Table of Contents

Chapter 1 Functions 1

Chapter 2 One-to-One and Onto Functions 2

Chapter 3 Inverse Functions 3

Chapter 4 Derivatives of Inverse Functions 6

Chapter 5 Inverse Trigonometric Functions 8

Chapter 6 Exponential Functions 12

Chapter 7 Logarithmic Functions 14

Chapter 8 Exponential Growth and Decay 16

Chapter 9 Derivatives of Inverse Trigonometric Functions 18

Chapter 10 Derivatives of Exponential Functions 20

Chapter 11 Derivatives of Logarithmic Functions 23

Chapter 12 Logarithmic Differentiation 24

Chapter 13 Hyperbolic Functions 25

Chapter 14 Integration by Parts 27

Chapter 15 Integrals of Trigonometric Functions 30

Chapter 16 Trigonometric Substitution 33

Chapter 17 Partial Fraction Decomposition 36

i
Chapter 18 Improper Integrals 40

Chapter 19 Sequences 45

Chapter 20 Induction 46

Chapter 21 Convergent Sequences 47

Chapter 22 Monotonic Sequences 50

Chapter 23 The Binomial Theorem 52

Chapter 24 Euler’s Number 54

Chapter 25 Proof of Lemma 69 56

Chapter 26 Series 58

Chapter 27 The Divergence Test 61

Chapter 28 The Comparison Test 62

Chapter 29 Geometric Series 63

Chapter 30 The Limit Comparison Test 65

Chapter 31 The Integral Test 67

Chapter 32 Absolute Convergence 69

Chapter 33 The Alternating Series Test 70

Chapter 34 The Ratio Test 72

Chapter 35 The Root Test 74

Chapter 36 Power Series 76

Chapter 37 Taylor Polynomials 80

Chapter 38 Taylor and Maclaurin Series 82

Chapter 39 Fourier Series 85

ii
Chapter 40 Separable Differential Equations 88

Chapter 41 Arc Length 90

Chapter 42 Solids of Revolution and Surface Area 91

Chapter 43 Centers of Mass 93

Chapter 44 Hydrostatic Pressure 94

Chapter 45 Parametric Equations 98

Chapter 46 Parametric Equations and Tangents 100

Chapter 47 Parametric Equations and Area 102

Chapter 48 Parametric Equations and Arc Length 103

Chapter 49 Parametric Equations and Surface Area 104

Chapter 50 Polar Coordinates 105

Chapter 51 Polar Curves 110

Chapter 52 Polar Curves and Tangents 113

Chapter 53 Polar Curves and Area 114

Chapter 54 Polar Curves and Arc Length 116

Chapter 55 Parabolas 117

Chapter 56 Ellipses 119

Chapter 57 Hyperbolas 120

Chapter 58 Rotations of Axes 122

Chapter 59 Conic Sections and Polar Coordinates 124

Chapter 60 Planetary Motion 128

iii
 Chapter 1: Functions
Definition 1: Let X and Y be sets.

(i) A function from X to Y is a rule that assigns to each element of X a unique element of Y .

(ii) The set X is the domain of f .

(iii) The set Y is the codomain of f .

Notation 2: Let X and Y be sets.

(i) If f is a function from X to Y , we write f : X → Y .

(ii) Y X = {f | f : X → Y }.

Definition 3: Suppose that f : X → Y and x ∈ X.

(i) The unique element of Y that f assigns to x is the function value of x or the image of x under

f and is usually denoted f (x).

(ii) The range of f is the set f (X) = {f (x) | x ∈ X}.

Notation 4: For a function f : X → Y ,

(i) dom(f ) will denote the domain of f ;

(ii) cod(f ) will denote the codomain of f ;

(iii) ran(f ) will denote the range of f .

Example 5: Define f : R → R by f (x) = x2 .

(a) f (3) = 9;

(b) f (−3) = 9;

(c) dom(f ) = R;

(d) cod(f ) = R;

(e) ran(f ) = [0, ∞).

Definition 6: The natural domain of a function f : R → R is the set {x ∈ R | f (x) is defined}.

Note 7: Throughout this text, the term “domain” will refer to the natural domain unless otherwise

stated.
√
Example 8: Set f (x) = 2x + 1.
 
1
(a) dom(f ) = − , ∞
2
(b) ran(f ) = [0, ∞).

1
 Chapter 2: One-to-One and Onto Functions
Theorem 9: The following are equivalent for any f : X → Y and any x1 , x2 ∈ X.

(i) f (x1 ) = f (x2 ) if and only if x1 = x2 ;

(ii) If x1 ̸= x2 , then f (x1 ) ̸= f (x2 );

(iii) If f (x1 ) = f (x2 ), then x1 = x2 .

Proof: Let f : X → Y .

((i) =⇒ (ii)) Suppose that (i) is satisfied and let x1 , x2 ∈ X with x1 ̸= x2 . Since x1 ̸= x2 , it is

clear that f (x1 ) ̸= f (x2 ).

((ii) =⇒ (iii)) Suppose (ii) is satisfied. To see that f (x1 ) = f (x2 ) implies that x1 = x2 , suppose

x1 , x2 ∈ X such that f (x1 ) = f (x2 ). If x1 ̸= x2 , then f (x1 ) ̸= f (x2 ) by the assumption. Since

f (x1 ) = f (x2 ), it must be that x1 = x2 .

((iii) =⇒ (i)) Suppose that (iii) is satisfied. To see that f (x1 ) = f (x2 ) if and only if x1 = x2 ,

suppose x1 , x2 ∈ X. If x1 = x2 , then f (x1 ) = f (x2 ) since f is a function. Also, if f (x1 ) = f (x2 ),

then by the assumption, we obtain x1 = x2 . Therefore, (i) is satisfied, and the theorem is proved.

Definition 10: A function f : X → Y is one-to-one if it satisfies any of the conditions in the

theorem above.

Notation 11: We sometimes write “1-1” instead of “one-to-one.”

Example 12: Determine whether the following functions are one-to-one.

(a) f (x) = x

To see that f is one-to-one, note that for x1 ̸= x2 , then f (x1 ) = x1 ̸= x2 = f (x2 ). Thus, f is

one-to-one.

(b) g(x) = x2

To see that f is not one-to-one, note that g(−3) = 9 = g(3). Thus, g is not one-to-one.

Theorem 13 (Horizontal Line Test): Let X ⊆ R. Then a function f : X → R is one-to-one if

and only if every horizontal line intersects the graph of the function in at most one point.

Proof: (=⇒) Suppose that f is one-to-one and y = c is a horizontal line. Attempting a contradiction,

suppose the line y = c intersects the graph of f at two points, (x1 , c) and (x2 , c). Since (x1 , c) lies

on the graph of f , f (x1 ) = c. Likewise, f (x2 ) = c. This is a contradiction since f is one-to-one.

(⇐=) Suppose that every horizontal line intersects the graph of f in at most one point. To see that

f is one-to-one, suppose x1 , x2 ∈ X with x1 ̸= x2 . Let c = f (x1 ) and consider the horizontal line

y = c. By the assumption, the line y = c intersects the graph of f only at the point (x1 , c). This

2
implies the point (x2 , f (x2 )) is not on the line y = c which means f (x2 ) ̸= c = f (x1 ). Since x1 and

x2 were chosen arbitrarily, f is one-to-one. ■

Proposition 14: A strictly increasing or strictly decreasing function is one-to-one.

Proof: Let f : X → R be either strictly increasing or strictly decreasing.

Case 1: f is strictly increasing

To see that f is one-to-one, suppose x1 , x2 ∈ X with x1 ̸= x2 . Without loss of generality, suppose

x1 < x2 . Since f is strictly increasing, f (x1 ) < f (x2 ), which implies that f (x1 ) ̸= f (x2 ). Therefore,

f is one-to-one.

Case 2: f is strictly decreasing

To see that f is one-to-one, use an argument similar to the one above. ■

Definition 15: A function f : X → Y is onto if for each y ∈ Y , there is an x ∈ X such that

f (x) = y.

Theorem 16: A function f : X → Y is onto if and only if ran(f ) = cod(f ).

Example 17:

(a) The function f : R → R defined by f (x) = x2 is not onto since ran(f ) = [0, ∞) ̸= R = cod(f ).

(b) The function f : R → [0, ∞) defined by f (x) = x2 is onto since ran(f ) = [0, ∞) = cod(f ).

Example 18: Define f : R → R by f (x) = x5 + x3 + x + 1. Is f one-to-one? If f is onto?

Recall from calculus 1 that f is strictly increasing. So f is one-to-one. Also, recall that lim f (x) =
x→−∞

−∞, lim f (x) = ∞, and f (x) is continuous. Therefore, f is onto.

x→∞ h π πi
Example 19: Define f : − , → [−1, 1] by f (θ) = sin θ. Is f one-to-one? Is f onto?
2 2 h π πi
To see that f is one-to-one, recall that f is strictly increasing on − , and apply the proposition
 π  π2 2
above. Now to see that f is onto, note that f − = −1, f = 1, and f is continuous on
h π πi 2 2
− , . So, by the Intermediate Value Theorem, f is onto.
2 2

Chapter 3: Inverse Functions

Definition 20: A function is a bijection if it is one-to-one and onto.

Definition 21: Let f : X → Y be a function. Then a function g : ran(f ) → dom(f ) is called the

inverse of f if (f ◦ g)(x) = (g ◦ f )(x) = x.

Definition 22: A function which has an inverse is called invertible.

Notation 23: If f is an invertible function, then the inverse of f is usually denoted f −1 .

Note 24: To find the inverse of an invertible function y = f (x), we proceed as follows.

(i) Interchange x and y;

3
(ii) Solve for y.

Example 25: Calculate the inverse of the given function.

(a) f : [0, ∞) → [−4, ∞); f (x) = x2 − 4

Consider y = x2 − 4. Interchanging x and y, we obtain the following.

x = y2 − 4

⇐⇒ x + 4 = y 2
√
⇐⇒ y = x + 4.

(b) g : [−3, ∞) → [−4, ∞); g(x) = x2 + 6x + 5

Consider y = x2 + 6x + 5. Interchanging x and y, we obtain the following.

x = y 2 + 6y + 5

⇐⇒ x = (y + 3)2 − 4

⇐⇒ x + 4 = (y + 3)2
√
⇐⇒ y = −3 + x + 4
x+1
Example 26: h : R \ {1} → R \ {3}; h(x) =
x−3
x+1
Consider y = . Interchanging x and y, we obtain the following.
x−3
y+1
x=
y−3
⇐⇒ x(y − 3) = y + 1

⇐⇒ xy − 3x = y + 1

⇐⇒ xy − y = 3x + 1

⇐⇒ y(x − 1) = 3x + 1
3x + 1
⇐⇒ y = .
x−1
Theorem 27: Let X, Y ⊆ R and suppose f : X → Y is invertible. Then the graph of f −1 is

obtained by reflecting the graph of f about the line y = x.

Proof: Note that (u, v) ∈ R2 lies on the graph of f if and only if the point (v, u) lies on the graph of

inv(f ). If u = v, then the point (u, v) = (v, u) lies on the line y = x. So suppose u ̸= v. Recall that
 
u+v u+v
the midpoint of the line segment connecting the points (u, v) and (v, u) is , , which
2 2
is on the line y = x. Also, recall that the slope of the line containing the points (u, v) and (v, u) is

−1. So, the line containing the points (u, v) and (v, u) is perpendicular to the line y = x. Therefore,

(u, v) is the reflection of (v, u) through the line y = x. ■

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Example 28: Given the graph of a function f , sketch a graph of f −1 .

The inverse of f is given below.

Proposition 29: If X, Y ⊆ R and f : X → Y is invertible and odd, then f −1 is odd.

Proof: Let y ∈ Y . Since f is odd,

f [−f −1 (y)]

= −f [f −1 (y)]

= −y

= f [f −1 (−y)].

Since f [−f −1 (y)] = f [f −1 (−y)] and f is one-to-one, f −1 (−y) = −f −1 (y). Therefore, f −1 is odd. ■

5
 Chapter 4: Derivatives of Inverse Functions
Lemma 30: Suppose that a, b ∈ R with a < b and f : [a, b] → R is one-to-one and continuous.

Then f (x) is strictly between f (a) and f (b) for all x ∈ (a, b).

Proof: Let x ∈ (a, b). Since f is one-to-one, f (a) ̸= f (b). This implies that either f (a) < f (b) or

f (b) < f (a).

Case 1: f (a) < f (b)

Toward a contradiction, suppose that f (x) < f (a). By the Intermediate Value Theorem, there

is a y ∈ [x, b] such that f (y) = f (a). This is a contradiction since f is one-to-one. Similarly, if

f (x) > f (b), then by the Intermediate Value Theorem, there is a y ∈ [a, x] such that f (y) = f (b),

which is a contradiction since f is one-to-one. Therefore, f (a) < f (x) < f (b).

Case 2: f (b) < f (a)

Exercise. ■

Proposition 31: A continuous 1-1 function on a closed interval is either strictly increasing or

strictly decreasing.

Proof: Suppose a, b ∈ R, a < b, and f : [a, b] → R is 1-1 and continuous. Then, since f is 1-1,

f (a) ̸= f (b). This means either f (a) < f (b) or f (b) < f (a).

Case 1: f (a) < f (b).

Proving that f is strictly increasing is left as an exercise to the reader.

Case 2: f (b) < f (a).

To see that f is strictly decreasing, let x1 , x2 ∈ [a, b] such that x1 < x2 .

If a = x1 < x2 = b, then f (x1 ) = f (a) > f (b) = f (x2 ).

If a = x1 < x2 < b, then, by the lemma above, f (x1 ) = f (a) > f (x2 ) < f (b).

If a < x1 < x2 = b, then, by the lemma above, f (a) > f (x1 ) > f (b) = f (x2 ).

If a < x1 < x2 < b, then, by the lemma above, f (a) > f (x1 ) > f (b).

Since f : [x, b] → R is continuous and f (x1 ) < f (b), we get f (x1 ) < f (x2 ) < f (b) by the lemma

above. Thus, f is strictly decreasing. ■

Theorem 32 (Continuous Inverse Theorem): Suppose f : X → Y is invertible, c ∈ X, and f

is continuous on an open interval containing c. Then f −1 is continuous at f (c).

Proof: To see that f −1 is continuous at f (c), let ε > 0 be given. By the hypothesis, there is an

open interval I such that c ∈ I and f is continuous on I. Let a, b ∈ R such that c ∈ (a, b) and

[a, b] ⊆ I ∩ (c − ε, c + ε). By the previous proposition, f ([a, b]) is either [f (a), f (b)] or [f (b), f (a)]

and f (c) is strictly between f (a) and f (b). Choose δ > 0 such that (f (c) − δ, f (c) + δ) ⊆ f ([a, b]).

6
Then,

f −1 [(f (c) − δ, f (c) + δ)]

⊆ f −1 [f ([a, b])]

⊆ [a, b]

⊆ I ∩ (c − ε, c + ε)

= I ∩ (f −1 [f (c)] − ε, f −1 [f (c)] + ε)

as desired. ■

Theorem 33: Suppose X, Y ⊆ R, f : X → Y is invertible, and c ∈ X. Further, suppose f is differ-

entiable and continuous on an open interval containing c, and f ′ (c) ̸= 0. Then f −1 is differentiable
1
at f (c) and (f −1 )′ [f (c)] = ′ .
f (c)
Proof: Since f is invertible, it is 1-1 and onto, and so, for each y ∈ y, there is a unique x ∈ X such
f −1 (y) − f −1 [f (c)] f −1 [f (x)] − f −1 [f (c)]
that y = f (x). So, lim can be written as lim =
y→f (c) y − f (c) f (x)→f (c) f (x) − f (c)
x−c x−c 1
lim . Since f is differentiable at c and f ′ (c) ̸= 0, lim = ′ . To
x→c f (x) − f (c) x→c f (x) − f (c) f (c)
x−c 1 x−c 1
see that lim = ′ , let ε > 0 be given. Since lim = ′ ,
f (x)→f (c) f (x) − f (c) f (c) x→c f (x) − f (c) f (c)
x−c 1
there is a δ > 0 such that − ′ < ε whenever |x − c| < δ. By the contin-
f (x) − f (c) f (c)
uous inverse theorem, f −1 is continuous at f (c). So, there is an α > 0 such that |f −1 (y) −

f −1 [f (c)]| < δ whenever |y − f (c)| < α. Thus, for x ∈ X, |f (x) − f (c)| < α, which implies
x−c 1 f −1 [f (x)] − f −1 [f (c)] 1
that |x − c| < δ, and so, − ′ = − ′ < ε. Therefore,
f (x) − f (c) f (c) f (x) − f (c) f (c)
f −1 [f (x)] − f −1 [f (c)] x−c 1
lim = lim = ′ , as desired. ■
x→c f (x) − f (c) f (x)→f (c) f (x) − f (c) f (c)
Example 34: Set f (x) = x3 + 7.

