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MATH 266 – CALCULUS II
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Jim Bailey, College of the Rockies
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David Guichard, Whitman College
APEX Calculus: Gregory Hartman, Virginia Military Institute
Joseph Ling, University of Calgary
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Adapted for Athabasca University, October 2017
• G. Hartman:
– Many new exercises are included, adapted by Lyryx from APEX Calculus. The following
exercises are from APEX: 1.2.2 to 1.2.21, 3.1.15 to 3.1.56, 3.2.13 to 3.2.35, 3.3.11 to 3.3.22,
3.4.11 to 3.4.33, 3.7.11 to 3.7.37, 4.2.10 to 4.2.26
– New content on Hyperbolic Functions 1.7 and Inverse Hyperbolic Functions 2.2 is included.
These sections were adapted by Lyryx from the section “Hyperbolic Functions” in APEX Calculus.
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• Lyryx:
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2017 A – Several examples and exercises from Chapter 15 and 16 have been rewritten or removed.
– Order and name of topics in Chapter 15 and Chapter 16 have been revised. • D. Guichard: New content developed for the Three Dimensions, Vector Functions, and Vector Calculus chapters.
2016 B • Lyryx: Exercise numbering has been updated to restart with each section.
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Continued on next page ... ii 2016 A • Lyryx: The layout and appearance of the text has been updated, including the title page and newly
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content and additional examples.
• M. Cavers: Addition of new material and images particularly in the Review chapter. 2014 A 2012 A • M. Blenkinsop: Addition of content including Linear and Higher Order Approximations section.
• Original text by D. Guichard of Whitman College, the single variable material is a modification and expansion of notes written and released by N. Koblitz of the University of Washington. That version also contains exercises and examples from Elementary Calculus: An
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for more information. Contents
Contents iii Introduction and Review 1 Unit 1:
1.1
1.2
1.3
1.4
1.5
1.6
1.7 Inverse Functions
Inverse Functions . . . . . . . . . . . . . . . . . . .
Derivatives of Inverse Functions . . . . . . . . . . .
Exponential Functions . . . . . . . . . . . . . . . .
Logarithms . . . . . . . . . . . . . . . . . . . . . .
Derivatives of Exponential & Logarithmic Functions
Logarithmic Differentiation . . . . . . . . . . . . . .
Hyperbolic Functions . . . . . . . . . . . . . . . . . Unit 2:
2.1
2.2
2.3
2.4 Inverse Trigonometric and Hyperbolic Functions; L’Hopital’s Rule
Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . .
Inverse Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . .
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Unit 3:
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8 Techniques of Integration
Integration by Parts . . . . . . . . .
Powers of Trigonometric Functions .
Trigonometric Substitutions . . . .
Rational Functions . . . . . . . . .
Riemann Sums . . . . . . . . . . .
Numerical Integration . . . . . . . .
Improper Integrals . . . . . . . . .
Additional Exercises . . . . . . . . Unit 4:
4.1
4.2
4.3
4.4 Applications of Integration
Volume . . . . . . . . . .
Arc Length . . . . . . . .
Surface Area . . . . . . .
Center of Mass . . . . . . .
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116 iv Contents Unit 5:
5.1
5.2
5.3
5.4 Differential Equations
First Order Differential Equations . . . . .
First Order Homogeneous Linear Equations
First Order Linear Equations . . . . . . . .
Approximation . . . . . . . . . . . . . . . Unit 6: Sequences and Infinite Series
6.1 Sequences . . . . . . . . . .
6.2 Series . . . . . . . . . . . .
6.3 The Integral Test . . . . . .
6.4 Alternating Series . . . . . .
6.5 Comparison Tests . . . . . .
6.6 Absolute Convergence . . .
6.7 The Ratio and Root Tests . .
6.8 Power Series . . . . . . . .
6.9 Calculus with Power Series .
6.10 Taylor Series . . . . . . . .
6.11 Taylor’s Theorem . . . . . . .
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169 Additional Material
175
Table of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
Selected Exercise Answers 183 Index 211 Introduction and Review
The emphasis in this course is on problems—doing calculations and story problems. To master problem
solving one needs a tremendous amount of practice doing problems. The more problems you do the
better you will be at doing them, as patterns will start to emerge in both the problems and in successful
approaches to them. You will learn quickly and effectively if you devote some time to doing problems
every day.
Typically the most difficult problems are story problems, since they require some effort before you can
begin calculating. Here are some pointers for doing story problems:
1. Carefully read each problem twice before writing anything.
2. Assign letters to quantities that are described only in words; draw a diagram if appropriate.
3. Decide which letters are constants and which are variables. A letter stands for a constant if its value
remains the same throughout the problem.
4. Using mathematical notation, write down what you know and then write down what you want to
find.
5. Decide what category of problem it is (this might be obvious if the problem comes at the end of a
particular chapter, but will not necessarily be so obvious if it comes on an exam covering several
chapters).
