐ฆ are ordinary functions of the variable ๐ก. A vector-valued function is just a convenient way to give a parametrically defined curve with a single function. The vector-valued function given in the problem is equivalent to the parametric equations ๐ฅ=
ln ๐ก ๐ก
, ๐ฆ = cos 2 ๐ก
Can you express the parametric equations given in problem 5 as a vector-valued function?
21
(2) The derivative of the vector-valued function ๐ญ which is defined by ๐ญ(๐ก) = โฉ๐ฅ(๐ก), ๐ฆ(๐ก)โช is the vector-valued function ๐ญโฒ which is defined by ๐ญโฒ(๐ก) = โฉ๐ฅโฒ(๐ก), ๐ฆโฒ(๐ก)โช. In other words, we simply differentiate each component. In this problem we have ๐ฅ(๐ก) =
ln ๐ก ๐ก
and ๐ฆ(๐ก) = cos 2 ๐ก.
Note also that ๐ญโฒโฒ(๐ก) = โฉ๐ฅโฒโฒ(๐ก), ๐ฆโฒโฒ(๐ก)โช. (3) Recall from problem 5 that
๐ ๐๐ฅ
1
[ln ๐ฅ] = . ๐ฅ
(4) We used the quotient rule to differentiate ๐ฅ and ๐ฅโฒ. See problem 6 for a detailed explanation of the quotient rule. (5) cos 2 ๐ก is an abbreviation for (cos ๐ก)2 . To differentiate ๐ฆ therefore required the chain rule. (6)To differentiate ๐ฆโฒ we used the product rule. (7) The following two identities can be useful: cos 2๐ก = cos 2 ๐ก โ sin2 ๐ก
sin 2๐ก = 2 sin ๐ก cos ๐ก
The second identity was used when simplifying ๐ฆโฒโฒ(๐ก). We could have used the first identity to write ๐ฆ โฒ (๐ก) = โ2 cos ๐ก sin ๐ก = โ2 sin ๐ก cos ๐ก = โ sin 2๐ก. Differentiating this last expression then gives ๐ฆ โฒโฒ (๐ก) = โ2 cos 2๐ก.
LEVEL 1: INTEGRATION 9.
โซ(3๐ฅ 2 โ 6โ๐ฅ + ๐ ๐ฅ ) ๐๐ฅ = (A) 6๐ฅ โ
3 โ๐ฅ
+ ๐๐ฅ + ๐ถ
(B) ๐ฅ 3 โ 4โ๐ฅ 3 + ๐ ๐ฅ + ๐ถ (C) ๐ฅ 3 โ 3๐ฅ + ๐ ๐ฅ + ๐ถ (D) ๐ฅ 3 โ 3๐ฅ + ๐ฅ๐ ๐ฅโ1 + ๐ถ
22
Solution: 2
๐ฅ
โซ(3๐ฅ โ 6โ๐ฅ + ๐ ) ๐๐ฅ = 3 โ
๐ฅ3 3
3
โ
6๐ฅ 2 3 2
+ ๐ ๐ฅ + ๐ถ = ๐ฅ 3 โ 4โ๐ฅ 3 + ๐ ๐ฅ + ๐ถ
This is choice (B). Notes: (1) If ๐ is any real number, then an antiderivative of ๐ฅ ๐ is Symbolically, โซ ๐ฅ ๐ ๐๐ฅ = For example, โซ ๐ฅ 2 ๐๐ฅ =
๐ฅ ๐+1 ๐+1 ๐ฅ3 3
๐ฅ ๐+1 ๐+1
.
+ ๐ถ, where ๐ถ is an arbitrary constant.
+ ๐ถ.
As another example, 3
1
โซ โ๐ฅ๐๐ฅ = โซ ๐ฅ 2 ๐๐ฅ = (2) Since
๐ ๐๐ฅ
๐ฅ2 3 2
3
3
3
2
2
3
3
3
+ ๐ถ = ๐ฅ 2 รท + ๐ถ = ๐ฅ 2 โ
+ ๐ถ = ๐ฅ 2 + ๐ถ. 2
[๐ ๐ฅ ] = ๐ ๐ฅ , it follows that โซ ๐ ๐ฅ ๐๐ฅ = ๐ ๐ฅ + ๐ถ
(3) If ๐ and โ are functions, then โซ[๐(๐ฅ) + โ(๐ฅ)]๐๐ฅ = โซ ๐(๐ฅ)๐๐ฅ + โซ โ(๐ฅ)๐๐ฅ. In other words, when integrating a sum we can simply integrate term by term. Similarly, โซ[๐(๐ฅ) โ โ(๐ฅ)]๐๐ฅ = โซ ๐(๐ฅ)๐๐ฅ โ โซ โ(๐ฅ)๐๐ฅ. (4) If ๐ is a function and ๐ is a constant, then โซ ๐๐(๐ฅ)๐๐ฅ = ๐ โซ ๐(๐ฅ)๐๐ฅ ๐ฅ3
For example, โซ 3๐ฅ 2 ๐๐ฅ = 3 โซ ๐ฅ 2 ๐๐ฅ = 3 ( ) + ๐ถ = ๐ฅ 3 + ๐ถ. 3
(5) In the given problem we integrate each of ๐ฅ 2 , โ๐ฅ, and ๐ ๐ฅ separately and then use notes (3) and (4) to write the final answer. (6) We do not need to include a constant ๐ถ for each individual integration since if we add or subtract two or more constants we simply get a new constant. This is why we simply add one constant ๐ถ at the end of the integration.
23
(7) It is also possible to solve this problem by differentiating the answer choices. For example, if we start with choice (C), then we have that ๐ (๐ฅ 3 โ 3๐ฅ + ๐ ๐ฅ + ๐ถ) = 3๐ฅ 2 โ 3 + ๐ ๐ฅ . So we can immediately see that ๐๐ฅ choice (C) is incorrect. When we differentiate choice (B) however, we get 3 ๐ 3 ๐ 3 [๐ฅ โ 4โ๐ฅ 3 + ๐ ๐ฅ + ๐ถ] = [๐ฅ โ 4๐ฅ 2 + ๐ ๐ฅ + ๐ถ] ๐๐ฅ ๐๐ฅ 3
1
= 3๐ฅ 2 โ 4 ( ๐ฅ 2 ) + ๐ ๐ฅ + 0 = 3๐ฅ 2 โ 6โ๐ฅ + ๐ ๐ฅ . 2
This is the integrand (the expression between the integral symbol and ๐๐ฅ) that we started with. So the answer is choice (B). (8) Note that the derivative of any constant is always 0, ie. 2
10. โซ0 (๐ฅ 2 โ 4๐ฅ)๐ 6๐ฅ (A) โ (B) (C) (D)
2 โ๐ฅ 3
๐ ๐๐ฅ
[๐ถ ] = 0.