(a) Calculate f −1 (y).

√
f −1 (y) = 3 y − 7.

(b) Calculate (f −1 )′ (y).

1
(f −1 )′ (y) = √ .
3( y − 7)2
3

1
(c) Calculate ′ −1
f [f (y)]
1 1
′ −1
= √ .
f [f (y)] 3( y − 7)2
3

7
 Chapter 5: Inverse Trigonometric Functions
Theorem 35: The following functions are bijections, and hence invertible.
h π πi
(i) f : − , → [−1, 1] defined by f (y) = x if and only if sin(y) = x.
2 2
(ii) g : [0, π] → [−1, 1] defined by g(y) = x if and only if cos(y) = x.
h π πi
(iii) h : − , → (−∞, ∞) defined by h(y) = x if and only if tan(y) = x.
2 2
(iv) j : [0, π] → (−∞, ∞) defined by j(y) = x if and only if cot(y) = x.
h π   3π 
(v) k : 0, ∪ π, → (−∞, −1] ∪ [1, ∞) defined by k(y) = x if and only if sec(y) = x.
2 2
 π i  3π 
(vi) ℓ : 0, ∪ π, → (−∞, −1] ∪ [1, ∞) defined by ℓ(y) = x if and only if csc(y) = x.
2 2
Proof: To see that f , g, h, and j are 1–1, calculate the derivative of each function. Note that

the derivative does not change sign on the domain of the function. So in all cases, the function is

either strictly increasing or strictly decreasing. Therefore, f , g, h, and j are all one-to-one by the

proposition above.
 
hπ 3π
Claim 1: The function k is one-to-one on the set 0, ∪ π, .
2 2h
π
Proof (Claim 1): Note that k ′ (θ) = sec θ tan θ ≥ 0 for every θ ∈ 0, . So k is strictly increasing
h π h π 2  
′ 3π
on 0, , and hence, one-to-one on 0, . Also, k (θ) = sec θ tan θ ≤ 0 for all θ ∈ π, . So k
2 2   2
3π
is strictly decreasing, and hence one-to-one, on the interval π, by the proposition above. Since
 2
h π 3π
k(θ) ≥ 1 for every θ ∈ 0, and k(θ) ≤ −1 for every θ ∈ π, , the function k is one-to-one on
  2 2
h π 3π
the set 0, ∪ π, . ■Claim 1
2 2
By an argument similar to the one above, the function ℓ is one-to-one. Note that all six functions

are continuous, so it remains to show that each function is onto.

Claim 2: The function f is onto.

 π π
Proof (Claim 2): Note that f − = −1 and f = 1. Hence, by the Intermediate Value
2 2
Theorem, f is onto. ■Claim 2

By an argument similar to the one above, the function g is onto.

Claim 3: The function h is onto.

Proof (Claim 3): Let y ∈ R. Recall that lim h(θ) = −∞ and lim h(θ) = ∞. So choose
θ→ π
2
+ θ→ π
2
−
 π π
θ1 , θ2 ∈ − , such that h(θ1 ) ≤ y ≤ h(θ2 ). By the Intermediate Value Theorem, h attains every
2 2
value between h(θ1 ) and h(θ2 ), which means that h is onto. ■Claim 3

By an argument similar to the one above, the function j is onto.

Claim 4: The function k is onto.

8
Proof (Claim 4): Let y ∈ (−∞, −1] ∪ [1, ∞). Then either y ≥ 1 or y ≤ −1.

Case 1: y ≥ 1.
h π
Recall that lim k(θ) = ∞. So choose a y ∈ 0, such that y ≤ k(θ). By the Intermediate Value
θ→ π
2
− 2
Theorem, k attains every value between 1 and k(θ).

Case 2: y ≤ −1.
 
3π
Recall that lim k(θ) = −∞. So choose a y ∈ π, such that k(θ) ≤ y. By the Intermediate
θ→ 3π
2
− 2
Value Theorem, k attains every value between k(θ) and −1. Therefore, k is onto. ■Claim 4

By an argument similar to the one above, the function ℓ is onto. ■

Definition 36: Define functions f, g, h, j, k, ℓ as in the theorem above.

(i) f −1 is called the inverse sine (arcsine) function and is denoted by arcsin(x) or sin−1 (x).

(ii) g −1 is called the inverse cosine (arccosine) function and is denoted by arccos(x) or cos−1 (x).

(iii) h−1 is called the inverse tangent (arctangent) function and is denoted by arctan(x) or

tan−1 (x).

(iv) j −1 is called the inverse cotangent (arccotangent) function and is denoted by arccot(x)

or cot−1 (x).

(v) k −1 is called the inverse secant (arcsecant) function and is denoted by arcsec(x) or sec−1 (x).

(vi) ℓ−1 is called the inverse cosecant (arccosecant) function and is denoted by arccsc(x) or

csc−1 (x).

Note 37: The graphs of the inverse trigonometric functions are given below.

9
10
Example 38: Calculate each of the following.
h  π i π
(a) sin−1 sin = .
√ 2 2
(b) sec−1 ( 2)
√ 1 π
Set θ = sec−1 ( 2). Then cos(θ) = √ , which means that θ = .
   2 4
1
(c) tan cos−1
2
 2 √
1 1 1 3 3
If x = , then + y 2 = 1, which means that + y 2 = 1, or y 2 = , or y = . Now,
2 √ 2 4 4 2
3
y √
tan(θ) = = 2 = 3.
x 1
2
(d) cos[sin−1 (x)]
q √
Set θ = sin−1 (x). Then x = sin(θ), and so, cos[sin−1 (x)] = cos(θ) = 1 − sin2 (θ) = 1 − x2 .

(e) tan[sin−1 (x)]

11
 sin(θ) x
Set θ = sin−1 (x). Then x = sin(θ), and so, tan[sin−1 (x)] = tan(θ) = =√ .
cos(θ) 1 − x2
(f ) sec2 [tan−1 (x)]

Set θ = tan−1 (x). Then tan(θ) = x, and so, sec2 (tan−1 (x)) = sec2 (θ) = tan2 (θ) + 1 = x2 + 1.

Chapter 6: Exponential Functions

Definition 39: Let a ∈ R+ \{1}. The function f : R → R+ defined by f (x) = ax is an exponential

function.

Example 40: The sketch of the function f (x) = 2x is given below.

Note 41: Define f : R → R+ by f (x) = bx for some b ∈ R+ \ {1}.

(i) If b > 1, then f is strictly increasing;

(ii) If b ∈ (0, 1), then f is strictly decreasing.

(iii) The graph of f has a horizontal asymptote at y = 0.

(iv) f is invertible.

Example 42: Solve the following equations.

12
  2x
x+1 1
(b) 7 =
(a) 23x+1 = 25 7
Consider the following.
Consider the following.  2x
1
7 x+1
= = 7−2x
23x+1 = 25 7
⇐⇒ 7x+1 = 7−2x
⇐⇒ 3x + 1 = 5
⇐⇒ x + 1 = −2x
⇐⇒ 3x = 4
4 ⇐⇒ 3x = −1
⇐⇒ x = .
3 −1
⇐⇒ x = .
3

Theorem 43: Exponential functions are continuous on R.

Exercises

Section 6.2: 7-19 odd.

13
 Chapter 7: Logarithmic Functions
Definition 44: Let a ∈ R+ \ {1}. The function loga : R+ → R defined by loga x = y if and only if

x = ay is called a logarithmic function of base a.

Proposition 45:

(i) Logarithmic functions are invertible.

(ii) Exponential and logarithmic functions are inverses.

Proof: Exercise. ■

Example 46: Calculate the following.

(a) log2 (8) = 3;

(g) loga (a5 ) = 5;
(b) log5 (25) = 2;
1 (h) loga (ax ) = x;
(c) log9 (3) = ;
2 (i) loga (a) = 1;
(d) log3 (9) = 2;
 
1 (j) loga (1) = 0;
(e) log3 = −2;
9 (k) loga (0) is undefined.
(f ) loga (a2 ) = 2;

Notation 47:

(i) The function log10 is denoted log;

(ii) The function loge is denoted ln.

Theorem 48 (Properties of Logarithms): Let a ∈ R+ \ {1}.

(i) loga (ax ) = x for all x ∈ R; (iv) loga (xy) = loga (x) + loga (y) for all x, y > 0;

(ii) aloga (x) = x for all x > 0; (v) loga (xr ) = r loga (x) for all x > 0 and all r ∈ R;
 
x
(iii) loga (1) = 0; (vi) loga = loga (x) − loga (y) for all x, y > 0.
y

Proof:

For (iii), consider the following. For (iv), consider the following.

loga (1) = x xy = aloga x · aloga y

⇐⇒ ax = 1 = aloga x+loga y

⇐⇒ x = 0. ⇐⇒ loga (xy) = loga x + loga y.

For (v), consider the following. For (vi), consider the following.
x aloga x
xr = aloga x

r = log y = aloga x · a− loga y
y a a
= ar loga x = aloga x−loga y
 
x
⇐⇒ loga (xr ) = r loga x. ⇐⇒ loga = loga x − loga y. ■
y

14
Example 49: Simplify the following.

(a) log6 (9) + log6 (4) = log6 (36) = 2.

√
 
1 1 1
(b) ln √ = ln(1) − ln( e) = − ln(e) = − .
e √ 2 2
(c) 10 logb (b3 ) − 4 logb ( b) = 30 − 2 = 28.

Theorem 50 (Change of Base for Logarithms): Let a, b ∈ R+ \ {1}. Then for any x > 0,
logb (x)
loga (x) = .
logb (a)
Proof: Consider the following.
aloga (x) = x ⇐⇒ (loga x) (logb a) = logb x
logb x
⇐⇒ logb aloga (x) = logb x

⇐⇒ loga x = . ■
log a b

Example 51: Calculate each of the following.

log27 (27) 1 ∼
(a) log4 (27) = = = 2.377443751.
log27 (4) log27 (4)
log18 (18) 1 ∼
(b) log3 (18) = = = 2.630929754.
log18 (3) log18 (3)
Example 52: Calculate each of the following without a calculator.
3
(a) e3 ln(x) = eln(x )
= x3 . (b) ln(e1000 ) = 1000.

Example 53: Solve the following equations for x.

(b) 5x+1 = 12
x
(a) 4 = 6
Consider the following.
Consider the following.
5x+1 = 12
4x = 6
⇐⇒ log5 (5x+1 ) = log5 (12)
⇐⇒ log4 (4x ) = log4 (6)
⇐⇒ x + 1 = log5 (12)
⇐⇒ x = log4 (6).
⇐⇒ x = log5 (12) − 1.
(c) e2x = 8x
⇐⇒ 2x − x ln(8) = 0
Consider the following.
⇐⇒ x(2 − ln(8)) = 0
e2x = 8x
⇐⇒ x = 0.
⇐⇒ ln(e2x ) = ln(8x )

⇐⇒ 2x = x ln(8)
Example 54: The bacteria population in a bottle at time t (in hours) has size P (t) = 1000e0.35t .

After how many hours will there be 5000 bacteria in the bottle?
ln(5) ∼
Solving 1000e0.35t = 5000 for t, we obtain 0.35t = ln(5), or t = = 4.6 hours.
0.35
Theorem 55: Logarithmic functions are continuous on R+ .

15
 Exercises

Section 6.3: 3-19 odd, 23-35 odd, 41-63 odd.

Chapter 8: Exponential Growth and Decay

Definition 56:

(i) A quantity Q grows exponentially if Q = Q0 ekt , where Q0 is the quantity at time t = 0 and

k > 0 is the growth constant.

(ii) A quantity Q decays exponentially if Q = Q0 e−kt , where Q0 is the quantity at time t = 0

and k > 0 is the decay constant.

Example 57: A bacteria culture containing 200 bacteria is growing at a rate proportional to its

size. In an hour the culture will contain 300 bacteria.

(a) Express the number of bacteria in the culture as a function of time.

Consider the following. 3

⇐⇒ = ek
2
Q(t) = Q0 ekt
 
3
⇐⇒ k = ln
2
⇐⇒ 300 = 200ek  t
3
⇐⇒ Q(t) = 200eln(3/2)t = 200 · .
2
(b) How many bacteria will the culture contain in three hours?
 3
3
Q(3) = 200 · = 675.
2
Definition 58: The half-life of a substance is the amount of time it takes for half of the amount

of the substance to decay.

Example 59: The half-life of Carbon-14 is 5730 years.

Theorem 60: Suppose the half-life of a substance is λ. If Q is the amount present at time t and
 t/λ
1
Q0 is the initial amount present at time t = 0, then Q = Q0 · = Q0 e−[ln(2)t]/λ .
2
Corollary 61: Suppose the half-life of a substance is λ and k is a decay constant. Then kλ = ln(2).
 t/λ
1
Proof: By the theorem above and the definition of exponential decay, Q(t) = Q0 = Q0 e−kt .
2
So,
 t/λ
1
Q0 = Q0 e−kt t
2 ⇐⇒ − ln(2) = −kt
 t/λ λ
1 ⇐⇒ ln(2) = kλ. ■
⇐⇒ = e−kt
2

Example 62: Find the half-life of a radioactive substance that decays from 100 g to 40 g in three

hours.

16
Consider the following.
 3/λ  
 t/λ 21 3 1
1 ⇐⇒ = = ln
Q(t) = Q(0) · 52  λ 2
2 1
 3/λ 3 ln
1 2 3 ln(2)
⇐⇒ 40 = 100 · ⇐⇒ λ =   = hours.
2 2 ln(5) − ln(2)
ln
5
Example 63: If you have 50 grams of Carbon-14 today, how much will be left after 100 years?

Let y(t) denote the mass of Carbon-14 in grams after t years. By a previous example, λ = 5730 years.
 t/5730  100/5730
1 1 ∼
Since y(0) = 50, y(t) = 50 by the theorem above. Hence, y(100) = 50 =
2 2
49.3988.
ln 2
Theorem 64: If a quantity is decaying exponentially with decay rate r, then the half-life is − .
r
Proof: Exercise. ■

Exercises

Section 6.5: 1, 3, 9-17 odd.

17
 Chapter 9: Derivatives of Inverse Trigonometric Functions
Theorem 65:
d 1
(i) sin−1 (x) = √ .
dx 1 − x2
d −1
(ii) cos−1 (x) = √ .
dx 1 − x2
d 1
(iii) tan−1 (x) = .
dx 1 + x2
d −1
(iv) cot−1 (x) = .
dx 1 + x2
d 1
(v) sec−1 (x) = √ .
dx |x| x2 − 1
d −1
(vi) csc−1 (x) = √ .
dx |x| x2 − 1
Proof: For (i), let f (x) = sin(x) and g(x) = sin−1 (x). Then,
d
sin−1 (x)
dx
1
=
cos(sin−1 (x))
1
=√
1 − x2
as desired.
π π
For (ii), recall that sin−1 (x) + cos−1 (x) = . This implies that cos−1 (x) = − sin−1 (x). So,
2 2
d −1
cos (x)
dx
d π
= [ − sin−1 (x)]
dx 2
d
= − sin−1 (x)
dx
−1
=√
1 − x2
as desired.

The proofs of (iii)-(vi) are left as exercises. ■

Example 66: Calculate the following.

d
(a) tan−1 (3x + 1)
dx
d 1 3
By the chain rule and the theorem above, tan−1 (3x + 1) = 2
·3= 2 .
dx 1 + (3x + 1) 9x + 6x + 2
d
(b) cos−1 (x−1 ) − sec−1 (x)
dx
By the theorem above,
d
cos−1 (x−1 ) − sec−1 (x)
dx
1 1 1
= −√ ·− 2 − √
1−x −2 x |x| x2 − 1
1 1
= √ − √
x2 1 − x−2 |x| x2 − 1
1 1
= r − √
2 |x| x 2−1
x −1
x2 2
x

18
 1 1
= √ − √
|x| x − 1 |x| x2 − 1
2

= 0.