6. Double check each step as you go along; don’t wait until the end to check your work.
7. Use common sense; if an answer is out of the range of practical possibilities, then check your work
to see where you went wrong. 1 Unit 1: Inverse Functions
1.1 Inverse Functions
In mathematics, an inverse is a function that serves to “undo” another function. That is, if f (x) produces
y, then putting y into the inverse of f produces the output x. A function f that has an inverse is called
invertible and the inverse is denoted by f −1 . It is best to illustrate inverses using an arrow diagram: →a
2 →b
3 →c
4 →d →1
b →2
c →3
d →4
a 1 Notice how f maps 1 to a, and f −1 undoes this, that is, f −1 maps a back to 1. Don’t confuse f −1 (x) with
1
.
exponentiation: the inverse f −1 is different from f (x)
Not every function has an inverse. It is easy to see that if a function f (x) is going to have an inverse,
then f (x) never takes on the same value twice. We give this property a special name.
A function f (x) is called one-to-one if every element of the range corresponds to exactly one element
of the domain. Similar to the Vertical Line Test (VLT) for functions, we have the Horizontal Line Test
(HLT) for the one-to-one property.
Theorem 1.1: The Horizontal Line Test A function is one-to-one if and only if there is no horizontal line that intersects its graph more than
once. Example 1.2: Parabola is Not One-to-one The parabola f (x) = x2 it not one-to-one because it does not satisfy the horizontal line test. For
example, the horizontal line y = 1 intersects the parabola at two points, when x = −1 and x = 1.
We now formally define the inverse of a function. 3 4 Unit 1: Inverse Functions
Definition 1.3: Inverse of a Function Let f (x) and g(x) be two one-to-one functions. If ( f ◦ g)(x) = x and (g ◦ f )(x) = x then we say that
f (x) and g(x) are inverses of each other. We denote g(x) (the inverse of f (x)) by g(x) = f −1 (x).
Thus, if f maps x to y, then f −1 maps y back to x. This gives rise to the cancellation formulas:
f −1 ( f (x)) = x,
f ( f −1 (x)) = x, for every x in the domain of f (x),
for every x in the domain of f −1 (x). Example 1.4: Finding the Inverse at Specific Values If f (x) = x9 + 2x7 + x + 1, find f −1 (5) and f −1 (1).
Solution. Rather than trying to compute a formula for f −1 and then computing f −1 (5), we can simply find a number c such that f evaluated at c gives 5. Note that subbing in some simple values (x =
−3, −2, 1, 0, 1, 2, 3) and evaluating f (x) we eventually find that f (1) = 19 +2(17 ) +1 +1 = 5 and f (0) = 1.
Therefore, f −1 (5) = 1 and f −1 (1) = 0.
♣
To compute the equation of the inverse of a function we use the following guidelines.
Guidelines for Computing Inverses
1. Write down y = f (x).
2. Solve for x in terms of y.
3. Switch the x’s and y’s.
4. The result is y = f −1 (x). Example 1.5: Finding the Inverse Function We find the inverse of the function f (x) = 2x3 + 1.
Solution. Starting with y = 2x3 + 1 we solve for x as follows:
3 y − 1 = 2x
Therefore, f −1 (x) = r
3 x−1
.
2 → y−1
= x3
2 → x= r
3 y−1
.
2
♣ This example shows how to find the inverse of a function algebraically. But what about finding the
inverse of a function graphically? Step 3 (switching x and y) gives us a good graphical technique to find
the inverse, namely, for each point (a, b) where f (a) = b, sketch the point (b, a) for the inverse. More
formally, to obtain f −1 (x) reflect the graph f (x) about the line y = x. 1.2. Derivatives of Inverse Functions 5 y x Exercises for 1.1
Exercise 1.1.1 Is the function f (x) = |x| one-to-one?
Exercise 1.1.2 Find a formula for the inverse of the function f (x) = x+2
.
x−2 1.2 Derivatives of Inverse Functions
Suppose we wanted to find the derivative of the inverse, but do not have an actual formula for the inverse
function? Then we can use the following derivative formula for the inverse evaluated at a.
Derivative of f −1 (a)
Given an invertible function f (x), the derivative of its inverse function f −1 (x) evaluated at x = a is:
′
f 1 (a) = 1
f ′ [ f −1 (a)] To see why this is true, start with the function y = f −1 (x). Write this as x = f (y) and differentiate both
sides implicitly with respect to x using the chain rule:
1 = f ′ (y) · dy
.
dx Thus,
dy
1
= ′ ,
dx
f (y) 6 Unit 1: Inverse Functions but y = f −1 (x), thus, At the point x = a this becomes: 1 ′
f −1 (x) = f ′ [ f −1 (x)] ′
f −1 (a) = f ′ [ f −1 (a)] . 1 Example 1.6: Derivatives of Inverse Functions
′
Suppose f (x) = x5 + 2x3 + 7x + 1. Find f −1 (1).