๐๐ฅ =
๐ 16 3
0 ๐ 16 3 1โ๐ 16 3
Solution: 2
โซ0 (๐ฅ 2 โ 4๐ฅ)๐ 6๐ฅ
2 โ๐ฅ 3
1
๐๐ฅ = โ ๐ 6๐ฅ 3
2 โ๐ฅ 3
2
1
1
โ๐ 16 +1
3
3
3
|0 = โ ๐ 16 โ (โ ๐ 0 ) =
This is equivalent to choice (D). 2
3
Notes: (1) To evaluate โซ(๐ฅ 2 โ 4๐ฅ)๐ 6๐ฅ โ๐ฅ ๐๐ฅ, we can formally make the substitution ๐ข = 6๐ฅ 2 โ ๐ฅ 3 . It then follows that ๐๐ข = (12๐ฅ โ 3๐ฅ 2 )๐๐ฅ = 3(4๐ฅ โ ๐ฅ 2 )๐๐ฅ = โ3(๐ฅ 2 โ 4๐ฅ)๐๐ฅ Uh oh! There is no factor of โ3 inside the integral. But constants never 1 pose a problem. We simply multiply by โ3 and โ at the same time. We 3
1
place the โ3 inside the integral where it is needed, and we leave the โ 3 outside of the integral sign as follows:
24
โซ(๐ฅ 2 โ 4๐ฅ)๐ 6๐ฅ
2 โ๐ฅ 3
1
๐๐ฅ = โ โซ(โ3)(๐ฅ 2 โ 4๐ฅ)๐ 6๐ฅ 3
2 โ๐ฅ 3
๐๐ฅ
1
We have this flexibility to place the โ3 and โ where we like because 3 multiplication is commutative, and constants can be pulled outside of the integral sign freely. We now have โซ(๐ฅ 2 โ 4๐ฅ)๐ 6๐ฅ
2 โ๐ฅ 3
1
๐๐ฅ = โ โซ(โ3)(๐ฅ 2 โ 4๐ฅ)๐ 6๐ฅ 3
1
1
1
= โ โซ ๐ ๐ข ๐๐ข = โ ๐ ๐ข + ๐ถ = โ ๐ 6๐ฅ 3 3 3
2 โ๐ฅ 3
2 โ๐ฅ 3
๐๐ฅ
+๐ถ
๐
(2) โซ๐ ๐(๐ฅ)๐๐ฅ = ๐น(๐) โ ๐น(๐) where ๐น is any antiderivative of ๐. 1
In this example, ๐น(๐ฅ) = โ ๐ 6๐ฅ ๐(๐ฅ) =
(๐ฅ 2
โ 4๐ฅ)๐
6๐ฅ 2 โ๐ฅ 3
2 โ๐ฅ 3
3
is an antiderivative of the function
. So
2
1
2 โ23
โซ0 ๐(๐ฅ)๐๐ฅ = ๐น(2) โ ๐น(0) = โ 3 ๐ 6โ
2
1
โ (โ ๐ 0 ). 3
๐
(3) We sometimes write ๐น(๐) โ ๐น(๐) as ๐น(๐ฅ) |๐ . This is just a convenient way of focusing on finding an antiderivative before worrying about plugging in the upper and lower limits of integration (these are the numbers ๐ and ๐, respectively). (4) If we are doing the substitution formally, we can save some time by changing the limits of integration. We do this as follows: 2
โซ0 (๐ฅ 2 โ 4๐ฅ)๐ 6๐ฅ 1
16
2 โ๐ฅ 3
1
2
๐๐ฅ = โ โซ0 (โ3)(๐ฅ 2 โ 4๐ฅ)๐ 6๐ฅ 3 1
16
1
= โ โซ0 ๐ ๐ข ๐๐ข = โ ๐ ๐ข |0 = โ (๐ 16 โ ๐ 0 ) = 3 3 3
2 โ๐ฅ 3
๐๐ฅ
โ๐ 16 +1 3
Notice that the limits 0 and 2 were changed to the limits 0 and 16, respectively. We made this change using the formula that we chose for the substitution: ๐ข = 6๐ฅ 2 โ ๐ฅ 3 . When ๐ฅ = 0, we have ๐ข = 0 and when ๐ฅ = 2, we have ๐ข = 6(2)2 โ 23 = 6 โ
4 โ 8 = 24 โ 8 = 16. 1
11. โซ ๐๐ฅ = ๐ฅ ln ๐ฅ Solution: โซ
1 ๐ฅ ln ๐ฅ
๐๐ฅ = ln|ln ๐ฅ| + ๐ถ.
25
Notes: (1) To evaluate โซ
1
๐๐ฅ, we can formally make the substitution
๐ฅ ln ๐ฅ
1
๐ข = ln ๐ฅ. It then follows that ๐๐ข = ๐๐ฅ. So we have ๐ฅ
1
1
1
1
โซ ๐ฅ ln ๐ฅ ๐๐ฅ = โซ ln ๐ฅ โ
๐ฅ ๐๐ฅ = โซ ๐ข ๐๐ข = ln|๐ข| + ๐ถ = ln|ln ๐ฅ| + ๐ถ. 1
1
1
1
1
To get the first equality we simply rewrote as โ
= โ
. This ๐ฅ ln ๐ฅ ๐ฅ ln ๐ฅ ln ๐ฅ ๐ฅ way it is easier to see exactly where ๐ข and ๐๐ข are. 1
To get the second equality we simply replaced ln ๐ฅ by ๐ข, and ๐๐ฅ by ๐๐ข. ๐ฅ
To get the third equality we used the basic integration formula 1
โซ ๐ฅ ๐๐ฅ = ln|๐ฅ| + ๐ถ. To get the last equality we replaced ๐ข by ln ๐ฅ (since we set ๐ข = ln ๐ฅ in the beginning). (2) Recall from problem 5 that 1
๐ ๐๐ฅ
1
[ln ๐ฅ] = . It therefore seems like it ๐ฅ
should follow that โซ ๐๐ฅ = ln ๐ฅ + ๐ถ. But this is not completely accurate. ๐ฅ
Observe that we also have
๐ ๐๐ฅ
[ln(โ๐ฅ)] =
1 โ๐ฅ
(โ1) = 1
1 ๐ฅ
(the chain rule
was used here). So it appears that we also have โซ ๐๐ฅ = ln(โ๐ฅ) + ๐ถ. ๐ฅ
How can the same integral lead to two different answers? Well it doesnโt. Note that ln ๐ฅ is only defined for ๐ฅ > 0, and ln(โ๐ฅ) is only defined for ๐ฅ < 0. Furthermore, observe that ln|๐ฅ| = {
ln ๐ฅ if ๐ฅ > 0 . ln(โ๐ฅ) if ๐ฅ < 0
It follows that 1
โซ ๐ฅ ๐๐ฅ = ln|๐ฅ| + ๐ถ. 12. โซ 5cot ๐ฅ csc 2 ๐ฅ ๐๐ฅ = Solution: โซ 5cot ๐ฅ csc 2 ๐ฅ ๐๐ฅ = โ
5cot ๐ฅ ln 5
+ ๐ถ.
Notes: (1) Recall from problem 6 that 5๐ฅ
๐ ๐๐ฅ
[5๐ฅ ] = 5๐ฅ (ln 5). It follows that
โซ 5๐ฅ ๐๐ฅ = ln 5 + ๐ถ.
26
To verify this, note that ๐
[
5๐ฅ
๐๐ฅ ln 5
+ ๐ถ] =
1
๐
ln 5 ๐๐ฅ
[5๐ฅ ] +
๐ ๐๐ฅ
[๐ถ ] =
1 ln 5
โ
5๐ฅ (ln 5) + 0 = 5๐ฅ .
More generally, we have that for any ๐ > 0, ๐ โ 1, โซ ๐ ๐ฅ ๐๐ฅ =
๐๐ฅ ln ๐
+ ๐ถ.
(2) As an alternative way to evaluate โซ 5๐ฅ ๐๐ฅ, we can rewrite 5๐ฅ as ๐ ๐ฅ ln 5 and perform the substitution ๐ข = ๐ฅ ln 5, so that ๐๐ข = (ln 5) ๐๐ฅ. So we have 1
1
โซ 5๐ฅ ๐๐ฅ = โซ ๐ ๐ฅ ln 5 ๐๐ฅ = ln 5 โซ ๐ ๐ฅ ln 5 (ln 5) ๐๐ฅ = ln 5 โซ ๐ ๐ข ๐๐ข =
1 ln 5
๐๐ข + ๐ถ =
1 ln 5
๐ ๐ฅ ln 5 + ๐ถ =
1 ln 5
5๐ฅ + ๐ถ =
5๐ฅ ln 5
+ ๐ถ.
(3) To evaluate โซ 5cot ๐ฅ csc 2 ๐ฅ ๐๐ฅ, we can formally make the substitution ๐ข = cot ๐ฅ. It then follows that ๐๐ข = โ csc 2 ๐ฅ ๐๐ฅ. So we have 5๐ข
โซ 5cot ๐ฅ csc 2 ๐ฅ ๐๐ฅ = โ โซ 5cot ๐ฅ (โcsc 2 ๐ฅ)๐๐ฅ = โ โซ 5๐ข ๐๐ข = โ ln 5 + ๐ถ =โ
5cot ๐ฅ ln 5
+ ๐ถ.
(4) As an alternative, we can combine notes (2) and (3) to evaluate the integral in a single step by rewriting 5cot ๐ฅ csc 2 ๐ฅ as ๐ (cot ๐ฅ)(ln 5) csc 2 ๐ฅ, and then letting ๐ข = (cot ๐ฅ)(ln 5), so that ๐๐ข = (โ csc 2 ๐ฅ)(ln 5)๐๐ฅ. I leave the details of this solution to the reader. 13. If ๐ is a continuous function for all real ๐ฅ, and ๐ is an 1 ๐+โ antiderivative of ๐, then lim โซ๐ ๐(๐ฅ) ๐๐ฅ is โโ0 โ
(A) ๐(0) (B) ๐โฒ(0) (C) ๐(๐) (D) ๐โฒ(๐) 1
๐+โ
Solution: lim โซ๐ โ โโ0
1
๐(๐ฅ) ๐๐ฅ = lim [๐(๐ฅ)]๐+โ ๐ โโ0 โ
= lim โโ0
๐(๐+โ)โ๐(๐) โ
This is choice (D).
27
= ๐โฒ(๐).