Recall from calculus 1 that this implies that cos−1 (x−1 ) − sec−1 (x) is a constant.

Corollary 67:
1 1 x
Z
(i) 2 2
dx = arctan + C;
Z x +a a a
1  x 
(ii) √ dx = arcsin + C;
2
Z a −x
2 a
1 1  x
(iii) √ dx = sec−1 + C.
x x2 − a2 a a
Proof: Exercise. ■

Example 68: Integrate.

1 1 x
Z
(a) 2
dx = arctan +C
Z x +4 2 2
1  x 
(b) √ dx = arcsin +C
9 − x 2 3  
1 1 x
Z
(c) √ dx = √ sec−1 √ + C.
x x2 − 3 3 3

Exercises

Section 6.6: 17-33 odd, 39, 43-45 odd, 49, 51, 53, 59-70 (omit 63, 65, 68).

1. Prove corollary 67.

19
 Chapter 10: Derivatives of Exponential Functions
Lemma 69: Let f (x) = ex . Then f ′ (0) = 1.

Proof: Proved in a later chapter. ■

d x
Theorem 70: e = ex .
dx
Proof: Note the following.
d x
e
dx
ex+h − ex
= lim
h→0 h
ex eh − ex
= lim
h→0 h
x eh − 1
= e lim
h→0 h
= ex f ′ (0)

= ex · 1

= ex . ■
d x
Corollary 71: Suppose a > 0 with a ̸= 1. Then a = ax ln(a).
dx
Proof: Note the following.
d x
a
dx
d ln(ax )
= e
dx
d x ln(a)
= e
dx
= ex ln(a) ln(a)

= ax ln(a). ■

Example 72: Differentiate the following.

(a) f (x) = 4x

By the corollary above, f ′ (x) = 4x ln(4).

3
+x2 +1
(b) g(x) = ex
3
+x2 +1
By the chain rule, g ′ (x) = ex · (3x2 + 2x).
2
ex + 1
(c) h(x) =
ex 2 2 2 2 2
ex [2xex ] − ex [ex + 1]
′ 2xex − ex − 1 ex (2x − 1) + 1
By the quotient and chain rules, h (x) = = = .
[ex ]2 ex ex
3
(d) f (x) = esin(x )
3 3
By two applications of the chain rule, f ′ (x) = esin(x ) · cos(x3 ) · 3x2 = 3x2 cos(x3 )esin(x ) .
2
Example 73: Sketch the graph of the function f (x) = e−x .

By the definition of f , the line y = 0 is a horizontal asymptote of the graph of f and there are
2
no vertical or horizontal asymptotes. Note that f ′ (x) = −2xe−x . If x ∈ [0, ∞), then j ′ (x) < 0,

20
which means that f is decreasing on [0, ∞). If x ∈ (−∞, 0), then f ′ (x) > 0, which means that f is

increasing on (−∞, 0). By the first derivative test, the point (0, 1) is a local maximum of f . Note
√ √
 
′′ −x2 1 1
that f (x) = 2e (x 2 + 1)(x 2 − 1). If x ∈ − √ , √ , then f ′′ (x) < 0, which means that
  2 2 
1 1 1 1
f is concave downward on − √ , √ . If x ∈ R \ − √ , √ , then f ′′ (x) > 0, which means
2 2  2 2    
1 1 1 1 1 1
that f is concave upward on R \ − √ , √ . Note that the point − √ , √ , √ , √ is an
2 2 2 e 2 e
inflection point. A sketch of the graph of f is below.

Example 74: Sketch the graph of the function f (x) = xex .

x 1
Note that lim xex = lim −x = lim = lim −ex = 0. Similarly, lim xex = ∞. By
x→−∞ x→−∞ e x→−∞ −e−x x→−∞ x→∞

the definition of horizontal asymptote, the line y = 0 is a horizontal asymptote of the graph of f .

Also, there are no vertical or oblique asymptotes. Note that f ′ (x) = xex + ex = ex (x + 1). If x ≤ −1,

then f ′ (x) ≤ 0, and so f is decreasing on the interval (−∞, −1]. If x ≥ −1, then f ′ (x) ≥ 0, and
 
1
so f is increasing on the interval [−1, ∞). Also, note that the point −1, − is a local minimum
e
and there are no local maxima. Note that f ′′ (x) = ex (x + 1) + ex = ex (x + 2). If x < −2, then

f ′′ (x) < 0, and so f is concave downward on the interval (−∞, −2). If x > −2, then f ′′ (x) > 0,
 
2
and so f is concave upward on the interval (−2, ∞). This implies that the point −2, − 2 is an
e
inflection point. A sketch of the graph of f is below.

21
 Exercises

Section 6.2: 31-61 odd, 67-75 odd, 83-93 odd.

Section 6.6: 68.

22
 Chapter 11: Derivatives of Logarithmic Functions
1
Theorem 75: Let f (x) = ln(x). Then f ′ (x) = .
x
Proof: We will use implicit differentiation. Let y = ln(x). Then,

ey = x
d y d
⇐⇒ ey y ′ = e = x=1
dx dx
1 1 1
⇐⇒ y ′ = y = ln(x) = . ■
e e x
1
Corollary 76: Let f (x) = ln |x|. Then f ′ (x) = .
x
d
Proof: If x > 0, then ln |x| = ln(x), and apply the theorem above. If x < 0, then ln |x| =
dx
d 1 d 1
ln(−x) = (−x) = . ■
dx −x dx x
1
Corollary 77: Let f (x) = loga (x). Then f ′ (x) = .
x ln(a)
d d ln(x) 1 d 1
Proof: Note that loga (x) = = · ln(x) = . ■
dx dx ln(a) ln(a) dx x ln(a)
Example 78: Differentiate the following.

(a) f (x) = log3 (x)

1
f ′ (x) = .
x ln(3)
(b) g(x) = ln |2x15 + x12 − 3|
30x14 + 12x11
g ′ (x) = 15 .
2x + x12 − 3
(c) h(x) = ln(sin2 (x))

h′ (x) = 2 csc2 (x) cos(x) = 2 csc(x) cot(x).

Exercises

Section 6.4: 3-37 odd, 41, 61-65 odd, 71-85 odd, 89.

Section 6.6: 63, 65, 68.

23
 Chapter 12: Logarithmic Differentiation
d f ′ (x)
Proposition 79: For any function f (x), ln[f (x)] = .
dx f (x)
Proof: Use the chain rule. ■

Example 80: Differentiate the following.

(x + 1)2 (2x2 − 3)
(a) f (x) = √
x2 + 1
Note the following.
(x + 1)2 (2x2 − 3)
f (x) = √
x2 + 1 √
⇐⇒ ln[f (x)] = ln[(x + 1)2 ] + ln(2x2 − 3) − ln( x2 + 1)
1
= 2 ln(x + 1) + ln(2x2 − 3) − ln(x2 + 1)
2
f ′ (x) 2 4x 2x
⇐⇒ = + 2 − 2
f (x) x + 1 2x − 3 2(x + 1)
2 4x x
= + 2 − 2
x + 1 2x − 3 x + 1
(x + 1)2 (2x2 − 3) 2 4x x
⇐⇒ f ′ (x) = √ [ + 2 − 2 ].
2
x +1 x + 1 2x − 3 x + 1
(b) g(x) = xx (for x > 0)

Note the following.

g(x) = xx

⇐⇒ ln[g(x)] = x ln(x)
g ′ (x)
⇐⇒ = 1 + ln(x)
g(x)
⇐⇒ g ′ (x) = xx (1 + ln(x)). ■
d f ′ (x) g ′ (x)
Proposition 81: If f, g are differentiable functions, then ln[f (x)g(x)] = + .
dx f (x) g(x)
Proof: Exercise. ■
d
Corollary 82 (Product Rule): If f, g are differentiable functions, then f (x)g(x) = f (x)g ′ (x) +
dx
f ′ (x)g(x).

Proof: Exercise. ■

Exercises

Section 6.4: 43-57 odd.

2. Prove proposition 81.

3. Prove corollary 82.

24
 Chapter 13: Hyperbolic Functions
Definition 83: The hyperbolic functions are defined as follows.
ex − e−x 1 2
(i) sinh x = ; (iv) csch(x) = = x ;
2 sinh(x) e − e−x
x −x 1 2
e +e
(ii) cosh x = ; (v) sech(x) = = x ;
2 cosh(x) e + e−x
x −x x −x
sinh(x) e −e 1 e +e
(iii) tanh x = = x ; (vi) coth(x) = = x .
cosh(x) e + e−x tanh(x) e − e−x

Proposition 84:

(i) The function sinh(x) is odd.

(ii) The function cosh(x) is even.

(iii) The function f (x) = 1 + tanh(x) is a logistic function.

Proof: Exercise. ■

Proposition 85:

(i) For any x ∈ R, sinh(x) + cosh(x) = ex .

(ii) For any x ∈ R, cosh(x) − sinh(x) = e−x .

Proof: Exercise. ■
2 2
Corollary 86: cosh (x) − sinh (x) = 1 for all x ∈ R.

Proof: By the proposition above, cosh(x) + sinh(x) = ex and cosh(x) − sinh(x) = e−x . Hence,

[cosh(x) + sinh(x)][cosh(x) − sinh(x)] = ex · e−x

⇐⇒ cosh2 (x) − sinh2 (x) = 1,

as desired. ■

Theorem 87: Let x, y ∈ R.

(i) sinh(x + y) = sinh(x) cosh(y) + cosh(x) sinh(y);

(ii) cosh(x + y) = cosh(x) cosh(y) + sinh(x) sinh(y);

tanh(x) + tanh(y)
(iii) tanh(x + y) = .
1 + tanh(x) tanh(y)
Proof: Exercise. ■

Corollary 88: Let x ∈ R.

(i) sinh(2x) = 2 sinh(x) cosh(x);

(ii) cosh(2x) = cosh2 (x) + sinh2 (x).

Proof: Exercise. ■

Theorem 89:
d
(i) sinh(x) = cosh(x);
dx
d
(ii) cosh(x) = sinh(x);
dx

25
 d
(iii) tanh(x) = sech2 (x);
dx
d
(iv) coth(x) = −csch2 (x);
dx
d
(v) sech(x) = −sech(x) tanh(x);
dx
d
(vi) csch(x) = −csch(x)coth(x).
dx

Exercises

Section 6.7: 1-19, 31-39 odd.

4. Prove proposition 84.

5. Prove proposition 85.

6. Prove theorem 87.

7. Prove corollary 88.

26
 Chapter 14: Integration by Parts
Theorem 90 (Integration by Parts): Let u = f (x), v = g(x) have continuous derivatives.
Z Z
(i) f (x)g (x) dx = f (x)g(x) − g(x)f ′ (x) dx;
′
Z Z
(ii) u dv = uv − v du.

Proof: For (i), by the product rule for derivatives and the Fundamental Theorem of Calculus, we

have the following.

d
f (x)g(x) = f (x)g ′ (x) + g(x)f ′ (x)
dx Z
d
Z Z
⇐⇒ f (x)g(x) = f (x)g (x) dx + g(x)f ′ (x) dx
′
dx Z Z
⇐⇒ f (x)g(x) = f (x)g (x) dx + g(x)f ′ (x) dx
′
Z Z
⇐⇒ f (x)g ′ (x) dx = f (x)g(x) − g(x)f ′ (x) dx

Now, for (ii), compute the differentials and apply (i). ■

Example 91: Evaluate the following.

Z
(a) x sin(x) dx.
Z
Set u = x. Then, du = dx and v = sin(x) dx = − cos(x). This implies the following.
Z
x sin(x) dx
Z
= −x cos(x) + cos(x) dx

= −x cos(x) + sin(x) + C.
Z
(b) x ln(x) dx.
1 1
Z
Set u = ln(x). Then du = dx and v = x dx = x2 . This implies the following.
Z x 2
x ln(x) dx
1 1
Z
= x2 ln(x) − x dx
2 2
1 1
= [x2 ln(x) − x2 ] + C.
2Z 2
x
(c) xe dx.
Z
Set u = x. Then du = dx and v = ex dx = ex . This implies the following.
Z
xex dx
Z
= xex − ex dx

= xex − ex + C

= ex [x − 1] + C.
Z
(d) x2 cos(x) dx.

27
 Z
Set u = x2 . Then du = 2x dx and v = cos(x) dx = sin(x). This implies the following.
Z Z
2 2
x cos(x) dx = x sin(x) − 2x sin(x) dx

= x2 sin(x) − 2[−x cos(x) + sin(x)] + C

= x2 sin(x) + 2x cos(x) − 2 sin(x) + C

= (x2 − 2) sin(x) + 2x cos(x) + C.

Z
(e) ln(x) dx
1
Z
Set u = ln(x). Then du = dx and v = 1 dx = x. This implies the following.
Z x
ln(x) dx
Z
= x ln(x) − 1 dx

= x ln(x) − x + C.
Zb
Proposition 92: Let u = f (x), v = g(x) have continuous derivatives on [a, b]. Then u dv =
a

b
Zb
uv − v du.
a
a
Example 93: Evaluate the following.
Z3
(a) ln(x) dx.
1
Z3 3
Z3
By the example and proposition above, ln(x) dx = x ln(x) − 1 dx = 3 ln(3) − 2.
1
Z 1 1
x
(b) e cos(x) dx.
Z
Set u = cos(x). Then du = − sin(x) dx and v = ex dx = ex . This implies the following.
Z Z
ex cos(x) dx = ex cos(x) + ex sin(x) dx
Z Z
x x
Set u = sin(x). Then du = cos(x) dx and v = e dx = e . This implies that ex sin(x) dx =
Z
e sin(x) − ex cos(x) dx.
x

Hence,
Z Z
e cos(x) dx = e cos(x) + e sin(x) − ex cos(x) dx
x x x
Z
⇐⇒ 2 ex cos(x) dx = ex cos(x) + ex sin(x)
1 1
Z
⇐⇒ ex cos(x) dx = [ex cos(x) + ex sin(x)] + C = ex [cos(x) + sin(x)] + C.
2 2
Proposition 94: Let n ∈ N. Then,
Z Z
(i) xn ex dx = xn ex − n xn−1 ex dx;
Z Z
(ii) (ln x)n dx = n[ln(x)]n − n [ln(x)]n−1 dx;

28
 Z Z
(iii) xn cos x dx = xn sin x − n xn−1 sin x dx;
Z Z
(iv) xn sin x dx = −xn cos x + n xn−1 cos x dx.
Z
Proof: For (i), set u = xn . Then du = nxn−1 dx and v = ex dx = ex . Now,
Z
xn ex dx
Z
= xn ex − nxn−1 ex dx
Z
= x e − n xn−1 ex dx.
n x

The proofs of (ii)-(iv) will be left as exercises for the reader. ■

Z
Example 95: Evaluate x3 ex dx.

By three applications of part (i) of the proposition above, we obtain the following.
Z
x3 ex dx
Z
= x e − 3 x2 ex dx
3 x
Z
= x3 ex − 3(x2 ex − 2 xex dx)
Z
= x3 ex − 3x2 ex − 6 xex dx

= x3 ex − 3x2 ex − 6[xex − ex ] + C

= x3 ex − 3x2 ex − 6xex − 6ex + C

= ex [x3 − 3x2 − 6x − 6] + C.

Exercises

Section 7.1: 1-23 odd.

8. Complete the proof of proposition 94.

Za
9. Prove that for any Riemann integrable function f and any a ∈ R, f (x) dx = af (a) −
0
Za
xf ′ (x) dx.
0
10. Let f be a twice differentiable function and suppose that f (0) = f (1) = 0. If f ′′ (x) = λf (x) for

some λ ∈ R, prove that λ < 0.

29
 Chapter 15: Integrals of Trigonometric Functions
Z
Example 96: Evaluate sin3 (x) dx.
Z Z
By trigonometry, sin3 (x) dx = sin(x)[1 − cos2 (x)] dx. Set u = − cos(x). Then du = sin(x) dx

and 1 − cos2 (x) = 1 − u2 . This implies the following.