Solution. First we should show that f −1 exists (i.e. that f is one-to-one). In this case the derivative
f ′ (x) = 5x4 + 6x2 + 7 is strictly greater than 0 for all x, so f is strictly increasing and thus one-to-one.
It’s difficult to find the inverse of f (x) (and then take the derivative). Thus, we use the above formula
evaluated at 1:
−1 ′
1
f
(1) = ′ −1
.
f [ f (1)]
Note that to use this formula we need to know what f −1 (1) is, and the derivative f ′ (x). To find f −1 (1)
we make a table of values (plugging in x = −3, −2, −1, 0, 1, 2, 3 into f (x)) and see what value of x gives
1. We omit the table and simply observe that f (0) = 1. Thus,
f −1 (1) = 0.
Now we have: And so, f ′ (0) = 7. Therefore, ′
f −1 (1) =
1
f ′ (0) . ′
1
f −1 (1) = .
7 ♣ Exercises for 1.2
Exercise 1.2.1 Suppose f (x) = x3 + 4x + 2. Find the slope of the tangent line to the graph of g(x) =
x f −1 (x) at the point where x = 7.
In the following, verify that the given functions are inverses.
Exercise 1.2.2 f (x) = 2x + 6 and g(x) = 21 x − 3
Exercise 1.2.3 f (x) = x2 + 6x + 11, x ≥ 3 and
√
g(x) = x − 2 − 3, x ≥ 2 1.2. Derivatives of Inverse Functions
Exercise 1.2.4 f (x) =
g(x) = 7 3
, x 6= 5 and
x−5 3 + 5x
, x 6= 0
x Exercise 1.2.5 f (x) = x+1
, x 6= 1 and g(x) = f (x)
x−1 In the following, an invertible function f (x) is given along with a point that lies on its graph. Evaluate
at the indicated value. ′
f −1 (x) Exercise 1.2.6 f (x) = 5x + 10
Point= (2, 20)
′
Evaluate f −1 (20) Exercise 1.2.9 f (x) = x3 − 6x2 + 15x − 2
Point= (1, 8)
′
Evaluate f −1 (8) Exercise 1.2.7 f (x) = x2 − 2x + 4, x ≥ 1
Point= (3, 7)
′
Evaluate f −1 (7) Exercise 1.2.10 f (x) = Exercise 1.2.8 f (x) = sin 2x, −π /4 ≤ x ≤ π /4
√
Point= (π /6, 3/2)
′ √
Evaluate f −1 ( 3/2) 1
,x≥0
1 + x2 Point= (1, 1/2)
′
Evaluate f −1 (1/2) Exercise 1.2.11 f (x) = 6e3x
Point= (0, 6)
′
Evaluate f −1 (6) In the following, compute the derivative of the given function. Exercise 1.2.12 h(t) = sin−1 (2t)
Exercise 1.2.13 f (t) = sec−1 (2t)
Exercise 1.2.14 g(x) = tan−1 (2x)
Exercise 1.2.15 f (x) = x sin−1 x
Exercise 1.2.16 g(t) = sint cos−1 t
Exercise 1.2.17 f (t) = lntet Exercise 1.2.18 h(x) = sin−1 x
cos−1 x √
Exercise 1.2.19 g(x) = tan−1 ( x)
Exercise 1.2.20 f (x) = sec−1 (1/x)
Exercise 1.2.21 f (x) = sin(sin−1 x) Exercises 1.2.2 to 1.2.21 were adapted by Lyryx from APEX Calculus, Version 3.0, written by G. Hartman. This material
is released under Creative Commons license CC BY-NC ( ). See
the Copyright and Revision History pages in the front of this text for more information. 8 Unit 1: Inverse Functions 1.3 Exponential Functions
An exponential function is a function of the form f (x) = ax , where a is a constant. Examples are 2x , 10x
and (1/2)x. To more formally define the exponential function we look at various kinds of input values.
It is obvious that a5 = a · a · a · a · a and a3 = a · a · a, but when we consider an exponential function ax
we can’t be limited to substituting integers for x. What does a2.5 or a−1.3 or aπ mean? And is it really true
that a2.5 a−1.3 = a2.5−1.3 ? The answer to the first question is actually quite difficult, so we will evade it; the
answer to the second question is “yes.”
We’ll evade the full answer to the hard question, but we have to know something about exponential
functions. You need first to understand that since it’s not “obvious” what 2x should mean, we are really
free to make it mean whatever we want, so long as we keep the behavior that is obvious, namely, when x
is a positive integer. What else do we want to be true about 2x ? We want the properties of the previous
two paragraphs to be true for all exponents: 2x 2y = 2x+y and (2x )y = 2xy .
After the positive integers, the next easiest number to understand is 0: 20 = 1. You have presumably
learned this fact in the past; why is it true? It is true precisely because we want 2a 2b = 2a+b to be true
about the function 2x ....

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