Notes: (1) The second Fundamental Theorem of Calculus says that if ๐ is a Riemann integrable function on [๐, ๐], then ๐
โซ๐ ๐(๐ฅ) ๐๐ฅ = ๐น(๐) โ ๐น(๐) where ๐น is any antiderivative of ๐. In this problem, since ๐ is an antiderivative of ๐, we have ๐+โ
โซ๐
๐(๐ฅ) ๐๐ฅ = ๐(๐ + โ) โ ๐(๐).
(2) We sometimes use the notation [๐น(๐ฅ)]๐๐ as an abbreviation for ๐น(๐) โ ๐น(๐). This is just a convenient way of focusing on finding an antiderivative before worrying about plugging in the upper and lower limits of integration (these are the numbers ๐ and ๐, respectively). In the problem above we have ๐+โ
โซ
๐(๐ฅ) ๐๐ฅ = [๐(๐ฅ)]๐+โ = ๐(๐ + โ) โ ๐(๐) ๐
๐
(3) If a function f is continuous on [๐, ๐], then ๐ is Riemann integrable on [๐, ๐]. (4) Recall the definition of the derivative: ๐(๐ฅ + โ) โ ๐(๐ฅ) โโ0 โ
๐โฒ (๐ฅ) = lim So we have ๐โฒ (๐) = lim
๐(๐+โ)โ๐(๐)
โโ0
โ
14. If the function ๐ given by ๐(๐ฅ) = โ๐ฅ 3 has an average value of 2 on the interval [0, ๐], then ๐ = 3
(A) 52 (B) 5 2
(C) 53 1
(D) 52 Solution: The average value of ๐ on [0, ๐] is 1
๐
3
2
5
๐
2
5
2
3
โซ ๐ฅ 2 ๐๐ฅ = 5๐ ๐ฅ 2 |0 = 5๐ โ
๐ 2 = 5 ๐ 2 . ๐โ0 0
28
3
2
2
3
So we have ๐ 2 = 2. Therefore ๐ 2 = 5, and so ๐ = 53 , choice (C). 5
Notes: (1) The average value of the function ๐ over the interval [๐, ๐] is ๐ 1 โซ ๐(๐ฅ) ๐๐ฅ. ๐โ๐ ๐
(2) Recall from problem 9 that for any real number ๐, we have โซ ๐ฅ ๐ ๐๐ฅ =
๐ฅ ๐+1 ๐+1
+ ๐ถ, where ๐ถ is an arbitrary constant. 5
3
For example, โซ ๐ฅ 2 ๐๐ฅ =
๐ฅ2 5 2
5
5
5
2
2
5
5
5
+ ๐ถ = ๐ฅ 2 รท + ๐ถ = ๐ฅ 2 โ
+ ๐ถ = ๐ฅ 2 + ๐ถ. 2
๐
(3) โซ๐ ๐(๐ฅ)๐๐ฅ = ๐น(๐) โ ๐น(๐) where ๐น is any antiderivative of ๐. 5
2
3
Here, ๐บ(๐ฅ) = ๐ฅ 2 is an antiderivative of the function ๐(๐ฅ) = ๐ฅ 2 . So 5
๐ โซ0 ๐(๐ฅ)๐๐ฅ 1
2
5
5
2
= ๐บ(๐) โ ๐บ(0) = ๐ 2 โ 0 = ๐ 2 . 5
5
5
5
5
2 5
3
(4) โ
๐ 2 = ๐ โ1 โ
๐ 2 = ๐ โ1+2 = ๐ โ2+2 = ๐ 2 . ๐
It follows that
5
2 5๐
2
1
5
๐
5
3
2
โ
๐2 = โ
โ
๐2 = ๐2. 2
5
3 2
(5) We solve the equation ๐ = 2 by first multiplying each side of the 5
5
5 2
2
2 5
3
5
equation by . Since โ
= 1, we get ๐ 2 = 2 ( ) = 5. 2
2
We then raise each side of this last equation to the power . Since 3 2 2 3
(๐ ) = ๐
32 โ
23
2 3
1
3
= ๐ = ๐, we get ๐ = 5 .
(6) See problem 3 for a review of the laws of exponents used in notes (4) and (5). โ
2
15. โซ0 2๐ฅ๐ โ๐ฅ ๐๐ฅ is (A) divergent (B) โ1 (C)
1 2
(D) 1
29
โ
2
2
Solution: โซ0 2๐ฅ๐ โ๐ฅ ๐๐ฅ = โ๐ โ๐ฅ |โ 0 = 0 โ (โ1) = 1, choice (D). Notes: (1) The given integral is an improper integral because one of the limits of integration is โ. This is actually a Type II improper integral. For an example of a Type I improper integral, see problem 45. โ
๐
(2) โซ0 ๐(๐ฅ) ๐๐ฅ is an abbreviation for lim โซ0 ๐(๐ฅ) ๐๐ฅ, and ๐น(๐ฅ) |โ 0 is an ๐โโ
abbreviation for lim ๐น(๐ฅ) |๐0 . ๐โโ
2
2
In this problem, ๐(๐ฅ) = 2๐ฅ๐ โ๐ฅ and ๐น(๐ฅ) = โ๐ โ๐ฅ . 2
(3) To evaluate the integral โซ 2๐ฅ๐ โ๐ฅ ๐๐ฅ, we can formally make the substitution ๐ข = โ๐ฅ 2 . It then follows that ๐๐ข = โ2๐ฅ๐๐ฅ. Uh oh! There is no minus sign inside the integral. But constants never pose a problem. We simply multiply by โ1 inside the integral where it is needed, and also outside of the integral sign as follows: 2
2
โซ 2๐ฅ๐ โ๐ฅ ๐๐ฅ = โ โซ โ2๐ฅ๐ โ๐ฅ ๐๐ฅ We have this flexibility to do this because constants can be pulled outside of the integral sign freely, and (โ1)(โ1) = 1, so that the two integrals are equal in value. We now have 2
2
โ โซ โ2๐ฅ๐ โ๐ฅ ๐๐ฅ = โ โซ ๐ ๐ข ๐๐ข = โ๐ ๐ข + ๐ถ = โ๐ โ๐ฅ + ๐ถ. We get the leftmost equality by replacing โ๐ฅ 2 by ๐ข, and โ2๐ฅ๐๐ฅ by ๐๐ข. We get the second equality by the basic integration formula โซ ๐ ๐ข ๐๐ข = ๐ ๐ข + ๐ถ. And we get the rightmost equality by replacing ๐ข with โ๐ฅ 2 . 2
(4) Note that the function ๐(๐ฅ) = ๐ โ๐ฅ can be written as the composition ๐(๐ฅ) = ๐(โ(๐ฅ)) where ๐(๐ฅ) = ๐ ๐ฅ and โ(๐ฅ) = โ๐ฅ 2 . Since โ(๐ฅ) = โ๐ฅ 2 is the inner part of the composition, it is natural to try the substitution ๐ข = โ๐ฅ 2 . Note that the derivative of โ๐ฅ 2 is โ2๐ฅ, so that ๐๐ข = โ2๐ฅ๐๐ฅ.
30
(5) With a little practice, we can evaluate an integral like this very quickly with the following reasoning: The derivative of โ๐ฅ 2 is โ2๐ฅ. So to 2 integrate โ2๐ฅ๐ โ๐ฅ we simply pretend we are integrating ๐ ๐ฅ but as we do it we leave the โ๐ฅ 2 where it is. This is essentially what was done in the above solution. Note that the โ2๐ฅ โgoes awayโ because it is the derivative of โ๐ฅ 2 . We need it there for everything to work. (6) If we are doing the substitution formally, we can save some time by changing the limits of integration. We do this as follows: โ
โ
2
2
โซ0 2๐ฅ๐ โ๐ฅ ๐๐ฅ = โ โซ0 โ2๐ฅ๐ โ๐ฅ ๐๐ฅ โโ
โโ ๐ข
๐ ๐๐ข = โ๐ ๐ข |0
= โ โซ0
= โ(0 โ ๐ 0 ) = โ(โ1) = 1.
Notice that the limits 0 and โ were changed to the limits 0 and โโ, respectively. We made this change using the formula that we chose for the substitution: ๐ข = โ๐ฅ 2 . When ๐ฅ = 0, we have that ๐ข = 0 and when ๐ฅ = โ, we have โ๐ข = โโ2 = โโ โ
โ = โโ.โ I used quotation marks in that last computation because the computation โ โ
โ is not really well-defined. What we really mean is that if we have two expressions that are approaching โ, then their product is approaching โ as well. For all practical purposes, the following computations are valid: โโ
โ=โ โ+โ=โ โโ โ โ = โโ For example, if lim ๐(๐ฅ) = โ and lim ๐(๐ฅ) = โ, then lim [๐(๐ฅ) โ
๐ฅโโ
๐ฅโโ
๐ฅโโ
๐(๐ฅ)] = โ and lim [๐(๐ฅ) + ๐(๐ฅ)] = โ. ๐ฅโโ
Note that the following forms are indeterminate: โ
0
โ
0
0โ
โ
โโโ
00
1โ
โ0
For example, if lim ๐(๐ฅ) = โ and lim ๐(๐ฅ) = โ, then in general we ๐ฅโโ
cannot say anything about lim
๐(๐ฅ)
๐ฅโโ
. The value of this limit depends on
๐ฅโโ ๐(๐ฅ)
the specific functions ๐ and ๐.