Z Z
sin3 (x) dx = 1 + u2 du
1
= u + u3 + C
3
1
= − cos(x) − cos3 (x) + C.
3 Z
Example 97: Evaluate sin4 (x) cos5 (x) dx.
Z Z
By trigonometry, sin4 (x) cos5 (x) dx = sin4 (x)(1 − sin2 (x))2 cos(x) dx. Set u = sin(x). Then

du = cos(x) dx. Now,

Z
sin4 (x) cos5 (x) dx
Z
= u4 (1 − u2 )2 du
Z
= u4 (1 − 2u2 + u4 ) du
Z
= u4 − 2u6 + u8 du
1 5 2 7 1 9
= u − u + u +C
5 7 9
1 2 1
= sin (x) − sin7 (x) + sin9 (x) + C.
5
5 7 9
Theorem 98 (Reduction Formulas for Sine and Cosine): Let n ∈ N.
1 n−1
Z Z
n n−1
(i) sin (x) dx = − sin (x) cos(x) + sinn−2 (x) dx
n n Z
1 n−1
Z
(ii) cosn (x) dx = cosn−1 (x) sin(x) + cosn−2 (x) dx.
n n
Proof: Exercise. ■
Z
Example 99: Evaluate sin4 (x) dx.

By two applications of the theorem above, we get the following.

Z
sin4 (x) dx
 
1 3 1 1
= − sin3 (x) cos(x) + − sin(x) cos(x) + x + C
4 4 2 2
1 3 3 3
= − sin (x) cos(x) − sin(x) cos(x) + x + C.
4 8 Z 8
2 4
Example 100: Evaluate sin (x) cos (x) dx.
Z Z Z
By trigonometry, sin2 (x) cos4 (x) dx = [1 − cos2 (x)] cos4 (x) dx = cos4 (x) − cos6 (x) dx.

Now, by several applications of the theorem above, we get the following.

Z
sin2 (x) cos4 (x) dx

30
 Z
= cos4 (x) − cos6 (x) dx
1 1 1 1
= − cos5 (x) sin(x) + cos3 (x) sin(x) + cos(x) sin(x) + x + C.
6 24 16 16
Proposition 101:
Z
(i) tan(x) dx = ln | sec(x)| + C;
Z
(ii) sec(x) dx = ln | sec(x) + tan(x)| + C.
sin(x)
Proof: For (i), recall that tan(x) = . Set u = cos x. Then du = − sin x dx, or −du = sin x dx.
cos(x)
Now,
Z
tan x dx
sin x
Z
= dx
Zcos x
1
=− du
u
= − ln |u| + C

= − ln | cos x| + C

= ln | sec x| + C.
sec(x) + tan(x) sec2 (x) + sec(x) tan(x)
For (ii), note that sec(x) = sec(x) · = .
sec(x) + tan(x) sec(x) + tan(x)
2
sec (x) + sec(x) tan(x)
Z Z
So, sec(x) dx = dx. Set u = sec(x) + tan(x). Then du = sec x tan x +
sec(x) + tan(x)
sec2 x dx. Now,
Z
sec(x) dx
sec2 (x) + sec(x) tan(x)
Z
= dx
sec(x) + tan(x)
1
Z
= du
u
= ln |u| + C

= ln | sec(x) + tan(x)| + C. ■
Zπ/4
Example 102: Evaluate tan3 (x) dx.
0
Zπ/4 Zπ/4 Zπ/4
3 2
By trigonometry, tan (x) dx = (sec (x) − 1) tan(x) dx = sec2 (x) tan(x) − tan(x) dx =
0 0 0
Zπ/4 Zπ/4
2
sec (x) tan(x) dx − tan(x) dx.
0 0
π
Set u = tan(x). Then du = sec2 (x). Also, if x = 0, then u = 0 and if x = , then u = 1.
4
This implies the following.
Zπ/4
tan(x) sec2 (x) dx
0

31
 Z1
= u du
0
1 21
= u
2 0
1
= .
2
Also, by the proposition above,
Zπ/4
tan(x) dx
0
π/4
= ln | sec(x)|
0
√
= ln( 2)
1
= ln(2).
2
Zπ/4
1 1 1
Thus, tan3 (x) dx = − ln(2) = [1 − ln(2)].
2 2 2
0
Proposition 103 (Reduction Formulas for Secant and Tangent): Let n ∈ Z.
tan(x) secn−2 (x) n − 2
Z Z
(i) secn (x) dx = + secn−2 (x) dx;
n − 1 n − 1
tann x
Z Z
(ii) tann (x) dx = − tann−2 x dx.
n−1
Proof: Exercise. ■
Z
Example 104: Evaluate tan2 (x) sec3 (x) dx.
Z Z Z
2 3 2 3
By trigonometry, tan (x) sec (x) dx = [sec (x) − 1] sec (x) dx = sec5 (x) − sec3 (x) dx =
Z Z
sec5 (x) dx− sec3 (x) dx. Now, by two applications of the proposition above, we get the following.
1 1 1
Z
tan2 (x) sec3 (x) dx = tan(x) sec3 (x) − tan(x) sec(x) − ln | sec(x) + tan(x)| + C.
4 8 8
Note 105: The following are other trigonometric integrals.
sinm+1 x cosn−1 x n−1 sinm−1 x cosn+1 x
Z Z
(i) sinm x cosn x dx = + sinm x cosn−2 x dx = − +
m+n m+n m+n
m−1
Z
sinm−2 x cosn x dx;
m +Zn
(ii) cot x dx = − ln | csc x| + C = ln | sin x| + C;
cotn−1 x
Z Z
(iii) cotn x dx = − − cotn−2 x dx (n ̸= 1);
Z n−1
(iv) csc(x) dx = ln | csc x − cot x| + C;
cot x cscn−2 x n − 2
Z Z
(v) cscn (x) dx = − + cscn−2 x dx (n ̸= 1);
n − 1 n − 1
sin(m − n)x sin(m + n)x
Z
(vi) sin(mx) sin(nx) dx = − + C (m ̸= ±n);
2(m − n) 2(m + n)
cos(m − n)x cos(m + n)x
Z
(vii) sin(mx) cos(nx) dx = − − + C (m ̸= ±n);
2(m − n) 2(m + n)
sin(m − n)x sin(m + n)x
Z
(viii) cos(mx) cos(nx) dx = + + C (m ̸= ±n).
2(m − n) 2(m + n)

32
 Exercises

Section 7.2: 1-49 odd, 55, 57.

11. Prove theorem 98.

12. Prove proposition 103.

13. (Wallis Product) In this exercise, you will verify a famous product called the Wallis product
π
by proving the intermediate claims. This product is a way to express as a product.
2
Zπ/2
For each m ∈ ω, set Im = sinm x dx.
0
π
Claim 1: I0 = and I1 = 1.
2
m−1
Claim 2: For each m ≥ 2, Im = · Im−2 .
m
Claim 3: For each m ∈ N, I2m+1 ≤ I2m ≤ I2m−1 .
I2m 1
Claim 4: For each m ∈ ω, 1 ≤ ≤1+ .
I2m+1 2m
I2m
By claim 4 and the Squeeze Theorem, lim = 1.
m→∞ I2m+1
m
π Y 4k 2
Claim 5: = lim .
2 m→∞ 4k 2 − 1
k=1

Chapter 16: Trigonometric Substitution

Note 106:

√you see you sub √ you use

2 2 2 2
√a − x x = a sin θ √ a − x = a cos θ
2 2 2 2
√x − a x = a sec θ √x − a = a tan θ
x2 + a2 x = a tan θ x2 + a2 = a tan θ

Example 107: Evaluate the following.

Z p
(a) 1 − x2 dx.

Set x = sin(θ). Then dx = cos(θ) dθ. This implies the following.

Z p Z q
1 − x2 dx = 1 − sin2 (θ) cos(θ) dθ
Z
= cos2 (θ) dθ
1 1
= θ − sin(θ) cos(θ) + C
2 2
1
= [θ − sin(θ) cos(θ)] + C
2
1
= [sin−1 (x) − x cos(sin−1 (x)] + C
2
1  −1 √ 
= sin (x) − x 1 − x2 + C.
2

33
 x2
Z
(b) dx.
(4 − x2 )3/2 √
Set x = 2 sin(θ). Then, dx = 2 cos(θ) dθ and note that 2 cos(θ) = 4 − x2 . This implies the follow-

ing.
x2
Z
dx
(4 − x2 )3/2
4 sin2 (θ)
Z
= 2 cos(θ) dθ
8 cos3 (θ)
2
sin (θ)
Z
= 2
dθ
Z cos (θ)
= tan2 (θ) dθ

= tan(θ) − θ + C
x x
=√ − sin−1 + C.
Z4 − x2 2
p
(c) 4x2 + 20 dx.
√ √
Z p Z p
Note that 2
4x + 20 dx = 2 x2 + 5 dx. Set x = 5 tan(θ). Then dx = 5 sec2 (θ) and note
√ √
that x2 + 5 = 5 sec(θ). This implies the following.
Z p
4x2 + 20 dx
Z √ h√ i
=2 5 sec(θ) 5 sec2 (θ) dθ
Z
= 10 sec3 (θ) dθ

= 5 tan(θ) sec(θ) + 5 ln[sec(θ) + tan(θ)] + C

= 5[tan(θ) sec(θ) + ln[sec(θ) + tan(θ)]] + C

             
x x x x
= 5 tan tan−1 √ sec tan−1 √ + ln sec tan−1 √ + tan tan−1 √ +
5 5 5 5
C " √ " √ ##
x x2 + 5 x + x2 + 5
=5 + ln √ +C
5 5
√ √ 1
= x x2 + 5 + 5 ln[x + x2 + 5] + ln 5 + C
√ √ 2
= x x2 + 5 + 5 ln[x + x2 + 5] + C.
1
Z
(d) √ dx.
x x2 − 9
2
√
Set x = 3 sec(θ). Then dx = 3 sec(θ) tan(θ) dθ and 3 tan(θ) = x2 − 9. This implies the following.
1 1
Z Z
√ dx = 2 (θ) 3 tan(θ)
3 sec(θ) tan(θ) dθ
2 2
xZ x − 9 9 sec
1 1
= dθ
9 Z sec(θ)
1
= cos(θ) dθ
9
1
= sin(θ) + C
9
1   x 
= sin sec−1 +C
9√ 3
1 x −92
= +C
9 x

34
 √
x2 − 9
= + C.
9x
1 1
Z Z
By the completing the square lemma, note that dx = dx. Set
(x − 6x + 11)2
2 [(x − 3)2 + 2]2
u = x − 3. Then du = dx. Then,
1
Z
2 + 2]2
dx
[(x − 3)
1
Z
= du.
[u2 √+ 2]2 √ √
Set u = 2 tan(θ). Then, dθ = 2 sec2 (θ) du and u2 + 2 = sec(θ). This implies the following.
1
Z
du
[u2√+ 2]2
2 sec2 (θ)
Z
= dθ
[2 sec2 (θ)]2
Z √
2 sec2 (θ)
= dθ
4 Zsec2 (θ)
1 1
= √ 2 (θ)
dθ
2 2Z sec
1
= √ cos2 (θ) dθ
2 2 
1 θ 1
= √ + sin(θ) cos(θ) + C
2 2 2 2
1
= √ [θ + sin(θ) cos(θ)] + C
4 2        
1 −1 u −1 u −1 u
= √ tan √ + sin tan √ cos tan √ +C
4 2" 2 √ # 2 2
x−3
 
1 u 2
= √ tan−1 √ + 2 +C
4 2 2 (u + 2)2
" √ #
− −
 
1 x 3 (x 3) 2
= √ tan−1 √ + +C
4 2 2 [(x − 3)2 + 2]2
x−3 x−3
 
1
= √ tan−1 √ + 2 − 6x + 11]2
+ C.
4 2 2 4[x

Exercises

Section 7.4: 1-37 odd.

35
 Chapter 17: Partial Fraction Decomposition
p(x)
Definition 108: A rational function r(x) = is proper if deg(p) < deg(q).
q(x)
Example 109:
x2 − 3x + 7
(a) is proper.
x4 − 16
x−2
(b) is not proper.
x−5
Note 110: By the Fundamental Theorem of Algebra, any polynomial g(x) can be expressed as a

product of distinct linear functions and irreducible monic quadratic functions.

p(x)
Definition 111: Let r(x) = be a rational function. A partial fraction decomposition
q(x)
m X ki n X si
X Aij X Bij x + Cij
of r(x) is the following double sum: r(x) = j
+ , where each
i=1 j=1
(x − r i ) i=1 j=1
[qi (x)]j
Aij , Bij , Cij ∈ R.
p(x)
Theorem 112: Let r(x) = be a rational function. Then r(x) has a unique partial fraction
q(x)
decomposition up to rearrangement of terms.

Example 113: For each of the following, find the partial fraction decomposition.
x + 11
(a) f (x) = 2 .
x − 2x − 15
Note that x2 − 2x − 15 = (x − 5)(x + 3). Then,
x + 11 A B
= +
x2 − 2x − 15 x−5 x+3
⇐⇒ x + 11 = A(x + 3) + B(x − 5).

If x = −3, then 8 = −8B, and so, B = −1.

If x = 5, then 16 = 8A, and so, A = 2.

x + 11 2 1
Thus, 2 = − .
x − 2x − 15 x−5 x+3
2
3x − 7x − 2
(b) g(x) = .
x3 − x
Note that x3 − x = x(x + 1)(x − 1). Then,
3x2 − 7x − 2 A B C
= + +
x3 − x x x+1 x−1
⇐⇒ 3x2 − 7x − 2 = A(x + 1)(x − 1) + Bx(x − 1) + Cx(x + 1).

If x = 0, then −2 = −A, and so, A = 2.

If x = 1, then −6 = 2C, and so, C = −3.

If x = −1, then 8 = 2B, and so, B = 4.

3x2 − 7x − 2 2 4 3
Thus, = + − .
x3 − x x x+1 x−1
2x3 − 4x2 − 15x + 5
(c) h(x) = .
x2 − 2x − 8
2x3 − 4x2 − 15x + 5 x+5
Note that by long division of polynomials, 2
= 2x + 2 . Now, x2 −
x − 2x − 8 x − 2x − 8

36
2x − 8 = (x − 4)(x + 2). Then,
x+5 A B
2
= +
x − 2x − 8 x−4 x+2
⇐⇒ x + 5 = A(x + 2) + B(x − 4).
1
If x = −2, then 3 = −6B, and so, B = − .
2
2
If x = 4, then 9 = 6A, and so, A = .
3
2x3 − 4x2 − 15x + 5 2/3 1/2
Thus, 2
= 2x + − .
x − 2x − 8 x−4 x+2
2x2 − 5x + 2
(d) Find the partial fraction decomposition of f (x) = .
x3 + x
Note that x3 + x = x(x2 + 1). Then,
2x2 − 5x + 2 A Bx + C
= + 2
x3 + x x x +1
⇐⇒ 2x − 5x + 2 = A(x2 + 1) + (Bx + C)x
2

= (A + B)x2 + Cx + A.

Matching coefficients, we get the following system.