31
๐๐ฆ
16. Let ๐ฆ = ๐(๐ฅ) be the solution to the differential equation = ๐๐ฅ arctan(๐ฅ๐ฆ) with the initial condition ๐(0) = 2. What is the approximation of ๐(1) if Eulerโs method is used, starting at ๐ฅ = 0 with a step size of 0.5? (A) 1 (B) 2 (C) 2 + (D) 2 +
๐ 8 ๐ 4
Solution: Letโs make a table: (๐ฅ, ๐ฆ)
๐๐ฅ
(0,2)
.5
(.5,2)
.5
๐๐ฆ
๐๐ฆ
๐๐ฅ ( ) = ๐๐ฆ
(๐ฅ + ๐๐ฅ, ๐ฆ + ๐๐ฆ)
0
0
(.5,2)
๐ 4
๐ 8
(1,2 + )
๐๐ฅ
๐๐ฅ
๐ 8
๐
From the last entry of the table we see that ๐(1) โ 2 + , choice (C). 8
Notes: (1) Eulerโs method is a procedure for approximating the solution of a differential equation. (2) To use Eulerโs method we must be given a differential equation ๐๐ฆ = ๐(๐ฅ, ๐ฆ), an initial condition ๐(๐ฅ0 ) = ๐ฆ0 , and a step size ๐๐ฅ. ๐๐ฅ
In this problem, we have
๐๐ฆ ๐๐ฅ
= arctan(๐ฅ๐ฆ), ๐(0) = 2, and ๐๐ฅ = 0.5.
(3) The initial condition ๐(๐ฅ0 ) = ๐ฆ0 is equivalent to saying that the point (๐ฅ0 , ๐ฆ0 ) is on the solution curve. So in this problem we are given that (0,2) is on the solution curve. (4) We can get an approximation to ๐(๐ฅ0 + ๐๐ฅ) by using a table (as shown in the above solution) as follows: In the first column we put the point (๐ฅ0 , ๐ฆ0 ) as given by the initial condition. In the second column we put the step size ๐๐ฅ.
32
In the third column we plug the point (๐ฅ0 , ๐ฆ0 ) into the differential ๐๐ฆ equation to get . ๐๐ฅ
In the fourth column we multiply the numbers in the previous two columns to get ๐๐ฆ. In the fifth column we add ๐๐ฅ to ๐ฅ0 and ๐๐ฆ to ๐ฆ0 to get the point (๐ฅ0 + ๐๐ฅ, ๐ฆ0 + ๐๐ฆ). This is equivalent to ๐(๐ฅ0 + ๐๐ฅ) = ๐ฆ0 + ๐๐ฆ. (5) We can now copy the point from the fifth column into the first column of the next row, and repeat this procedure to approximate ๐(๐ฅ0 + 2๐๐ฅ). In this problem, since ๐ฅ0 = 0 and ๐๐ฅ = 0.5, we have ๐ฅ0 + 2๐๐ฅ = 1, and so we are finished after the second iteration of the procedure. 17. The area of the region bounded by the lines ๐ฅ = 1, ๐ฅ = 4, and ๐ฆ = 0 and the curve ๐ฆ = ๐ 3๐ฅ is 1
(A) ๐ 3 (๐ 9 โ 1) 3
(B) ๐ 3 (๐ 9 โ 1) (C) ๐ 12 โ 1 (D) 3๐ 3 (๐ 9 โ 1) 4
1
4
1
1
1
3
3
3
Solution: โซ1 ๐ 3๐ฅ ๐๐ฅ = ๐ 3๐ฅ |1 = ๐ 12 โ ๐ 3 = ๐ 3 (๐ 9 โ 1). 3
This is choice (A). Notes: (1) To compute the area under the graph of a function that lies entirely above the ๐ฅ-axis (the line ๐ฆ = 0) from ๐ฅ = ๐ to ๐ฅ = ๐, we simply integrate the function from ๐ to ๐. In this problem, the function is ๐ฆ = ๐ 3๐ฅ , ๐ = 1, and ๐ = 4. Note that ๐ ๐ฅ > 0 for all ๐ฅ. So ๐ 3๐ฅ > 0 for all ๐ฅ. It follows that the graph of ๐ฆ = ๐ 3๐ฅ lies entirely above the ๐ฅ-axis. (2) Although it is not needed in this problem, here is a sketch of the area we are being asked to find.
33
(3) To evaluate โซ ๐ 3๐ฅ ๐๐ฅ, we can formally make the substitution ๐ข = 3๐ฅ. It then follows that ๐๐ข = 3๐๐ฅ. We place the 3 next to ๐๐ฅ where it is needed, and we leave the outside of the integral sign as follows:
1 3
1
โซ ๐ 3๐ฅ ๐๐ฅ = 3 โซ ๐ 3๐ฅ โ
3๐๐ฅ We now have 1
1
1
1
โซ ๐ 3๐ฅ ๐๐ฅ = 3 โซ ๐ 3๐ฅ โ
3๐๐ฅ = 3 โซ ๐ ๐ข ๐๐ข = 3 ๐ ๐ข + ๐ถ = 3 ๐ 3๐ฅ + ๐ถ. We get the second equality by replacing 3๐ฅ by ๐ข, and 3๐๐ฅ by ๐๐ข. We get the third equality by the basic integration formula โซ ๐ ๐ข ๐๐ข = ๐ ๐ข + ๐ถ. And we get the rightmost equality by replacing ๐ข with 3๐ฅ. (4) With a little practice, we can evaluate an integral like this very quickly with the following reasoning: The derivative of 3๐ฅ is 3. So we artificially 1 insert a factor of 3 next to ๐๐ฅ, and outside the integral sign. Now to 3 integrate 3๐ 3๐ฅ we simply pretend we are integrating ๐ ๐ฅ but as we do it we leave the 3๐ฅ where it is. This is essentially what was done in the above solution. Note that the 3 โgoes awayโ because it is the derivative of 3๐ฅ. We need it to be there for everything to work.
34
(5) If we are doing the substitution formally, we can save some time by changing the limits of integration. We do this as follows: 4
1
4
1
12
1
12
1
1
โซ1 ๐ 3๐ฅ ๐๐ฅ = 3 โซ1 ๐ 3๐ฅ โ
3๐๐ฅ = 3 โซ3 ๐ ๐ข ๐๐ข = 3 ๐ ๐ข |3 = 3 ๐ 12 โ 3 ๐ 3 . Notice that the limits 1 and 4 were changed to the limits 3 and 12. We made this change using the formula that we chose for the substitution: ๐ข = 3๐ฅ. When ๐ฅ = 1, we have ๐ข = 3(1) = 3. And when ๐ฅ = 4, we have ๐ข = 3(4) = 12. Note that this method has the advantage that we do not have to change back to a function of ๐ฅ at the end. 18. Which of the following integrals gives the length of the graph of ๐ฆ = ๐ 3๐ฅ between ๐ฅ = 1 and x = 2 ? 2
(A) โซ1 โ๐ 6๐ฅ + ๐ 3๐ฅ ๐๐ฅ 2
(B) โซ1 โ๐ฅ + 3๐ 3๐ฅ ๐๐ฅ 2
(C) โซ1 โ1 + 3๐ 3๐ฅ ๐๐ฅ 2
(D) โซ1 โ1 + 9๐ 6๐ฅ ๐๐ฅ Solution:
๐๐ฆ ๐๐ฅ
๐๐ฆ 2
= 3๐ 3๐ฅ , so that 1 + ( ) = 1 + 9๐ 6๐ฅ . It follows that the ๐๐ฅ
2
desired length is โซ1 โ1 + 9๐ 6๐ฅ ๐๐ฅ, choice (D). Notes: (1) The arc length of the differentiable curve with equation ๐ฆ = ๐(๐ฅ) from ๐ฅ = ๐ to ๐ฅ = ๐ is 2
๐ ๐๐ฆ Arc length = โซ๐ โ1 + ( ) ๐๐ฅ ๐๐ฅ
(2) By the chain rule, we have details.
๐๐ฆ ๐๐ฅ
= ๐ 3๐ฅ (3) = 3๐ 3๐ฅ . See problem 2 for
๐๐ฆ 2
(3) ( ) = (3๐ 3๐ฅ )2 = 32 (๐ 3๐ฅ )2 = 9๐ 3๐ฅโ
2 = 9๐ 6๐ฅ . See problem 3 for a ๐๐ฅ
review of the laws of exponents used here.
35
LEVEL 1: LIMITS AND CONTINUITY 19. lim
2๐ฅ 2 โ13๐ฅโ7
๐ฅโ7
๐ฅโ7
=
(A) โ (B) 0 (C) 2 (D) 15 Solution 1: lim
2๐ฅ 2 โ13๐ฅโ7 ๐ฅโ7
๐ฅโ7
= lim
(๐ฅโ7)(2๐ฅ+1)
๐ฅโ7
๐ฅโ7
= lim (2๐ฅ + 1) ๐ฅโ7
= 2(7) + 1 = 15. This is choice (D). Notes: (1) When we try to substitute 7 in for ๐ฅ we get the indeterminate 0 form . Here is the computation: 0
2(7)2 โ13(7)โ7 7โ7
=
98โ91โ7 7โ7
0
= . 0
This means that we cannot use the method of โplugging in the numberโ to get the answer. So we have to use some other method. (2) One algebraic โtrickโ that works in this case is to factor the numerator as 2๐ฅ 2 โ 13๐ฅ โ 7 = (๐ฅ โ 7)(2๐ฅ + 1). Note that one of the factors is (๐ฅ โ 7) which is identical to the factor in the denominator. This will always happen when using this โtrick.โ This makes factoring pretty easy in these problems. (3) Most important limit theorem: If ๐(๐ฅ) = ๐(๐ฅ) for all ๐ฅ in some interval containing ๐ฅ = ๐ except possibly at ๐ itself, then we have lim ๐(๐ฅ) = lim ๐(๐ฅ). ๐ฅโ๐
๐ฅโ๐
In this problem, our two functions are ๐(๐ฅ) =
2๐ฅ 2 โ13๐ฅโ7 ๐ฅโ7
and ๐(๐ฅ) = 2๐ฅ + 1.
Note that ๐ and ๐ agree everywhere except at ๐ฅ = 7. Also note that ๐(7) is undefined, whereas ๐(7) = 15.
36
(4) To compute a limit, first try to simply plug in the number. This will only fail when the result is an indeterminate form. The two basic 0 โ indeterminate forms are and (the more advanced ones are 0 โ
โ, 0 โ โ โ โ, 00 , 1โ , and โ0 , but these can always be manipulated into one of the two basic forms). If an indeterminate form results from plugging in the number, then there are two possible options: Option 1: Use some algebraic manipulations to create a new function that agrees with the original except at the value that is being approached, and then use the limit theorem mentioned in note (3). This is how we solved the problem above. Option 2: Try L'Hรดpital's rule (see solution 2 below). Solution 2: We use L'Hรดpital's rule to get lim
2๐ฅ 2 โ13๐ฅโ7 ๐ฅโ7
๐ฅโ7
= lim
๐ฅโ7
4๐ฅโ13 1
= 4(7) โ 13 = 15, choice (D).
Notes: (1) L'Hรดpital's rule says the following: Suppose that (i) ๐ and ๐ are differentiable on some interval containing ๐ (except possibly at ๐ itself). (ii) lim ๐(๐ฅ) = lim ๐(๐ฅ) = 0 or lim ๐(๐ฅ) = lim ๐(๐ฅ) = ยฑโ ๐ฅโ๐
(iii) lim
๐ฅโ๐
๐โฒ (๐ฅ)
๐ฅโ๐ ๐ โฒ (๐ฅ)
๐ฅโ๐
๐ฅโ๐
exists, and
(iv) ๐ โฒ (๐ฅ) โ 0 for all ๐ฅ in the interval (except possibly at ๐ itself). Then lim
๐(๐ฅ)
๐ฅโ๐ ๐(๐ฅ)
= lim
๐โฒ (๐ฅ)
๐ฅโ๐ ๐ โฒ (๐ฅ)
.
In this problem ๐(๐ฅ) = 2๐ฅ 2 โ 13๐ฅ โ 7 and ๐(๐ฅ) = ๐ฅ โ 7. (2) It is very important that we first check that the expression has the correct form before applying L'Hรดpital's rule. In this problem, note that when we substitute 7 in for ๐ฅ in the given 0 expression we get (see note 1 above). So in this case L'Hรดpital's rule 0 can be applied.
37
20. The graph of the function โ is shown in the figure above. Which of the following statements about โ is true? (A) lim โ(๐ฅ) = ๐ ๐ฅโ๐
(B) lim โ(๐ฅ) = ๐ ๐ฅโ๐
(C) lim โ(๐ฅ) = ๐ ๐ฅโ๐
(D) lim โ(๐ฅ) = โ(๐) ๐ฅโ๐
Solution: From the graph we see that lim โ(๐ฅ) = ๐, choice (C). ๐ฅโ๐
Notes: (1) The open circles on the graph at ๐ and ๐ indicate that there is no point at that location. The darkened circle at ๐ indicates โ(๐) = ๐. (2) limโ โ(๐ฅ) = ๐ and lim+ โ(๐ฅ) = ๐. Therefore lim โ(๐ฅ) does not ๐ฅโ๐
๐ฅโ๐
๐ฅโ๐
exist. (3) โ is not defined at ๐ฅ = ๐, ie. โ(๐) does not exist. In particular, โ is not continuous at ๐. So lim โ(๐ฅ) โ โ(๐). ๐ฅโ๐
21. What is lim
โโ0
๐ 4
๐ 4
tan( +โ)โtan( ) โ
?
(A) 0 (B) 1 (C) 2 (D) The limit does not exist.
38
Solution 1: If we let ๐(๐ฅ) = tan ๐ฅ, then ๐ โฒ (๐ฅ) = lim ๐ ๐ tan( +โ)โtan( ) 4 4
๐
๐ โฒ ( ) = lim 4
โ
โโ0
tan(๐ฅ+โ)โtan(๐ฅ) โ
โโ0
. So
.
Now, the derivative of tan ๐ฅ is sec 2 ๐ฅ. So we have ๐ 4
๐ 4
tan( +โ)โtan( )
lim
โ
โโ0
๐
๐
4
4
2
= ๐ โฒ ( ) = sec 2 ( ) = (โ2) = 2.
This is choice (C). Notes: (1) The derivative of the function ๐ is defined by ๐ โฒ (๐ฅ) = lim
๐(๐ฅ+โ)โ๐(๐ฅ) โ
โโ0
In this problem ๐(๐ฅ) = tan ๐ฅ, so that ๐ โฒ (๐ฅ) = lim
tan(๐ฅ+โ)โtan(๐ฅ) โ
โโ0
(2) See problem 1 for the basic trig derivatives. In particular, ๐ ๐๐ฅ ๐
1
4
โ2
(3) cos ( ) =
[tan ๐ฅ] = sec 2 ๐ฅ. ๐
1
4
cos( )
. Therefore sec ( ) =
๐
๐ 2
4
4
๐ 4
=1รท
1 โ2
= 1 โ
โ2 = โ2.
2
(4) sec 2 ( ) = (sec ) = (โ2) = 2. Solution 2: We use L'Hรดpital's rule to get lim
๐ 4
๐ 4
tan( +โ)โtan( ) โ
โโ0
= lim
๐ 4
sec2 ( +โ) 1
โโ0
๐
2
= sec 2 ( ) = (โ2) = 2. 4
This is choice (C). Note: See problem 19 for a detailed description of L'Hรดpital's rule. 22. What is lim
5โ๐ฅ 2 +3๐ฅ 3
๐ฅโโ ๐ฅ 3 โ2๐ฅ+3
?
(A) 1 (B)
5 3
(C) 3 (D) The limit does not exist.
39
Solution: lim
5โ๐ฅ 2 +3๐ฅ 3
= lim
๐ฅโโ ๐ฅ 3 โ2๐ฅ+3
3๐ฅ 3
๐ฅโโ ๐ฅ 3
= 3, choice (C).
Notes: (1) If ๐ and ๐ are polynomials, then lim
๐(๐ฅ)
๐ฅโโ ๐(๐ฅ) ๐โ1
= lim
๐๐ ๐ฅ ๐
๐ฅโโ ๐๐ ๐ฅ ๐
where
we have ๐(๐ฅ) = ๐๐ ๐ฅ ๐ + ๐๐โ1 ๐ฅ + โฏ + ๐1 ๐ฅ + ๐0 ๐ ๐(๐ฅ) = ๐๐ ๐ฅ + ๐๐โ1 ๐ฅ ๐โ1 + โฏ + ๐1 ๐ฅ + ๐0 . (2) If ๐ = ๐, then lim
๐๐ ๐ฅ ๐
๐ฅโโ ๐๐ ๐ฅ ๐
=
๐๐ ๐๐
and
.