A+B =2

C = −5

A=2
2x2 − 5x + 2 2 5
Note that from the first and last equations, we get B = 0. Thus, = − 2 .
x3 + x x x +1
Note 114: Recall that rational functions are integrable where they are continuous.
1
Z
Example 115: Evaluate dx.
x2 − 7x + 10
Note that x2 − 7x + 10 = (x − 5)(x − 2). Then,
1 A B
= +
x2 − 7x + 10 x−5 x−2
⇐⇒ 1 = A(x − 2) + B(x − 5).
1
If x = 2, then −3B = 1, which means B = − .
3
1
If x = 5, then 3A = 1, which means A = .
3
1 1/3 1/3
Now, 2 = − .
x − 7x + 10 Z x − 2 x − 5
1 1/3 1/3 1 1
Z
Integrating, we get 2
dx = − dx = ln |x − 2| − ln |x − 5| + C.
x − Z 7x 3+ 10 x−2 x−5 3 3
x +1
Example 116: Evaluate dx.
x2 − 4
x3 + 1 4x + 1 9/4 7/4
Note that 2 =x+ 2 =x+ + . This implies the following.
Z 3 x −4 x −4 x+2 x−2
x +1
2
dx
Zx − 4
9/4 7/4
= x+ + dx
x+2 x−2
1 9 7
= x2 + ln |x + 2| + ln |x − 2| + C.
2 4 4 Z
18
Example 117: Evaluate dx.
(x + 3)(x2 − 9)

37
 18 1/2 1/2 3
Note that 2
= − − . This implies the following.
(x + 3)(x − 9) x − 3 x + 3 (x + 3)2
18
Z
2
dx
Z(x + 3)(x − 9)
1/2 1/2 3
= − − dx
x − 3 x + 3 (x + 3)2
1 1
= ln |x − 3| − ln |x + 3| + 3(x + 3)−1 + C.
2 2
18
Z
Example 118: Evaluate dx.
(x + 3)(x2 + 9)
18 1 3−x
Note that 2 + 9)
= + 2 . This implies the following.
(x + 3)(x x + 3 x +9
18
Z
dx
(x + 3)(x2 + 9)
1 x
= ln |x + 3| − ln(x2 + 9) + tan−1 + C.
2 3
4−x
Z
Example 119: Evaluate dx.
x(x2 + 2)2
4−x 1 1 2x − 5
Note that 2 2
= − 2 + 2 . This implies the following.
(x + 2) x x + 2 (x + 2)2
4−x
Z
2 + 2)2
dx
x(x
1 1 2x − 5
Z
= − + 2 dx
x x2 + 2 (x  + 2)2 Z
1 x 2x 1
Z
= ln |x| − √ tan−1 √ + 2 + 2)2
dx − 5 2 + 2)2
dx
2  2  (x (x
1 x 1 1
Z
= ln |x| − √ tan−1 √ − 2 −5 2 + 2)2
dx
2 2 x + 2 (x
  "   √ #
1 x 1 5 x x 2
= ln |x| − √ tan−1 √ − 2 − √ tan−1 √ + 2 + C.
2 2 x +2 4 2 2 (x + 2)2
Example 120: Integrate.
Z √
x−1
(a) dx
2x
√
Set u = x − 1. Then u2 = x − 1, x = u2 + 1, and dx = 2u du. So,
Z √
x−1
dx
Z 2x
u
= 2 + 1)
2u du
2(u
u2
Z
= du.
u2 + 1
Set u = tan θ. Then u2 = tan2 θ, u2 + 1 = sec2 θ, and du = sec2 θ dθ. Hence,
u2
Z
2
du
Zu + 12
tan θ
= 2
· sec2 θ dθ
Z sec θ
= tan2 θ dθ

= tan2 θ − θ + C

= [tan(tan−1 u)]2 − tan−1 u + C

= u2 − tan−1 u + C
√
= x − 1 − tan−1 ( x − 1) + C.

38
 1
Z
(b) √ dx
x√+4+4 x+1
Set u = x + 1. Then u2 = x + 1, u2 + 3 = x + 4, x = u2 − 1, and dx = 2u du. Hence,
1
Z
√ dx
Zx + 4 + 4 x + 1
1
= 2
· 2u du
Z u + 4u + 3
2u
= 2 + 4u + 3
du
u
2u
Z
= du
(u + 1)(u + 3)
1 3
Z
= − +
u+1 u+3
= − ln |u + 1| + 3 ln |u + 3| + C
√ √
= − ln | x + 1 + 1| + 3 ln | x + 1 + 3| + C.
x
Z
(c) √ dx
√+1
x
Set u = x + 1. Then x = u2 − 1 and dx = 2u du. So,
x
Z
√ dx
Z + x 1
u2 − 1
= du
u
1
Z
= u − du
u
1
= u2 − ln |u| + C
2
1 1
= (x + 1) − ln |x + 1| + C.
2 2

Exercises

Section 7.1: 37-41 odd.

Section 7.4: 1-51 odd.

Section 7.5: 1-83 odd.

p(x)
14. Suppose that q(x) = (x − a)(x − b) where a, b ∈ R with a ̸= b and let
be a proper rational
q(x)
p(x) A B p(a) p(b)
function such that = + . Prove that A = ′ and B = ′ .
q(x) x−a x−b q (a) q (b)

39
 Chapter 18: Improper Integrals
Zt
Definition 121: Let a ∈ R and suppose that f : [a, ∞) → R such that f (x) dx exists for every
 t a 
Z 
t ∈ [a, ∞). The improper integral of f (x) over [a, ∞) is the set f (x) dx | t ∈ [a, ∞) .
 
a
Z∞
Notation 122: The improper integral of f (x) over [a, ∞) is usually denoted f (x) dx.
a
Zt
Definition 123: Let a ∈ R and suppose that f : [a, b] → R such that f (x) dx exists for
a
Zt Z∞
every t ∈ [a, ∞). If lim f (x) dx = ℓ ∈ R, then the improper integral f (x) dx is said to be
t→∞
a a
Z∞
convergent. Otherwise, the improper integral f (x) dx is said to be divergent.
a
Definition 124:
Zt Z∞
(i) If lim f (x) dx = ℓ ∈ R, then the integral f (x) dx converges to ℓ.
t→∞
a a
Zt Z∞
(ii) If lim f (x) dx = ∞, then the integral f (x) dx diverges to ∞.
t→∞
a a
Zt Z∞
(iii) If lim f (x) dx = −∞, then the integral f (x) dx diverges to −∞.
t→∞
a a
Za
Definition 125: Let a ∈ R and suppose that f : (−∞, a] → R such that f (x) dx exists for every
 at 
Z 
t ∈ (−∞, a]. The improper integral of f (x) over (−∞, a] is the set f (x) dx | t ∈ (−∞, a] .
 
t
Za
Notation 126: The improper integral of f (x) over (−∞, a] is usually denoted f (x) dx.
−∞
Za
Definition 127: Let a ∈ R and suppose that f : (−∞, a] → R such that f (x) dx exists for every
t
t ∈ (−∞, a].
Za Za
(i) If lim f (x) dx = ℓ ∈ R, then the improper integral f (x) dx is said to be convergent.
t→∞
t −∞
Za Za
(ii) If lim f (x) dx does not exist, then the improper integral f (x) dx is said to be divergent.
t→∞
t −∞
Definition 128:
Za Za
(i) If lim f (x) dx = ℓ ∈ R, then the integral f (x) dx converges to ℓ.
t→−∞
t −∞

40
 Za Za
(ii) If lim f (x) dx = ∞, then the integral f (x) dx diverges to ∞.
t→−∞
t −∞
Za Za
(iii) If lim f (x) dx = −∞, then the integral f (x) dx diverges to −∞.
t→−∞
t −∞
Example 129: Determine whether the following integrals are convergent or divergent.
Z∞ Z0 Z∞
1 1
(a) dx (b) ex dx (c) 2+4
dx
e x x
1 1
−∞
Zt 0 Zt
Z 1
= lim e−x dx x = lim dx
t→∞
= lim e dx t→∞ x 2+4
t→−∞
1 1
t  t
t 0 1 −1 x
= lim −e −x x = lim tan
t→∞
= lim e t→∞ 2 2 1
1 t→−∞ t  
−t −1 1 −1 t −1 1
= lim −e + e = lim 1 − e t = lim tan − tan
t→∞ t→−∞ 2 t→∞
2  2
1 1 π 1
= . = 1. = − tan −1
.
e 2 2 2
The integral converges. The integral converges. The integral converges.
Zt
Definition 130: Let a, b ∈ R such that f : [a, b) → R such that f (x) dx exists for all t ∈ [a, b).
 t a 
Z 
Then the improper integral of f over [a, b) is the set f (x) dx | t ∈ [a, b) .
 
a
Zb
Notation 131: The improper integral of f (x) over [a, b) is usually denoted f (x) dx.
a
Zt
Definition 132: Let a, b ∈ R and suppose that f : [a, b) → R such that f (x) dx exists for all
a
t ∈ [a, b).
Zt Zb
(i) If lim f (x) dx = ℓ ∈ R, then the improper integral f (x) dx is said to be convergent.
t→b−
a a
Zt Zb
(ii) If lim− f (x) dx does not exist, then the improper integral f (x) dx is said to be divergent.
t→b
a a
Definition 133:
Zt Zb
(i) If lim− f (x) dx = ℓ ∈ R, then the integral f (x) dx converges to ℓ.
t→b
a a
Zt Zb Zb
(ii) If lim− f (x) dx = ∞, then the integral f (x) dx diverges to ∞ and write f (x) dx = ∞.
t→b
a a a
Zt Zb Zb
(iii) If lim f (x) dx = −∞, then the integral f (x) dx diverges to −∞ and write f (x) dx =
t→b−
a a a
−∞.
Zb
Definition 134: Let a, b ∈ R such that f : (a, b] → R such that f (x) dx exists for all t ∈ (a, b].
t

41
  b 
Z 
Then the improper integral of f over (a, b] is the set f (x) dx | t ∈ (a, b] .
 
t
Zb
Notation 135: The improper integral of f (x) over (a, b] is usually denoted f (x) dx.
a
Zb
Definition 136: Let a, b ∈ R and suppose that f : (a, b] → R such that f (x) dx exists for all
t
t ∈ (a, b].
Zb Zb
(i) If lim+ f (x) dx = ℓ ∈ R, then the improper integral f (x) dx is said to be convergent.
t→a
t a
Zb Zb
(ii) If lim+ f (x) dx does not exist, then the improper integral f (x) dx is said to be divergent.
t→a
t a
Definition 137:
Zb Zb
(i) If lim+ f (x) dx = ℓ ∈ R, then the integral f (x) dx converges to ℓ.
t→a
t a
Zb Zb Zb
(ii) If lim f (x) dx = ∞, then the integral f (x) dx diverges to ∞ and write f (x) dx = ∞.
t→a+
t a a
Zb Zb Zb
(iii) If lim+ f (x) dx = −∞, then the integral f (x) dx diverges to −∞ and write f (x) dx =
t→a
t a a
−∞.
Zt
Theorem 138: Let a, b ∈ R and suppose that f : [a, b) → R such that f (x) dx for all x ∈ [a, b).
 a

f (x)
 if x ̸= b; Zb
Define g : [a, b] → R by g(x) = Then f (x) = ℓ if and only if the improper

0
 if x = b. a

Zb
integral g(x) = ℓ.
a
Zb
Theorem 139: Let a, b ∈ R and suppose that f : (a, b] → R such that f (x) dx for all x ∈ (a, b].
 t

f (x)
 if x ̸= a; Zb
Define g : [a, b] → R by g(x) = Then f (x) = ℓ if and only if the improper

0
 if x = a. a

Zb
integral g(x) = ℓ.
a
Z1
1
Example 140: Determine whether the integral dx converges or diverges.
x
0

42
 Z1 Z1 1
1 1
Note that dx = lim dx = lim ln x = ∞. Therefore, the integral diverges.
x t→0+ x t→0+ t
0 t
Theorem 141 (Comparison Theorem for Improper Integrals): Let a ∈ R and suppose that

f, g : [a, t] → R is integrable for all t ≥ a. Also, suppose that 0 ≤ f (x) ≤ g(x) for all x ≥ a.
Z∞ Z∞ Z∞ Z∞
(i) If g(x) dx converges, then f (x) dx converges and f (x) dx ≤ g(x) dx.
a a a a
Z∞ Z∞
(ii) If f (x) dx diverges, then g(x) dx diverges.
a a
Theorem 142 (Comparison Theorem for Improper Integrals): Let a ∈ R and suppose that

f, g : [a, t] → R is integrable for all t ∈ [a, b). Also, suppose that 0 ≤ f (x) ≤ g(x) for all x ∈ [a, b).
Zb Zb Zb Zb
(i) If g(x) dx converges, then f (x) dx converges and f (x) dx ≤ g(x) dx.
a a a a
Zb Zb
(ii) If f (x) dx diverges, then g(x) dx diverges.
a a
Theorem 143 (Comparison Theorem for Improper Integrals): Let a ∈ R and suppose that

f, g : [a, t] → R for all t ∈ (a, b]. Also, suppose that 0 ≤ f (x) ≤ g(x) for all x ∈ (a, b].
Zb Zb Zb Zb
(i) If g(x) dx converges, then f (x) dx converges and f (x) dx ≤ g(x) dx.
a a a a
Zb Zb
(ii) If f (x) dx diverges, then g(x) dx diverges.
a a
Z∞
1
Theorem 144: The integral dx converges if and only if p > 1.
xp
1
Z∞
1
Proof: Let J denote the p-integral dx. Then,
xp
a
J
Zt
1
= lim dx
t→∞ xp
a
x1−p t
= lim
t→∞ 1 − p a
t1−p a1−p
= lim − .
t→∞ 1 − p 1−p
a1−p
If p > 1, then 1 − p < 0 and lim t1−p = 0, which means that J = .
t→∞ p−1
If p < 1, then 1 − p > 0 and lim t1−p = ∞, which means that J diverges.
t→∞
Z∞
1
If p = 1, then J diverges since dx = lim ln t − ln a = ∞. ■
x t→∞
a
Example 145: Determine whether each of the following is convergent or divergent.
Z∞
1
(a) √ dx
3
x +1
1

43
 Z∞ Z∞
1 1 1 1
Note that √ < √ and the integral √ dx converges. So the integral √ dx
x3 + 1 x3 x3 x3 + 1
1 1
converges.
Z∞
x+1
(b) dx
x2 + 5x
1
Z∞
x+1 x 1 1
Note that 2 > 2 = and recall that the integral dx diverges. So the
x + 5x x + 5x x+5 x+5
1
Z∞
x+1
integral dx diverges.
x2 + 5x
1

Exercises

Section 7.8: 5-39 odd, 49-53 odd.

Z∞
Definition: The Laplace transform L (s) of a function f is defined by L (s) = f (x) · e−sx dx,
0
provided the integral converges.

15. Find the Laplace transform of each of the following functions.

(a) f (x) = c for any c ∈ R;

(b) g(x) = sin(αx);

(c) h(x) = cos(αx) for any s > 0;

(d) j(x) = eαx for any s > α.

Z∞
Definition: The Gamma function is defined to be Γ(n) = tn−1 e−t dt.
0
16.

(a) Show that the function Γ(n) converges for all n ≥ 1.

Note: The function Γ(n) converges for all n > 0.

(b) Show that Γ(n + 1) = nΓ(n).

(c) If n ∈ N, show that Γ(n) = n!.

Note: Part (c) implies that the Gamma function can be used to define the factorial of any real

number.

(d) Use the above to calculate the Laplace transform of the function f (x) = xn .

44
 Chapter 19: Sequences
Definition 146: A sequence in R is a function ϕ : N → R.

Notation 147:

(i) We usually write xn instead of ϕ(n) = xn .

(ii) A sequence is usually denoted {xn }∞

n=1 instead of ϕ : N → R.

Example 148: {1, −2, 3, −4, 5, . . .} = {(−1)n+1 · n}∞

n=1 .

Example 149 (Fibonacci Sequence): {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .}

Set x1 = 1 and x2 = 1. For n ≥ 2, define xn+1 = xn + xn−1 .

Example 150: {1, 2, 4, 8, 16, . . .} − {2n }∞

n=0 .

Exercises

Section 11.1: 3-18.

45
 Chapter 20: Induction
Note 151: Let P (n) denote a statement about n ∈ N.

(i) (Basis Step) Show that P (1) is true.

(ii) (Induction Hypothesis) Suppose that P (k) is true and use P (k) to show that P (k + 1) is

true.
n
X n(n + 1)
Example 152: Show that i= for every n ∈ N.
i=1
2
n
X n(n + 1)
Proof: We will proceed by induction on n. Let P (n) denote the statement “ i = .”
i=1
2
1
X 1+1
Since i=1= , P (1) is true. Now suppose that k ∈ N such that P (k) is true. To see that
i=1
2
P (k + 1) is true, consider the following.
k+1
X
i
i=1
k
X
=k+1+ i
i=1
k(k + 1)
=k+1+
2
2(k + 1) k(k + 1)
= +
2 2
2(k + 1) + k(k + 1)
=
2
(k + 2)(k + 1)
= .
2
Therefore, P (k + 1) is true, and so, the claim is proved by the Principle of Mathematical Induction.

Definition 153: Let n ∈ N. Then the factorial of n is n! = n(n − 1)(n − 2) . . . (3)(2)(1).

Example 154: Prove that for any n ∈ N with n > 4, 2n < n!.

Proof: We will proceed by induction on n. For each n ∈ N, let P (n) be the statement “2n < n!.”

Note that P (4) is true since 24 = 16 < 24 = 4!. Now suppose that P (k) is true for some k ∈ N. To

see that P (k + 1) is true, consider the following.

2k < k!