(3) Combining notes (1) and (2), we could have gotten the answer to this problem immediately by simply taking the coefficients of ๐ฅ 3 in the numerator and denominator and dividing. The coefficient of ๐ฅ 3 in the numerator is 3, and the coefficient of ๐ฅ 3 in 3 the denominator is 1. So the final answer is = 3. 1
(4) If ๐ > 0, then lim
1
๐ฅโโ ๐ฅ ๐
= 0.
(5) For a more rigorous solution, we can multiply both the numerator 1 and denominator of the fraction by 3 to get ๐ฅ
5โ๐ฅ 2 +3๐ฅ 3 ๐ฅ 3 โ2๐ฅ+3
It follows that lim
1
=
5โ๐ฅ 2 +3๐ฅ 3
๐ฅโโ ๐ฅ 3 โ2๐ฅ+3
( 3 ) (5โ๐ฅ 2 +3๐ฅ 3 ) ๐ฅ 1 โ
( 3 ) (๐ฅ 3 โ2๐ฅ+3)
=
๐ฅ
1
=
5 1 โ +3 ๐ฅ3 ๐ฅ 2 3 1โ 2 + 3 ๐ฅ ๐ฅ
.
1
5 lim ( 3 )โ lim ( )+ lim 3 ๐ฅโโ ๐ฅ ๐ฅโโ ๐ฅ ๐ฅโโ 1
1
lim 1โ2 lim ( 2 )+3 lim ( 3 ) ๐ฅโโ ๐ฅโโ ๐ฅ ๐ฅโโ ๐ฅ
=
5โ
0โ0+3 1โ2โ
0+3โ
0
= 3.
(6) L'Hรดpital's rule can also be used to solve this problem since the limit โ has the form : โ
lim
5โ๐ฅ 2 +3๐ฅ 3
๐ฅโโ ๐ฅ 3 โ2๐ฅ+3
= lim
๐ฅโโ
โ2๐ฅ+9๐ฅ 2 3๐ฅ 2 โ2
= lim
๐ฅโโ
โ2+18๐ฅ 6๐ฅ
= lim
18
๐ฅโโ 6
= 3.
Observe that we applied L'Hรดpital's rule three times. Each time we differentiated the numerator and denominator with respect to ๐ฅ to get โ another expression of the form . โ
See problem 19 for a detailed description of L'Hรดpital's rule.
40
23. lim
1
ln(
โโ0 โ
(A)
10+โ 10
) is equal to
1 10
(B) 10 (C) ๐ 10 (D) The limit does not exist. Solution 1: If we let ๐(๐ฅ) = ln ๐ฅ, then ๐ โฒ (๐ฅ) = lim
ln(๐ฅ+โ)โln(๐ฅ) โ
โโ0
So ๐ โฒ (10) = lim
1
โโ0 โ
ln(
10+โ 10
= lim
1
โโ0 โ
ln(
๐ฅ+โ ๐ฅ
).
). 1
Now, the derivative of ln ๐ฅ is . So we have ๐ฅ
lim
1
โโ0 โ
ln(
10+โ 10
) = ๐ โฒ (10) =
1
, choice (A).
10
Notes: (1) The derivative of the function ๐ is defined by ๐ โฒ (๐ฅ) = lim
๐(๐ฅ+โ)โ๐(๐ฅ) โ
โโ0
In this problem ๐(๐ฅ) = ln ๐ฅ, so that ๐ โฒ (๐ฅ) = lim
ln(๐ฅ+โ)โln(๐ฅ) โ
โโ0
.
๐
๐ฅ+โ
๐
๐ฅ
(2) Recall that ln ๐ โ ln ๐ = ln ( ). So ln(๐ฅ + โ) โ ln(๐ฅ) = ln ( and therefore
ln(๐ฅ+โ)โln(๐ฅ) โ
1
1
๐ฅ+โ
โ
๐ฅ
= [ln(๐ฅ + โ) โ ln(๐ฅ)] = ln ( โ
),
).
(3) See the notes at the end of problem 3 for a review of the laws of logarithms. Solution 2: We use L'Hรดpital's rule to get
lim
1
โโ0 โ
ln(
10+โ 10
) = lim
โโ0
ln(
10+โ ) 10
โ
1
1 10
( 10+โ )( )
= lim
โโ0
10
1
=
1
, choice (A).
10
Note: (1) See problem 19 for more information on L'Hรดpital's rule. (2) To apply L'Hรดpital's rule we separately took the derivative of 10+โ ๐(๐ฅ) = ln( ) and ๐(๐ฅ) = โ. 10
41
10+โ
(3) ๐(๐ฅ) = ln( ) is a composition of the functions ln ๐ฅ and 10 therefore need to use the Chain Rule to differentiate it. The first part of the Chain Rule gives us
1
10+โ 10
. We
.
10+โ 10
10+โ
1
For the second part, it may help to rewrite as (10 + โ). It is now 10 10 easy to see that the derivative of this expression with respect to โ is 1 1 (0 + 1) = . 10
10
sin 7๐ฅ
24. lim
๐ฅโ0 sin 4๐ฅ
Solution: lim
=
sin 7๐ฅ
๐ฅโ0 sin 4๐ฅ
7
= ( lim
4 7๐ฅโ0
sin 7๐ฅ 7๐ฅ
= lim
7โ
4๐ฅ sin 7๐ฅ
7
= lim
๐ฅโ0 4โ
7๐ฅ sin 4๐ฅ
)( lim
4๐ฅ
7
4๐ฅโ0 sin 4๐ฅ
sin 7๐ฅ
4 ๐ฅโ0
) = (lim
sin ๐ข
4 ๐ขโ0
โ
7๐ฅ
๐ข
)
4๐ฅ sin 4๐ฅ
1 sin ๐ฃ (lim ) ๐ฃโ0 ๐ฃ
7
1
๐
4
1
๐
= โ
1โ
= .
Notes: (1) A basic limit worth memorizing is lim
sin ๐ฅ ๐ฅ
๐ฅโ0
= 1.
(2) The limit in note (1) is actually very easy to compute using L'Hรดpital's rule: lim
๐ฅโ0
sin ๐ฅ ๐ฅ
= lim
๐ฅโ0
cos ๐ฅ 1
= cos 0 = 1
(3) It is not hard to see that ๐ฅ โ 0 if and only if 4๐ฅ โ 0 if and only if 7๐ฅ โ 0. This is why we can replace ๐ฅ by 4๐ฅ and 7๐ฅ in the subscripts of the limits above. (4)
sin 7๐ฅ sin 4๐ฅ
can be rewritten as
7โ
4๐ฅ sin 7๐ฅ 4โ
7๐ฅ sin 4๐ฅ
It follows that we can rewrite
sin 7๐ฅ sin 4๐ฅ
. 7 sin 7๐ฅ
as โ
4
7๐ฅ
โ
4๐ฅ sin 4๐ฅ
(5) Using the substitution ๐ข = 7๐ฅ, we have lim
sin 7๐ฅ
7๐ฅโ0
7๐ฅ
= lim
๐ขโ0
sin ๐ข ๐ข
.
Using the substitution ๐ฃ = 4๐ฅ, we have lim
4๐ฅ
4๐ฅโ0 sin 4๐ฅ
=
1 sin 4๐ฅ 4๐ฅโ0 4๐ฅ
lim
42
=
1 sin ๐ฃ ๐ฃโ0 ๐ฃ
lim
.
(6) We can also solve this problem using L'Hรดpital's rule as follows: lim
sin 7๐ฅ
๐ฅโ0 sin 4๐ฅ
25.
๐ฅ lim (๐ฅโ11)2 ๐ฅโ11
= lim
7 cos 7๐ฅ
๐ฅโ0 4cos 4๐ฅ
=
7(1) 4(1)
7
= . 4
= ๐ฅ
Solution: The function ๐(๐ฅ) = (๐ฅโ11)2 has a vertical asymptote of ๐ฅ = 11. If ๐ฅ is โnearโ 11, then ๐ฅ lim ๐ฅโ11 (๐ฅโ11)2
๐ฅ (๐ฅโ11)2
is positive. It follows that
= +โ . ๐ฅ
Notes: (1) When we substitute 11 in for ๐ฅ into f (๐ฅ) = (๐ฅโ11)2 , we get 11 0
. This is not an indeterminate form. ๐
For a rational function, the form where ๐ is a nonzero real number 0 always indicates that ๐ฅ = ๐ is a vertical asymptote. This means that at least one of limโ ๐(๐ฅ) or lim+ ๐(๐ฅ) is + or โโ. If both limits agree, ๐ฅโ๐
๐ฅโ๐
then lim ๐(๐ฅ) is the common value. If the two limits disagree, then ๐ฅโ๐
lim ๐(๐ฅ) does not exist.
๐ฅโ๐
(2) A nice visual way to find the left hand and right hand limits is by creating a sign chart. We split up the real line into intervals using the ๐ฅ-values where the numerator and the denominator of the fraction are zero, and then check the sign of the function in each subinterval formed.
In this case we split up the real line into three pieces. Notice that the cutoff points are 0 and 11 because the numerator of the function is zero when ๐ฅ = 0, and the denominator of the function is zero when ๐ฅ = 11.
43
We then plug a real number from each of these three intervals into the function to see if the answer is positive or negative. For example, 5 ๐(5) = (5โ11)2 > 0. Note that we do not need to finish the computation. We only need to know if the answer is positive or negative. Since there are + signs on both sides of ๐ฅ = 11, we have that ๐ฅ lim (๐ฅโ11)2 = +โ. ๐ฅโ11
(3) We actually do not care about the minus sign to the left of 0. We could have left that part out of the sign chart. It is however important that we include the zero as a cutoff point. This tells us that we can test any value between 0 and 11 to find limโ ๐(๐ฅ). ๐ฅโ๐
26. Let ๐ be the function defined by 5๐ ๐ฅโ7 , ๐ฅโค7 1 + ln|๐ฅ โ 8| ๐(๐ฅ) = 15 cos(๐ฅ โ 7) , ๐ฅ>7 { sin(7 โ ๐ฅ) + 3 Show that ๐ is continuous at ๐ฅ = 7. Solution: limโ ๐(๐ฅ) =
๐ฅโ7
lim ๐(๐ฅ) =
๐ฅโ7+
5๐ 7โ7 1+ln|7โ8|
15 cos(7โ7)
=
sin(7โ7)+3
=
5๐ 0 1+ln 1
15 cos 0 sin 0+3
=
=
5 1+0
15(1) 0+3
= 5. =
15 3
= 5.
So lim ๐(๐ฅ) = 5. ๐ฅโ7
Also, ๐(7) =
5๐ 7โ7 1+ln|7โ8|
= 5. So, lim ๐(๐ฅ) = ๐(7). ๐ฅโ7
It follows that ๐ is continuous at ๐ฅ = 7.
LEVEL 1: SERIES 5
15
7
28
27. The sum of the infinite geometric series +
+
45 112
+ โฏ is 5
Solution: The first term of the geometric series is ๐ = , and the common ratio is ๐ =
15 28
5
15
7
28
รท =
7
3
5
4
7
โ
= . It follows that the sum is
44
๐ 1โ๐
=
5 7
3 1โ 4
5
1
5
4
๐๐
7
4
7
1
๐
= รท = โ
=
.
Notes: (1) A geometric sequence is a sequence of numbers such that the quotient ๐ between consecutive terms is constant. The number ๐ is called the common ratio of the geometric sequence. For example, consider the sequence 5 15 45
,
,
,โฆ
7 28 112
We have
15 28
5
15
7
28
รท =
7
3
5
4
โ
=
and
45 112
รท
15 28
=
45 112
โ
28 15 3
3
= . It follows that 4
the sequence is geometric with common ratio ๐ = . 4
(2) A geometric series is the sum of the terms of a geometric sequence. The series in this problem is an infinite geometric series. (3) The sum ๐บ of an infinite geometric series with first term ๐ and common ratio ๐ with โ1 < ๐ < 1 is ๐ฎ =
๐ ๐โ๐
Note that if the common ratio ๐ is greater than 1 or less than โ1, then the geometric series has no sum. (4) As we saw in note (1), we can get the common ratio ๐ of a geometric series, by dividing any term by the term which precedes it. 28. Which of the following series converge? I. โโ ๐=1
1 ๐
II. โโ ๐=1 III. โโ ๐=1
๐3 2๐3 +5 cos(๐๐) ๐
(A) I only (B) II only (C) III only (D) I and III only Solution: The first series is the harmonic series which diverges.
45
lim
๐3
๐โโ 2๐3 +5
1
๐3
2
2๐3 +5
= and so โโ ๐=1
โโ ๐=1
diverges by the divergence test.
cos(๐๐) ๐
= โโ ๐=1
(โ1)๐ ๐
1
Since ( ) is a decreasing sequence with lim โโ ๐=1
๐ (โ1)๐ ๐
1
= 0, the series
๐โโ ๐
converges by the alternating series test.
So the answer is choice (C). 1
1
1
1
Notes: (1) โโ ๐=1 ๐ = 1 + 2 + 3 + 4 + โฏ is called the harmonic series. This series diverges. It is not at all obvious that this series diverges, and one of the reasons that it is not obvious is because it diverges so slowly. The advanced student might want to show that given any ๐ > 0, there 1 1 1 is a positive integer ๐ such that 1 + + + โฏ + > ๐. This would give 2 3 ๐ a proof that the harmonic series diverges. (2) The divergence test or ๐th term test says: (i) if โโ ๐=1 ๐๐ converges, then lim ๐๐ = 0, or equivalently ๐โโ
(ii) if lim ๐๐ โ 0, then โโ ๐=1 ๐๐ diverges. ๐โโ
Note that statements (i) and (ii) are contrapositives of each other, and are therefore logically equivalent. It is usually easier to apply the divergence test by using statement (ii). In other words, simply check the limit of the underlying sequence of the series. If this limit is not zero, then the series diverges. 1
In this problem, the limit of the underlying sequence is . Since this is not 2 zero, the given series diverges. A common mistake is to infer from lim
๐3
๐โโ 2๐3 +5
converges to 1
=
1 2
that the series
1
๐3
2
2๐3 +5
. This is of course not true: the sequence (
)
converges to , but the corresponding series diverges by the divergence 2 test.
46
(3) The converse of the divergence test is false. In other words, if lim ๐๐ = 0, it does not necessarily follow that โโ ๐=1 ๐๐ converges. ๐โโ
Students make this mistake all the time! It is absolutely necessary for lim ๐๐ = 0 for the series to have any chance of converging. But it is not ๐โโ
enough! A simple counterexample is the harmonic series. To summarize: (a) if lim ๐๐ โ 0, then โโ ๐=1 ๐๐ diverges. ๐โโ
(b) if lim ๐๐ = 0, then โโ ๐=1 ๐๐ may converge or diverge. ๐โโ
(4) To see that cos(๐๐) = (โ1)๐ , first note that cos(0๐) = cos 0 = 1. It follows that cos(2๐๐) = cos(0 + 2๐๐) = 1 for all integers ๐, or equivalently, cos(๐๐) = 1 whenever ๐ is even. Next note that cos(1๐) = cos ๐ = โ1. It then follows that cos((2๐ + 1)๐) = cos(๐ + 2๐๐) = cos ๐ = โ1 for all integers ๐, or equivalently, cos(๐๐) = โ1 whenever ๐ is odd. Finally note that (โ1)๐ = {
1 if ๐ is even โ1 if ๐ is odd
๐ (5) An alternating series has one of the forms โโ ๐=1(โ1) ๐๐ or ๐+1 โโ ๐๐ where ๐๐ > 0 for each positive integer ๐. ๐=1(โ1)
For example, the series given in III is an alternating series since it is equal (โ1)๐
to โโ ๐=1 integers ๐.
1
1
๐
๐
๐ = โโ ๐=1(โ1) ( ) , and ๐๐ =
๐
> 0 for all positive
(6) The alternating series test says that if (๐๐ ) is a decreasing sequence ๐ with lim ๐๐ = 0, then the alternating series โโ ๐=1(โ1) ๐๐ or ๐โโ
๐+1 โโ ๐๐ converges. ๐=1(โ1)
Since for all positive integers ๐, ๐ < ๐ + 1, it follows that 1
1
๐
๐โโ ๐
the sequence ( ) is decreasing. Also it is clear that lim that โโ ๐=1
(โ1)๐ ๐
1 ๐
>
1
, and
๐+1
= 0. It follows
converges by the alternating series test. 1
(7) Another way to check that the sequence ( ) is decreasing is to note that
๐
1
[ ]=
๐๐ ๐ฅ
๐ ๐๐
[๐ฅ โ1 ] = โ1๐ฅ โ2 = โ
1 ๐ฅ2
๐
1
< 0. So the function ๐(๐ฅ) = is 1
๐ฅ
a decreasing function, and therefore the sequence ( ) is also ๐ decreasing.