⇐⇒ 2 · 2k < (k + 1)k!

⇐⇒ 2k+1 < (k + 1)!.

Therefore, P (k + 1) is true, and so, the claim is proved by the Principle of Mathematical Induction.

46
 Exercises
n
X 1 − xn+1
17. For any n ∈ ω and any x ∈ R \ {1}, show that xi = .
i=0
1−x
18. For each n ∈ N, let Fn denote the nth term of the Fibonacci sequence.
n
X
(a) Show that Fi = Fn+2 − 1.
i=1
Xn
(b) Show that Fi2 = Fn+1 · Fn .
i=1 √
n n
R+ − R− 1± 5
(c) Show that Fn = √ , where R± = .
5 2
(d) Show that Fn+1 · Fn−1 = Fn2 + (−1)n .
n
X n!
19. Show that = 2n for each n ∈ N.
k!(n − k)!
k=0
n
X n!
20. Show that (−1)k = 0 for each n ∈ N.
k!(n − k)!
k=0
21. Show that for any x ∈ R and any n ∈ N, | sin(nx)| ≤ n| sin(x)|.

Definition: Let f : X → R be any function where X ⊆ R. The first difference of f is defined to

be δ(f ) = f (x + 1) − f (x).

22.

(a) Calculate δ(x2 ).

(b) For any two functions f and g and any c ∈ R, show that δ(f +g) = δ(f )+δ(g) and δ(cf ) = cδ(f ).

(c) Prove that for any k ∈ N, there exists a polynomial P (x) of degree k+1 such that δ(P ) = (x+1)k

and P (0) = 0.

Chapter 21: Convergent Sequences

Definition 155: A sequence {xn }∞
n=1 converges to a number ℓ ∈ R if for every ε > 0, there is an

N ∈ N such that |xn − ℓ| < ε for every n ≥ N .

Notation 156: If a sequence {xn }∞

n=1 converges to a number ℓ ∈ R, then we write lim xn = ℓ or
n→∞

xn → ℓ.

Definition 157:

(i) A sequence that converges to a real number is called convergent;

(ii) A sequence that does not converge is called divergent.

1
Example 158: lim = 0.
n→∞ n

Example 159: The sequence {(−1)n }∞ n=1 diverges.

n! n(n − 1)(n − 2) . . . (3)(2)(1) 1
Example 160: lim = lim = lim = 0.
n→∞ (n + 1)! n→∞ (n + 1)(n)(n − 1)(n − 2) . . . (3)(2)(1) n→∞ n + 1

Theorem 161: Suppose {xn }∞ ∞

n=1 , {yn }n=1 ⊆ R, xn → x, yn → y, and c ∈ R. Then,

47
(i) (xn + yn ) → x + y

(ii) (xn − yn ) → x − y

(iii) cxn → cx

(iv) xn yn → xy
1 1
(v) → (provided no division by zero)
xn x
xn x
(vi) → (provided no division by zero)
yn y
Theorem 162 (Squeeze Theorem for Sequences): Suppose {xn }∞ ∞ ∞
n=1 , {yn }n=1 , {zn }n=1 ⊆ R

such that xn ≤ yn ≤ zn for every n ∈ N. If xn → ℓ and zn → ℓ, then yn → ℓ.

sin n + n
Example 163: lim
n→∞ n
−1 + n sin n + n 1+n −1 + n 1+n
Recall that ≤ ≤ and lim = 1 = lim . Therefore, by the
n n n n→∞ n n→∞ n
sin n + n
Squeeze Theorem for Sequences, we obtain that lim = 1.
n→∞ n
1
Example 164: lim =0
n→∞ n!

Theorem 165: Suppose that {xn }∞ n=1 ⊆ R and define f : R → R by f (n) = xn . If lim f (x) = ℓ,
x→∞

then lim xn = ℓ.
n→∞

Note 166: The converse to the theorem above is not true. 


1
 if x ∈ N;
Example 167: Consider the sequence {1, 1, 1, 1, . . .} and the function f (x) =

x if x ∈
 / N.
arctan n π
Example 168: lim = 0 since arctan n ≤ .
n→∞ n 2
Theorem 169: For any {xn }∞ n=1 ⊆ R, lim x n = 0 if and only if lim |xn | = 0.
n→∞ n→∞
n1
Example 170: lim (−1)
n→∞ n
n1 1 1
Note that lim (−1) = lim = 0. By the theorem above, lim (−1)n = 0.
n→∞ n n→∞ n
( n   )∞
n→∞ n
X 1 i
Example 171: The sequence converges.
i=0
2
n=0
n  i
X 1
Proof: For every n ∈ ω, set sn = .
i=0
2
 n
1
Claim: For every n ∈ ω, sn = 2 − .
2  n
1
Proof (Claim): For each n ∈ ω, let P (n) denote the statement “sn = 2 − .” Note that
2
 0  0
1 1
s0 = = 1 = 2− , which means that P (0) is true. So suppose that P (k) is true for some
2 2
k ∈ ω. Then,
 k+1
1
sk +
2
 k  k+1
1 1
=2− +
2 2

48
  k  k+1
1 1
⇐⇒ sk+1 = 2 − +
2 2
 k  
1 1
=2+ −1 +
2 2
 k+1
1
=2− .
2
Hence, P (k + 1) is true, and so, the claim is proved by the Principle of Mathematical Induction.

■Claim

Now,

lim sn
n→∞  n
1
= lim 2 −
n→∞ 2
=2−0

= 2. ■

Exercises

Section 11.1: 23-55 odd.

49
 Chapter 22: Monotonic Sequences
Definition 172: Let {xn }∞ ∞
n=1 ⊆ R. Then {xn }n=1 is

(i) increasing if xn ≤ xn+1 for every n ∈ N;

(ii) decreasing if xn ≥ xn+1 for every n ∈ N;

(iii) monotonic if it is either increasing or decreasing.

Example 173:

(a) The sequence {n}∞ is increasing.

 n=1
∞
1
(b) The sequence is decreasing.
n n=1
 ∞
1
(c) The sequence (−1)n · is not monotonic.
n n=1
Definition 174: A sequence {xn }∞n=1 ⊆ R is

(i) bounded above if there is an m ∈ R such that xn ≤ m for all n ∈ N;

(ii) bounded below if there is an m ∈ R such that xn ≥ m for all n ∈ N.

(iii) bounded if it is bounded above and bounded below.

Theorem 175 (Monotone Convergence Theorem): Every monotonic bounded sequence con-

verges. )∞
n
(
X 1
Example 176: Prove that the sequence converges.
k!
( n
)∞k=1 n=1
X 1
Proof: Note that the sequence is increasing. So it remains to show that the sequence
k!
)∞ k=1 n=1
n n n
(
X 1 X 1 X 1
is bounded above. For each n ∈ N, set sn = i
and t n = .
k! i=0
2 k!
k=1 n=1 k=1
Claim: For all n ≥ 4, tn < sn .

Proof (Claim): For each n ∈ N, let P (n) denote the statement “tn < sn .” To see that P (4) is
31 41
true, note that s4 = and t4 = . Hence, P (4) is true. Now suppose that P (k) is true for some
16 24
k ∈ N. To see that P (k + 1) is true, recall example 146 and consider the following.

tk < sk
1 1
⇐⇒ tk + < sk + k+1
(k + 1)! 2
⇐⇒ tk+1 < sk+1 .

Hence, P (k + 1) is true, and so, the claim is proved by the Principle of Mathematical Induction.

■Claim )∞ )∞
n n
( (
X 1 X 1
Since the sequence i
is bounded above by 2, the sequence is bounded
i=0
2 k!
n=1 k=1 n=1
above by 2. Therefore, the sequence converges by the monotone convergence theorem (theorem

167). ■

50
 ( n
)∞
X 1
Note 177: The proof above shows that the sequence is bounded above by 3.
k!
k=1 n=1 )∞
n
(
X 1
Note 178: When we study power series, we will see that the sequence the sequence
k!
k=1 n=1
converges to e.

Exercises

Section 11.1: 69-83 odd, 90.

23. Set an = Hn − ln n, where Hn is the nth harmonic number.

(a) Show that for every an ≥ 1 for every n ∈ N.

(b) Show that the sequence {an }∞

n=1 is decreasing.

(c) Conclude that the sequence {an }∞

n=1 is convergent.

Note: The limit above, denoted γ, is called Euler’s constant. It appears in many areas of math-

ematics, including analysis and number theory, and has been calculated to more than 100 million

decimal places, but it is still not known whether it is rational or not. The first ten digits of γ are

0.5772156649.
√ √
24. Set a1 = 2. For each n ∈ N with n > 1, set an+1 = 2 + an .

(a) Show that an < 2 for all n ∈ N.

(b) Show that the sequence {an } is increasing.

(c) Conclude that the sequence {an } is convergent and compute the limit.

51
 Chapter 23: The Binomial Theorem
n
X n!
Theorem 179 (Binomial Theorem): For every a, b ∈ R and every n ∈ N, (a+b)n = an−k bk .
k!(n − k)!
k=0
n
X n!
Proof: For each n ∈ N, let P (n) denote the statement “(a + b)n = an−k bk .” Note
k!(n − k)!
k=0
that

(a + b)1

=a+b
1! 1−0 0 1! 0 1
= a b + a b
0!1! 1!0!
1
X 1!
= a1−k bk .
k!(1 − k)!
k=0
Hence, P (1) is true. So suppose P (ℓ) is true for some ℓ ∈ N. To see that P (ℓ + 1) is true, consider

the following.

(a + b)ℓ+1

= (a + b)(a + b)ℓ
ℓ
X ℓ!
= (a + b) aℓ−k bk
k![(ℓ − k)!]
k=0
ℓ ℓ
X ℓ! X ℓ!
=a aℓ−k bk + b aℓ−k bk
k!(ℓ − k)! k!(ℓ − k)!
k=0 k=0
ℓ ℓ
X ℓ! X ℓ!
= aℓ+1−k bk + aℓ−k bk+1
k!(ℓ − k)! k!(ℓ − k)!
k=0 k=0
ℓ ℓ+1
X ℓ! X ℓ!
= aℓ+1−k bk + aℓ−(k−1) bk
k!(ℓ − k)! (k − 1)![ℓ − (k − 1)]!
k=0 k=1
ℓ ℓ
X ℓ! X ℓ!
= aℓ+1 + aℓ+1−k bk + aℓ−(k−1) bk + bℓ+1
k!(ℓ − k)! (k − 1)![ℓ − (k − 1)]!
k=1 k=1
ℓ ℓ
X ℓ! X ℓ!
= aℓ+1 + aℓ+1−k bk + aℓ−(k−1) bk + bℓ+1
k!(ℓ − k)! (k − 1)!(ℓ − k + 1)!
k=1 k=1
ℓ  
X ℓ! ℓ!
= aℓ+1 + aℓ+1−k bk + aℓ−(k−1) bk + bℓ+1
k!(ℓ − k)! (k − 1)!(ℓ − k + 1)!
k=1
ℓ  
X ℓ! ℓ!
= aℓ+1 + aℓ+1−k bk + + bℓ+1
k!(ℓ − k)! (k − 1)!(ℓ − k + 1)!
k=1
ℓ
ℓ!(ℓ − k + 1)
 
ℓ+1
X
ℓ+1−k k kℓ!
=a + a b + + bℓ+1
k!(ℓ − k)!(ℓ − k + 1) k(k − 1)!(ℓ − k + 1)!
k=1
ℓ
ℓ!(ℓ − k + 1)
 
ℓ+1
X
ℓ+1−k k kℓ!
=a + a b + + bℓ+1
k!(ℓ − k + 1)! k!(ℓ − k + 1)!
k=1
ℓ
ℓ+1−k k ℓ!(ℓ + 1 − k) + kℓ!
X  
ℓ+1
=a + a b + bℓ+1
k!(ℓ + 1 − k)!
k=1

52
 ℓ
ℓ!(ℓ + 1) − kℓ! + kℓ!
X  
= aℓ+1 + aℓ+1−k bk + bℓ+1
k!(ℓ + 1 − k)!
k=1
ℓ  
X ℓ!(ℓ + 1)
= aℓ+1 + aℓ+1−k bk + bℓ+1
k!(ℓ + 1 − k)!
k=1
ℓ  
X (ℓ + 1)!
= aℓ+1 + aℓ+1−k bk + bℓ+1
k!(ℓ + 1 − k)!
k=1
ℓ  
X (ℓ + 1)!
= aℓ+1 + aℓ+1−k bk aℓ+1−k bk + bℓ+1
k!(ℓ + 1 − k)!
k=1
ℓ  
(ℓ + 1)! X (ℓ + 1)! (ℓ + 1)!
= aℓ+1−0 b0 + aℓ+1−k bk aℓ+1−k bk + aℓ+1−(ℓ+1) bℓ+1
0!(ℓ + 1 − 0)! k!(ℓ + 1 − k)! (ℓ + 1)![ℓ + 1 − (ℓ + 1)]!
k=1
ℓ+1
X (ℓ + 1)!
= aℓ+1−k bk .
k!(ℓ + 1 − k)!
k=0
Therefore, P (ℓ+1) is true, and so, the theorem is proved by the Principle of Mathematical Induction.

Exercises
 4
1
25. Calculate x+ .
x

5
26. In the expression x3 + x , what is the coefficient of the term containing x9 ?

27. For each n ∈ N, prove that the first two terms in the expansion of (x − y)n are xn − nyxn−1 .
b
28. Consider the polynomial an xn + an−1 xn−1 + . . . + a1 x + a0 . Set x = y − . Prove that this
na
substitution results in a polynomial in y of degree n such that the (n − 1)st degree.

Definition: A polynomial as in the result of the exercise above is called a depressed polynomial.

53
 Chapter 24: Euler’s Number
Lemma 180: For k, n ∈ N, (n + 1)k − nk ≤ k(n + 1)k−1 .

Proof: Note that both (n + 1)k − nk and k(n + 1)k−1 are polynomials of degree k − 1. Suppose
k!
0 ≤ j ≤ k−1. By the binomial theorem, the coefficient of the nj term of (n+1)k −nk is and
j!(k − j)!
(k − 1)!
the coefficient for the nj term of k(n + 1)k−1 is . Then, since (k − 1 − j)! ≤ (k − j)!,
j!(k − 1 − j)!
k! k!
≤ . Also, since the coefficient of the nj term of (n + 1)k − nk is less
j!(k − j)! j!(k − 1 − j)!
than or equal to the coefficient for the nj term of k(n + 1)k−1 for all 0 ≤ j ≤ k − 1, we get

(n + 1)k − nk ≤ k(n + 1)k−1 . ■

  n ∞
1
Theorem 181: The sequence 1+ is strictly increasing.
n n=1
Proof: Suppose n ∈ N and 1 ≤ k ≤ n. By the lemma above, (n + 1)k − nk ≤ k(n + 1)k−1 , which
n + 1)k − k(n + 1)k−1
means (n + 1)k − k(n + 1)k−1 ≤ nk , and so, ≤ 1. Now,
nk
1
n + 1)k − k(n + 1)k−1
≥
nk
(n + 1)k−1 (n + 1 − k)
=
nk
(n + 1)k (n + 1 − k)
=
(n + 1)nk
n!(n + 1)k (n + 1 − k)
=
(n + 1)!nk
n!(n + 1)k (n − k + 1) n!(n + 1)k (n − k + 1)!
= =
(n + 1)!nk (n + 1)!(n − k)!nk
k
n!(n + 1) (n + 1 − k)!
=
(n + 1)!(n − k)!nk
n! (n + 1)!
⇐⇒ ≤
(n − k)!nk (n + 1 − k)!(n + 1)k
n! (n + 1)!
⇐⇒ ≤
k!(n − k)!nk k!(n + 1 − k)(n + 1)k
n n
X n! X (n + 1)!
⇐⇒ ≤
k!(n − k)!nk k!(n + 1 − k)(n + 1)k
k=1 k=1
n n
X n! X (n + 1)!
⇐⇒ 1 + k
≤ 1 +
k!(n − k)!n k!(n + 1 − k)(n + 1)k
k=1 k=1
n n
X n! X (n + 1)!
⇐⇒ k
≤
k!(n − k)!n k!(n + 1 − k)(n + 1)k
k=0 k=0
n n+1
X n! X (n + 1)!
⇐⇒ ≤
k!(n − k)!nk k!(n + 1 − k)(n + 1)k
k=0 k=0
 n  n+1
1 1
⇐⇒ 1 + ≤ 1+ .
n n+1
Therefore, the sequence is strictly increasing. ■
  n ∞
1
Theorem 182: The sequence 1+ is bounded above.
n n=1

54
Proof: Note that for n ∈ N and 0 ≤ k ≤ n, we have the following.
n! n(n − 1)(n − 2) . . . (n − k + 1) n n−1 n−2 n−k+1
k
= k
= · · · ... · ≤ 1.
(n − k)!n n n n n n
 n n n
1 X n! X 1
This implies that for all n ∈ N, 1 + = ≤ . By a previous note, the
n k!(n − k)!nk k!
)∞ k=0 k=0
( n
X 1   n ∞
1
sequence is bounded above by 3. Therefore, the sequence 1+ is bounded
k! n n=1
k=0 n=0
above by 3. ■
  n ∞
1
Corollary 183: The sequence 1+ converges.
n n=1 ∞
1
Proof: By the theorems above, the sequence (1 + )n is increasing and bounded above.
n n=1
Therefore, it converges by the monotone convergence theorem. ■
 n
1
Definition 184 (Euler’s Number): The number e = lim 1 + is called Euler’s number.
n→∞ n

Exercises

29. Show that e is the absolute maximum of the function f (x) = x1/x .

55
 Chapter 25: Proof of Lemma 69
 n k 
lim (1 + ) −1
Lemma 185: lim  k→∞ k
 = 1.
 
n→∞ 1/n

 Note thenfollowing.
Proof: 
lim (1 + )k − 1
lim  k→∞
 k 
1/n

n→∞

n
= lim (n lim [(1 + )k − 1])
n→∞ k→∞ k
k
" #
 k X k! 
= lim n lim 1 + + −1
n→∞ k→∞ nk i=2 i!(k − i)!k i ni
k
" #
 X k! 
= lim n lim 1 +
n→∞ k→∞
i=2
i!(k − i)!k i ni
k
" #
X k! 
= 1 + lim lim
n→∞ k→∞
i=2
i!(k − i)!k i ni−1
≥ 1.