47
29. Let ๐ be a decreasing function with ๐(๐ฅ) โฅ 0 for all positive ๐ real numbers ๐ฅ. If lim โซ1 ๐(๐ฅ) ๐๐ฅ is finite, then which of the ๐โโ
following must be true? (A) โโ ๐=1 ๐(๐) converges (B) โโ ๐=1 ๐(๐) diverges 1
(C) โโ ๐=1
๐(๐)
(D) โโ ๐=1
๐(๐)
1
converges diverges
Solution: By the integral test, โโ ๐=1 ๐(๐) converges, choice (A). Notes: (1) The integral test says the following: Let ๐ be a continuous, positive, decreasing function on [๐, โ). Then โ โโ ๐=๐ ๐(๐) converges if and only if โซ๐ ๐(๐ฅ) ๐๐ฅ converges. โ
๐
(2) โซ1 ๐(๐ฅ) ๐๐ฅ = lim โซ1 ๐(๐ฅ) ๐๐ฅ. ๐โโ
(3) The integral test cannot be used to evaluate โโ ๐=1 ๐(๐). In general โ โ โ๐=1 ๐(๐) โ โซ1 ๐(๐ฅ) ๐๐ฅ. (4) The condition of ๐ decreasing can actually be weakened to ln ๐ฅ
๐ โeventually decreasing.โ For example, ๐(๐ฅ) = is not decreasing on ๐ฅ [1, โ), but is decreasing eventually. This can be verified by using the first derivative test. See problem 37 for details on how to apply this test. I leave the details to the reader. Now,
โ ln ๐ฅ
โซ1
โ ln ๐ฅ โซ1 ๐ฅ ๐๐ฅ
๐ฅ
1
โ
๐๐ฅ = (ln ๐ฅ)2 |1 = lim (ln ๐)2 = โ. 2
๐โโ
diverges. By the integral test โโ ๐=1
(5) For details on how to integrate โซ 5.
ln ๐ฅ ๐ฅ
ln ๐ ๐
It
follows
that
diverges.
๐๐ฅ, see the solution to problem
48
30. Which of the following series converge to โ1 ? I. โโ ๐=1
3 (โ2)๐
II. โโ ๐=1 III. โโ ๐=1
1โ3๐2 3๐2 +2 1 ๐(๐+1)
(A) I only (B) II only (C) III only (D) I and III only Solution: The first series is geometric with first term ๐ = โ 1
3
2
(โ2)๐
common ratio ๐ = โ . So the sum is โโ ๐=1 lim
๐โโ
1โ3๐2 3๐2 +2
= โ1 and so โโ ๐=1 โโ ๐=1
1โ3๐2 3๐2 +2
1 ๐(๐+1)
2
and
= โ1.
diverges by the divergence test. 1
1
๐
๐+1
= โโ ๐=1 ( โ
1
1
1
1
1
2
2
3
๐
๐+1
= lim [(1 โ ) + ( โ ) + โฏ + ( โ ๐โโ
=
3 โ 2 1 1+ 2
3
)
)] = lim (1 โ ๐โโ
1 ๐+1
) = 1.
So the answer is choice (A). Notes: (1) See problem 27 for more information on infinite geometric series. (2) For the first series it might help to write out the first few terms: โ
โ ๐=1
3 3 3 3 3 = โ + โ +โฏ+ +โฏ ๐ (โ2)๐ (โ2) 2 4 8
It is now easy to check that the series is geometric by checking the first 3
3
3
2
1
3
3
3
4
1
4
2
4
3
2
8
4
8
3
2
two quotients: รท (โ ) = โ
(โ ) = โ , โ รท = โ โ
= โ . 1
So we see that the series is geometric with common ratio ๐ = โ . It is 3
also quite clear that the first term is ๐ = โ . 2
49
2
๐ (3) A geometric series has the form โโ ๐=0 ๐๐ . In this form, the first term is ๐ and the common ratio is ๐. I wouldnโt get too hung up on this form though. Once you recognize that a series is geometric, itโs easy enough to just write out the first few terms and find the first term and common ratio as we did in note (2) above.
If we were to put the given series in this precise form it would look like 3
1 ๐
2
2
this: โโ ๐=0 (โ ) (โ ) . But again, this is unnecessary (and confusing). (4) See problem 28 for more information on the divergence test. (5) The third series is a telescoping sum. We can formally do a partial 1 1 1 ) . We fraction decomposition to see that โโ = โโ ๐=1 ๐=1 ( โ ๐(๐+1)
1
start by writing
๐(๐+1)
๐ด
๐ต
๐
๐+1
= +
๐
๐+1
. Now multiply each side of this
equation by ๐(๐ + 1) to get 1 = ๐ด(๐ + 1) + ๐ต๐ = ๐ด๐ + ๐ด + ๐ต๐. So we have 0๐ + 1 = (๐ด + ๐ต)๐ + ๐ด. Equating coefficients gives us ๐ด + ๐ต = 0 and ๐ด = 1, from which we also get ๐ต = โ1. So
1 ๐(๐+1)
๐ด
๐ต
๐
๐+1
= +
1
(โ1)
๐
๐+1
= +
1
1
๐
๐+1
= โ
.
(6) Another way to find ๐ด and ๐ต in the equation 1 = ๐ด(๐ + 1) + ๐ต๐ is to substitute in specific values for ๐. Two good choices are ๐ = 0 and ๐ = โ1. ๐ = 0: 1 = ๐ด(0 + 1) + ๐ต(0) = ๐ด. So ๐ด = 1. ๐ = โ1: 1 = ๐ด(โ1 + 1) + ๐ต(โ1). So 1 = โ๐ต, and ๐ต = โ1. 31. Which of the following series diverge? I. โโ ๐=1 II.
๐๐
๐2 +1 99 โ โ๐=1( )๐ 100
III. โโ ๐=1
2๐ ๐!
(A) I only (B) II only (C) III only (D) I and III only
50
Solution: lim
๐๐
๐โโ ๐2 +1
= โ and so โโ ๐=1
๐๐ ๐2 +1
diverges by the divergence
test. 99 ๐ ) 100 99 ๐ โโ ๐=1(100)
โโ ๐=1(
lim |
๐โโ
2๐+1 (๐+1)! 2๐ ๐!
is geometric with common ratio ๐ =
99 100
< 1, and
so
converges. 2๐+1
| = lim |(๐+1)! โ
๐โโ
๐!
๐ | = lim
2
2
๐โโ ๐+1
= 0 < 1,
and
โโ ๐=1
so
2๐ ๐!
converges by the ratio test. Therefore the answer is choice (A). Notes: (1) See problems 27 and 30 for more information on infinite geometric series, and see problem 28 for more information on the divergence test. โ (2) We say that the series โโ ๐=0 ๐๐ converges absolutely if โ๐=0|๐๐ | converges. If a series converges absolutely, then it converges.
A series which is convergent, but not absolutely convergent is said to converge conditionally. (3) The Ratio Test: For the series โโ ๐=0 ๐๐ , define ๐ฟ = lim | ๐โโ
๐๐+1 ๐๐
|.
If ๐ฟ < 1, then the series converges absolutely, and therefore converges. If ๐ฟ > 1, then the series diverges. If ๐ฟ = 1, then the ratio test fails. For the series โโ ๐=1 ๐๐+1 =
2๐+1 . (๐+1)!
2๐ ๐!
given in this problem, we have ๐๐ =
2๐ ๐!
, and so
32. What are all values of ๐ฅ for which the series โโ ๐=1 converges? (A) All ๐ฅ except ๐ฅ = 0 (B) |๐ฅ| <
1 5
1
1
5 1
5 1
5
5
(C) โ โค ๐ฅ < (D) โ โค ๐ฅ โค
51
5๐ ๐ฅ ๐ ๐
Solution: lim | ๐โโ
5๐+1 ๐ฅ๐+1 (๐+1) 5๐ ๐ฅ๐ ๐
| = lim |
5๐+1 ๐ฅ ๐+1 ๐+1
๐โโ
โ
๐ 5๐ ๐ฅ ๐
| = 5|๐ฅ|. So by the ratio
test, the series converges for all ๐ฅ such that 5|๐ฅ| < 1, or equivalently 1 1 1 |๐ฅ| < . Removing the absolute values gives โ < ๐ฅ < . 5
5
5
1
We still need to check the endpoints. When ๐ฅ = , we get the divergent 1
harmonic series โโ ๐=1 , and when ๐ฅ = โ alternating series
๐ ๐1 โโ ๐=1(โ1) ๐. 1
1 5
5
we get the convergent
So the series diverges at ๐ฅ =
1 5
and
converges at ๐ฅ = โ . 5
The answer is therefore choice (C). ๐ Notes: (1) A power series about ๐ฅ = 0 is a series of the form โโ ๐=1 ๐๐ ๐ฅ . To determine where a power series converges we use the ratio test. In other words, we compute
๐ฟ = lim | ๐โโ
๐๐+1 ๐ฅ ๐+1 ๐๐ ๐ฅ ๐
| = lim | ๐โโ
๐๐+1 ๐๐
| |๐ฅ|.
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