Now, note the following. 

n
lim (1 + )k − 1
 k→∞
lim  k 
1/n

n→∞

n
= lim (n lim [(1 + )k − 1])
n→∞ k→∞ k
k
" #
 k X k! 
= lim n lim 1 + + −1
n→∞ k→∞ nk i=2 i!(k − i)!k i ni
k
" #
1 X k! 
= lim n lim +
n→∞ k→∞ n
i=2
i!(k − i)!k i ni
k
" #
1 X k! 
= lim n + lim
n→∞ n k→∞ i=2 i!(k − i)!k i ni
k
" #
1 X 1 k!
= lim n + ·
n→∞ n i=2 ni · i! (k − 1)!k i
k
" #
1 X 1
≤ lim n +
n→∞ n i=2 ni · i!
k
X 1
= lim 1 +
n→∞
i=2
ni−1 · i!
k
X 1
≤ lim 1 +
n→∞
i=2
ni−1
k
X 1
= lim 1 + i
n→∞
i=1
n
1
= lim 1 + −1
n→∞ 1
1−
n

56
 1
= lim
n→∞ 1
1−
n
= 1.  n k   n 
lim (1 + ) −1 lim (1 + )k − 1
Since 1 ≤ lim  k→∞ k  k→∞ k
 ≤ 1, we obtain lim   = 1, as desired. ■
  
n→∞ 1/n n→∞ 1/n

Lemma 69: Let f (x) = ex . Then f ′ (0) = 1.

Proof: Note the following.

eh − e0
f ′ (0) = lim
h→0 h
eh − 1
= lim
h→0 h
e1/n − 1
= lim
n→∞ 1/n
  m 1/n
1
lim 1 + −1
m→∞ m
= lim
n→∞
" 1/n
 m/n #
1
lim 1 + −1
m→∞ m
= lim
n→∞ 1/n
  n k 
lim 1 + −1
= lim  k→∞
 k 
1/n

n→∞

=1
m
where k = , as desired. ■
n

Exercises

30. Prove that the function f (x) = ex is not a polynomial function.

31. Set f (x) = xex . Form a conjecture for the nth derivative of f (x) and prove your conjecture.
ex
32. Prove that for each n ∈ N, lim n = ∞.
n→∞ x

57
 Chapter 26: Series
Definition 186: Suppose {xn }∞
n=1 ⊆ R is a sequence. The n
th
partial sum of {xn }∞
n=1 is
n
X
sn = xi .
i=1
∞
X
Definition 187: Suppose {xn }∞
n=1 ⊆ R is a sequence and for all n ∈ N, sn = xi . Then, the
n=1
series generated by the sequence {xn }∞ ∞
n=1 is the sequence {sn }n=1 .
∞
X
Notation 188: If {xn }∞
n=1 ⊆ R is a sequence, then we usually write xn instead of {sn }∞
n=1 to
n=1
denote the series generated by {xn }∞
n=1 .
∞
X
Definition 189: Suppose {xn }∞
n=1 ⊆ R is a sequence and for all n ∈ N, sn = xi . If lim sn = s,
n→∞
n=1
∞
X
then xn is said to converge to s.
n=1
Definition 190: A series that converges is called convergent.

Definition 191: A series that does not converge is called divergent.

Note 192: s is called the sum of the series.

Note 193: A series can converge or diverge, but not both.

Notation 194: Let {xn }∞ n=1 ⊆ R be a sequence and for each n ∈ N, and sn denote the n
th
partial
∞
X ∞
X
sum. If the series xn converges to s ∈ R, then we usually write xn = s instead of lim sn = s
n→∞
n=0 n=0
∞
X
or lim xn = s.
n→∞
n=0
∞
X
Note 195: The notation xn is (unfortunately, but usually not ambiguously) used to denote
n=1
both a series and its sum, if it exists.
∞  n
X 1
Example 196: = 2.
n=0
2
( n   )∞
X 1 i
Proof: Recall that the sequence converges to 2 by a previous example. Therefore,
i=0
2
n=0
∞  n
X 1
= 2 by the definition of series. ■
n=0
2
∞
X 1
Definition 197: The series is called the harmonic series.
n=1
n
Theorem 198: The harmonic series diverges.
n
X 1
Proof: For each n ∈ N, set sn = .
k
k=1
n
Claim: For each n ∈ N, s2n ≥ 1 + .
2
n
Proof (Claim): For each n ∈ N, let P (n) denote the statement “s2n ≥ 1 + .” To see that P (1)
2
1 1
is true, note that s2 = 1 + ≥ 1 + . So suppose that P (k) is true for some k ∈ N. To see that
2 2
P (k + 1) is true, consider the following.

58
s2k+1
1 1 1
= s2k + + k + . . . + k+1
2k
+1 2 +2 2
k 1 1 1
≥1+ + k + + . . . + k+1
2 2 + 1 2k + 2 2
k 1 1 1
≥ 1 + + k+1 + k+1 + . . . + k+1
2 2 2 2
k 2k
= 1 + + k+1
2 2
k 1
=1+ +
2 2
k+1
=1+ .
2
Therefore, P (k + 1) is true, and so, the claim is proved by the Principle of Mathematical Induction.

■Claim

By the claim, lim s2n = ∞. Also, note that {xn }∞ ∞

n=1 is increasing. So, since {xn }n=1 is increasing
n−→∞

and lim s 2n = ∞, lim sn = ∞, and so the harmonic series diverges. ■

n−→∞ n−→∞

Example 199: For each of the following, determine whether the given series converges or diverges.
∞
X 1
(a) 2−n
n=2
n
n
1 1 1 X 1 1 1
Recall that 2 = − . Set sn = − and note that sn = 1 − . Now,
n −n n−1 n k−1 k n
k=1
∞
X 1
lim sn = 1, and so, 2−n
= 1.
n→∞
n=2
n
∞
X 1
(b) 2
n=1
n +n
n
1 1 1 X 1 1 1
Recall that 2 = − . Set sn = − and note that sn = 1 − . Now,
n +n n n+1 k k+1 n+1
k=1
∞
X 1
lim sn = 1, and so, 2+n
= 1.
n→∞
n=1
n
∞
X 1
(c)
n=1
n2
∞ ∞ ∞
X 1 X 1 X 1 1
Recall that 2−n
= 1 and 2+n
= 1, which means that 2+n
= . Now, note the
n=2
n n=1
n n=2
n 2
following.
1 1 1
≤ 2 ≤ 2
n2 + n n n −n
∞ ∞ ∞
X 1 X 1 X 1
⇐⇒ 2+n
≤ 2
≤ 2−n
n=2
n n=2
n n=2
n
∞
1 X 1
⇐⇒ ≤ ≤1
2 n=2 n2
∞
3 X 1
⇐⇒ ≤ ≤ 2.
2 n=1 n2
∞
X 1 π2
Note 200: It can be shown that 2
= .
n=1
n 6

59
 n
X n
X
Theorem 201: Suppose that xi = x, yi = y, and c ∈ R. Then
i=1 i=1
n
X
(i) xi + yi = x + y;
i=1
Xn
(ii) xi − yi = x − y;
i=1
Xn
(iii) cxi = cx.
i=1

Exercises

Section 11.2: 5-8, 43-48.

n
X n
X n
X
33. If a series ai converges and a series bi diverges, show that ai + bi diverges.
i=1 i=1 i=1

60
 Chapter 27: The Divergence Test
∞
X
Theorem 202 (Divergence Test): If the series xn converges, then lim xn = 0.
n→0
n=1
n
P ∞
P
Proof: For all n ∈ N, set sn = xi and s = xn . Then,
i=1 n=1
lim xn
n→∞

= lim xn+1
n→∞

= lim sn+1 − sn
n→∞

=s−s

= 0. ■

Note 203: The converse of the Divergence Test is not true. The harmonic series is a counter-

example.

Example 204: For each of the following, determine whether the series converge or diverge.
∞
X n!
(a)
n=1
n2
n! n(n − 1)(n − 2) n(n − 1)(n − 2)
Note that for n ≥ 3, 2 ≥ 2
. Also, recall that lim = ∞. So,
n n n→∞ n2
∞
n! X n!
lim = ∞, which means that the series diverges by the Divergence Test.
n→∞ n2 n2
n=1
∞  √n
X n
(b)
n=1
n+1
 n  n
n+1 n 1
Recall that lim = e. This implies that lim = . Hence,
n→∞ n n→∞ n+1 e
  √n
n
n+ 1 
n
n
≥
n+1
1
=
e
> 0.
∞  √n
X n
Therefore, the series diverges by the Divergence Test.
n=1
n+1

Exercises

Section 11.2: 27-30, 37, 39.

61
 Chapter 28: The Comparison Test
∞
X ∞
X
Theorem 205 (Comparison Test): Suppose xn and yn are series such that 0 ≤ xn ≤ yn
n=1 n=1
for all n ∈ N.
X∞ ∞
X
(i) If yn converges, then xn converges.
n=1 n=1
∞
X ∞
X
(ii) If xn diverges, then yn diverges.
n=1 n=1
Proof: We will prove (i) and leave the proof of (ii) as an exercise for the reader. For each n ∈ N,
n
X Xn ∞
X
set sn = xi and tn = yi . Since yn converges, there is a t ∈ R such that lim tn = t. Also,
n→∞
i=1 i=1 n=1
since yn ≥ 0 for every n ∈ N, the sequence {tn }∞
n=1 is increasing. So, tn ≤ t for all n ∈ N. Since

xn ≥ 0 for every n ∈ N, the sequence {xn }∞

n=1 is increasing. So, we obtain that 0 ≤ sn ≤ tn ≤ t for

every n ∈ N, which means that the sequence {sn }∞ ∞

n=1 is increasing. Hence, the sequence {sn }n=1
∞
X
converges, and so, the series xn converges. ■
n=1
Note 206: In the theorem above, (ii) is the contrapositive of (i).

Example 207: For each of the following, determine whether the given series converges or diverges.
∞
X 3
(a)
n=1
n
∞ ∞
3 1 X 1 X 3
Note that ≥ and recall that the series diverges. Therefore, the series diverges by
n n n=1
n n=1
n
the Comparison Test.
∞  n
X 1
(b)
n=1
3
 n  n ∞  n
1 1 X 1
Note that ≤ and recall that converges by a previous example. Therefore,
3 2 n=1
2
∞  n
X 1
the series converges by the Comparison Test.
n=1
3

Exercises

Section 11.4: 3, 4, 5, 9, 10, 14, 15, 17, 18, 19, 28.

62
 Chapter 29: Geometric Series
∞
X
Definition 208: A geometric series is a series of the form arn where a, r ∈ R \ {0}.
n=0
Theorem 209: Let a, r ∈ R \ {0}.
∞
X a
(i) If |r| < 1, then arn = ;
n=0
1 − r
∞
X
(ii) If |r| ≥ 1, then arn diverges.
n=0
∞  n
X 1
Example 210: Find the sum of the series 3 .
n=0
2
3 3
By the theorem above, the sum is given by = = 6.
1 1
1−
2 2
∞
X 2n
Example 211: Determine whether the series n+1
converges or diverges.
n=1
5 +n
 n
2n 2n 2
Note that n+1 ≤ n = and by the theorem above,
5 +n 5 5
∞
X 2   n

n=1
5
"∞   #
X 2 n
= −1
n=1
5
1
= −1
2
1−
5
1
= −1
3
5
5
= −1
3
2
= .
3
∞  n ∞
X 2 X 2n
Therefore, the series converges, which means the series converges by the
n=1
5 n=1
5n+1 + n
Comparison Test.

63
 Exercises

Section 11.2: 17-41 odd.

Section 11.4: 7, 8, 15, 19, 20, 22-24, 27-30.

n−1
X 1 − rn
34. (Gregory-Leibniz Formula) Consider the sum rk = .
1−r
k=0
1 2 4 n−1 2n−2 (−1)n x2n
(a) Show that = 1 − x + x − . . . + (−1) x + .
1 + x2 1 + x2
1
π 1 1 1 (−1)n−1 x2n
Z
(b) Show that = 1 − + − + . . . + + (−1)n dx.
4 3 5 7 2n − 1 1 + x2
0
Z1 2n
x 1
(c) Show that 0 ≤ dx ≤ .
1 + x2 2n + 1
0
∞
X (−1)n+1
π
(d) Conclude that = .
4 n=1
2n + 1
35. (Cantor’s Disappearing Table) Consider a table of length L. To construct Cantor’s Disap-
L
pearing Table, we remove a section of length n for each n ∈ N. For example, at stage 1, we remove
4
L
the section of length centered at the midpoint of the table. At stage 2, remove the sections of
4
L 1
length . (After this stage, of the table is removed.) Now, four sections remain, each of which
16 8
L L
have length less than . For stage 3, remove the four central regions of length . Continue this
4 64
process indefinitely.
L
(a) Show that at the nth stage, each remaining section has length less than n and that the total
2
n
!
X 1
amount of table removed is L .
2k+1
k=1
(b) As n approaches ∞, show that one-half of the table remains.

Note: The result in part (b) is curious since there are no nonzero intervals of table left (hence the
L
name “disappearing table”). However, we can place any object of length greater than on the
4
table and it will not fall through.

36. (Koch’s Snowflake) The Koch snowflake, described in 1904 by Swiss mathematician Helge

von Koch, is a jagged fractal curve obtained by a limit of polygonal curves. It is continuous but has

no tangent line at any point. Begin with an equilateral triangle, call this stage 0. For stage 1, replace

each edge of the triangle with four edges of one-third the length. Continue this process indefinitely.

For example, at stage n, replace each edge with four edges of one-third the length. Let Pn and An

denote the perimeter and area, respectively, of the polygon at stage n of this construction.
4
(a) Show that Pn = Pn−1 .
3
(b) Show that lim Pn = ∞.
n→∞

Note: Part (b) implies that the Koch snowflake has infinite length.

64
(c) Consider A0 . For each n ∈ N, show that at stage n of this construction, 3 · 4n−1 triangles of
A0
area n are added to the original equilateral triangle.
9
8
(d) Show that the total area of the Koch snowflake is · A0 .
5

Chapter 30: The Limit Comparison Test

∞
X ∞
X
Theorem 212 (Limit Comparison Test): Suppose xn and yn are series such that xn , yn ≥
n=1 n=1
∞
xn X
0 for all n ∈ N. If lim = L ∈ R with L > 0, then the series xn converges if and only if
n→∞ yn
n=1
∞
X
yn .
n=1
L 3L xn
Proof: Since L > 0, <L< for all n > N . Since lim = L, there is an N ∈ N such that
2 2 n→∞ yn
L xn 3L L 3L
< < for all n > N . Thus, for all n > N , · yn < xn < · yn .
2 yn 2 2 2
∞ ∞
X X L
(=⇒) We prove the contrapositive. If the series yn diverges, then the series · yn diverges,
n=1 n=1
2
X∞
which means that the series xn diverges by the Comparison Test.
n=1
∞ ∞
X X 3L
(⇐=) If the series yn converges, then the series · yn converges, which means that the
n=1 n=1
2
∞
X
series xn converges by the Comparison Test. ■
n=1
Example 213: For each of the following, determine whether the given series converges or diverges.
∞
X 3
(a)
n=1
n
3
∞ ∞
n
X 1 X 3
Note that lim = 3 > 0 and recall that the series diverges. Therefore, the series
n→∞ 1 n n
n=1 n=1
n
diverges by the Limit Comparison Test.
∞  
X n+1
(b) ln
n=1
n
Consider the following.

n+1
ln
n
lim
n→∞ 1
 n
−1

n
· 2
n+1 n
= lim
n→∞ −1
n2
n
= lim
n→∞ n + 1

=1

65
> 0.
∞ ∞  
X 1 X n+1
Also, recall that the series diverges. Therefore, the series ln diverges by the
n=1
n n=1
n
Limit Comparison Test.

Exercises

Section 11.4: 3, 4, 17, 18, 22, 24, 31.

66
 Chapter 31: The Integral Test
Theorem 214 (Integral Test): Suppose that f : [1, ∞) → R is a continuous decreasing function
∞
X
with f (x) > 0 for all x ∈ [1, ∞). Then the series f (n) converges if and only if the improper
n=1
Z∞
integral f (x) dx converges.
1
Example 215: Use the integral test to show that the harmonic series diverges.
1
Proof: Recall that the function f : [1, ∞) → R defined by f (x) = is continuous and decreasing
x
Z∞
1
and f (x) > 0 for all x ∈ [1, ∞). Also, recall that the improper integral dx diverges. Therefore,
x
1
the harmonic series diverges by the Integral Test. ■
∞
X 1
Example 216: Does the series 2
converge or diverge?
n=1
n
1
Recall that the function f : [1, ∞) → R defined by f (x) = is continuous and decreasing and
x2
Z∞
1
f (x) > 0 for all x ∈ [1, ∞). Also, by a previous theorem, the improper integral dx converges.
x2
1
∞
X 1
Hence, the series 2
converges by the Integral Test.
n=1
n
∞
X 1
Definition 217: A series of the form is called a p-series.
n=1
np
Corollary 218: A p-series converges if and only if p > 1.
∞ √
X n
Example 219: Determine whether the series 2
converges or diverges.
n=1
n + 2n + 5
√ √ ∞
n n X 1
Note that 2 ≤ 2 = n−3/2 . By the corollary above, the series 3/2
converges, and
n + 2n + 5 n n=1
n
∞ √
X n
so, the series 2
converges by the Comparison Test.
n=1
n + 2n + 5
Theorem 220 (Remainder Estimate for the Integral Test): Suppose that f : [1, ∞) → R is a
Z∞
continuous decreasing function with f (x) > 0 for all x ∈ [1, ∞) and the improper integral f (x) dx
1
Z∞ ∞
X n
X Z∞
converges. Then for each n ∈ N, f (x) dx ≤ f (n) − f (k) ≤ f (x) dx.
n+1 n=1 k=1 n
Note 221: In the theorem above,
Z∞ n
X ∞
X Z∞ n
X
f (x) dx + f (k) ≤ f (n) ≤ f (x) dx + f (k).
n+1 k=1 n=1 n k=1
∞
X 1
Example 222: Consider the series 2
.
n=1
n
Note the following.

67
 Z∞ 1
1 9991−2 999 1
(i) dx = = = .
x2 2−1 1 999
999
Z∞ 1
1 10001−2 1
(ii) dx = = 1000 = .
x2 2−1 1 1000
1000
Now, using a calculator, we obtain the following.
999
X 1
(i) 2
= 1.643933567
n=1
n
999
X 1 1
(ii) 2
+ = 1.644933567
n=1
n 1000
999
X 1 1
(iii) 2
+ = 1.644934568
n=1
n 999
π2 ∼
(iv) = 1.644934067.
6

Exercises

Section 11.3: 3 – 35 odd, 34.

Section 11.4: 3 – 31 odd, 32.

68
 Chapter 32: Absolute Convergence
∞
X ∞
X
Definition 223: A series xn is absolutely convergent if the series |xn | converges.
n=1 n=1
Definition 224: A series is conditionally convergent if it is not absolutely convergent.
∞
X 1
Example 225: The series (−1)n+1 is conditionally convergent.
n=1
n
Theorem 226: An absolutely convergent series is convergent.
∞
X
Proof: Let xn be an absolutely convergent series. Then for each n ∈ N, xn + |xn | ≤ 2|xn |.
n=1
∞
X ∞
X
Since the series xn is absolutely convergent, the series 2|xn | converges, and so, the series
n=1 n=1
∞
X ∞
X ∞
X
(xn +|xn |) converges by the Comparison Test. Therefore, the series xn = (xn +|xn |)−|xn
n=1 n=1 n=1
converges. ■
∞
X sin n
Example 227: Determine whether the series converges or diverges.
n=1
n2
∞ ∞
sin n 1 X 1 X sin n
Since ≤ and the series converges, the series converges by the Compar-
n2 n2 n=1
n 2
n=1
n2
∞
X sin n
ison Test, and so, the series converges absolutely. Hence, by the theorem above, the series
n=1
n2
∞
X sin n
converges.
n=1
n2

69
 Chapter 33: The Alternating Series Test
∞
X 1
Definition 228: The series (−1)n+1 is called the alternating harmonic series.
n=1
n
Theorem 229: The alternating harmonic series converges.
1 1 1 1 2 1 1 1 1 2
Proof: For all n ∈ N, s2n = 1− + − +. . .+ − = (1− )+( − )+. . .+( − ) > 0.
2 3 4 2n − 1 n 2 3 4 2n − 1 n
1 1 1 1 2 1
Also, s2n+1 = 1 − + − + . . . + − + < 1. So, for all n ∈ N, we obtain
2 3 4 2n − 1 n 2n + 1
1
0 < s2n < s2n + = s2n+1 < 1, which means that lim s2n and lim s2n+1 exist. Now, since
2n + 1 n−→∞ n−→∞
1 1
s2n+1 = s2n + and lim = 0, we get lim s2n = lim s2n+1 . Therefore, lim sn
2n + 1 n−→∞ 2n + 1 n−→∞ n−→∞ n−→∞
∞
X 1
exists, which means that (−1)n+1 converges. ■
n=1
n
Note 230:

(i) It can be proven that the alternating harmonic series converges to ln(2).

(ii) The alternating harmonic series is conditionally convergent.

∞
X ∞
X
Definition 231: An alternating series is a series that can be written as (−1)n xn or (−1)n+1 xn ,
n=1 n=1
where xn > 0 for each n ∈ N.

Theorem 232 (Alternating Series Test): Let {xn }∞ +

n=1 ⊆ R be a decreasing sequence. Then
∞
X ∞
X
the series (−1)n xn and (−1)n+1 xn converge if and only if lim xn = 0.
n→∞
n=1 n=1
n
X
Proof: For each n ∈ N, set sn = xi .
i=1
Claim 1: The sequence {s2n }∞
n=1 converges.

Proof (Claim 1): Note that for each n ∈ N, s2n = (x1 − x2 ) + (x3 − x4 ) + . . . + (xn−1 − xn ) ≥ 0

and s2(n+1) = s2n+2 = s2n + (x2n+1 − x2n+2 ) ≥ s2n . Therefore, {s2n }∞

n=1 is an increasing sequence.

Now to see that the sequence {s2n }∞

n=1 is bounded, note that s2n = x1 − (x2 − x3 ) − (x4 − x5 ) −

. . . − (x2n−2 − x2n−1 ) − x2n ≤ x1 , which means that the sequence {s2n }∞

n=1 is bounded. Therefore,

the sequence {s2n }∞

n=1 converges by the monotone convergence theorem (theorem 167). ■Claim 1

By claim 1, there is an s ∈ R such that lim s2n = s.

n→∞

Claim 2: lim s2n+1 = s.

n→∞

Proof (Claim 2): Consider the following.

lim s2n+1
n→∞

= lim (s2n + x2n+1 )

n→∞

= lim s2n + lim x2n+1

n→∞ n→∞

=s+0

= s. ■Claim 2

70
Since lim s2n = s = lim s2n+1 , we obtain that lim sn = s, as desired. ■
n→∞ n→∞ n→∞

Example 233: Use the Alternating Series Test to show that the alternating harmonic series con-

verges.
1
Proof: Note that lim = 0. Therefore, the alternating harmonic series converges by the Alter-
n→∞n
nating Series Test. ■
∞
X 1
Example 234: Determine whether the series (−1)n 2 converges or diverges.
n=1
n
∞
1 X 1
Note that lim 2 = 0. Therefore, the series (−1)n 2 converges by the Alternating Series Test.
n→∞ n n
n=1
Note that this series is absolutely convergent, but the alternating harmonic series is not.

Theorem 235 (Remainder Estimate for Alternating Series Test): Let {xn }∞ n=1 ⊆ R
+
n
X
be a decreasing sequence that converges to zero. For each n ∈ N, set sn = (−1)n+1 xi and
i=1
∞
X
s= (−1)n+1 xn . Then for all n ∈ N, |s − sn | ≤ xn+1 .
n=1
∞  n
X 1
Example 236: Show that the series − is convergent. How many terms is needed to add
n=1
n
in order to find the sum to be within 0.00005?
∞  n  n
X 1 1 1 1
To see that the series − is convergent, since ≤ and lim = 0, we obtain that
n=1
n n n n→∞ n
 n ∞  n
1 X 1
lim = 0, which means that the series − is convergent by the Alternating Series
n→∞ n n
n=1
Test.
 n+1  5
1 1 1
It remains to find how many terms is needed so that < . Note that =
20000 n+1 5
0.000032 < 0.00005. So it suffices to consider four terms.

71
 Exercises

Section 11.5: 3-31 odd.

37. In this exercise, you will prove the following corollary of the Alternating Series Test.

If {xn }∞
n=1 ⊆ R is a positive and decreasing sequence with lim xn = 0, then the series a1 + a2 −
n→∞

2a3 + a4 + a5 − 2a6 + . . . converges.

38. In 1829, Lejeune Dirichlet pointed out that the great French mathematician Augustin Louis

Cauchy made a mistake in a published paper by improperly assuming the Limit Comparison Test

to be valid for non-positive series. Here are the two series that Dirichlet came up with:
∞ ∞
(−1)n (−1)n (−1)n
X X  
√ and √ 1+ √ .
n=1
n n=1
n n
How do these series contradict the Limit Comparison Test when the series are not assumed to be

positive?

Chapter 34: The Ratio Test

xn+1
Theorem 237 (Ratio Test): Let {xn }∞
n=1 ⊆ R be a sequence such that ρ = lim exists.
n→∞ xn
∞
X
(i) If ρ < 1, then the series xn converges absolutely.
n=1
∞
X
(ii) If ρ > 1, then the series xn diverges.
n=1
xn+1
Proof: For (i), suppose that ρ < 1 and choose an r ∈ (ρ, 1). Since lim = ρ, there is an
n→∞ xn
xn+1
M ∈ N such that < r for all n ≥ M .
xn
Claim: |xM +n | < rn |xM | for all n ∈ N.

Proof (Claim 1): For each n ∈ N, let P (n) denote the statement “|xM +n | < rn |xM |.” To see that

P (1) is true, consider the following.

xM +1
<r
xM
⇐⇒ |xM +1 | < r|xM |.

So, P (1) is true. Now suppose that P (k) is true for some k ∈ N. To see that P (k + 1) is true,

consider the following.

|xM +k+1 |

< r|xM +k |

< r(rk |xM +k |)

= rk+1 |xM |.

Therefore, P (k + 1) is true, and so, the claim is proved by the Principle of Mathematical Induction.

72
■Claim 1

Thus, by the claim,

X∞
|xn |
n=M
∞
X
= |xM +n |
n=0
X∞
≤ |xM |rn
n=0
∞
X
= |xM | rn .
n=0
∞
X ∞
X
Since r ∈ (0, 1), the series rn converges, and so, the series |xn | converges by the Comparison
n=0 n=M
X∞
Test. Therefore, the series xn converges absolutely.
n=1
xM +1
For (ii), suppose that ρ > 1 and choose an r ∈ (1, ρ). Then there is an M ∈ N such that >r
xM
for all n ≥ M . By an argument similar to the one above, |xM +n | ≥ rn |xM |. Since lim rn = ∞, the
n→∞
∞
X
terms lim xM +n ̸= 0, and so, the series xn diverges. ■
n→∞
n=1
Example 238: For each of the following, determine whether the series is absolutely convergent,

conditionally convergent, or divergent.

∞
X 3n
(a)
n=1
n!
Consider the following.
3n+1 n!
lim ·
n→∞ (n + 1)! 3n
3
= lim
n→∞ n + 1

= 0 < 1.
∞
X 3n
Therefore, the series is absolutely convergent by the Ratio Test.
n=1
n!
∞
X n!
(b) n
n=1
2
Consider the following.
(n + 1)! 2n
lim ·
n→∞ 2n+1 n!
n+1
= lim
n→∞ 2
= ∞ > 1.

73
 ∞
X n!
Therefore, the series n
is divergent by the Ratio Test.
n=1
2

Chapter 35: The Root Test

Theorem 239 (Root Test): Let {xn }∞
p
n=1 ⊆ R be a sequence such that ℓ = lim |xn | exists.
n
n→∞
∞
X
(i) If ℓ < 1, then the series xn converges absolutely.
n=1
∞
X
(ii) If ℓ > 1, then the series xn diverges.
n=1
Proof: Exercise. ■

Example 240: For each of the following, determine whether the series is absolutely convergent,

conditionally convergent, or divergent.

∞  n
X n+1
(a) (−1)n
n=1
n2
Consider
s the following.
 n
n n
n+1
lim (−1)
n→∞ n2
n+1
= lim
n→∞ n2

=0

< 1.
∞  n
X n+1
Therefore, the series (−1)n is absolutely convergent by the Root Test.
n=1
n2
2
∞  n
X n
(b)
n=1
n+1
Consider
s the following.
 n2
n n
lim
n→∞ n+1
 n
n
= lim
n→∞ n + 1
1
=
e
< 1.
∞  n2
X n
Hence, the series is absolutely convergent by the Root Test.
n=1
n+1

74
 Exercises

Section 11.6: 3-37 odd.

Section 11.7: 1-37 odd.

∞ 2
X 2n
39. Show that the series diverges. (Hint: Show that n! ≤ n for every n ∈ N and recall that
n=1
n!
2
2n = (2n )n .)
∞
X cn n!
40. Set S = , where c ∈ R.
n=1
nn
(a) Show that S converges absolutely if and only if |c| < e and diverges otherwise.
en n! √
(b) Verify numerically that lim n+1/2 = 2π.
n→∞ n

(c) If c = e, use the Limit Comparison Test to show that S diverges.

∞
1 X (4k)! · (1103 + 26390k)
41. In 1910, the mathematician Srinivasa Ramanujan discovered the formula = .
π n=0 (k!)4 · 3964n
(a) Approximate the series with only the n = 0 term. How many correct digits of π does this ap-

proximation give?

(b) Approximate the series with the n = 0 and n = 1 terms. How many correct digits of π does

this approximation give?

Note: In general, each term of this series increases the accuracy of the approximations of π by 8

digits.

(c) Show that Ramanujan’s series converges.

42. In this exercise, you will provide a proof of the root test.
∞
X
(a) Show that the series xn converges for ℓ < 1. (Hint: Choose ℓ < R < 1 and show that
n=1
∞
X
n
xn ≤ R for all n ∈ N. Then compare with the geometric series Rn .)
n=1
∞
X
(b) Show that the series xn diverges for ℓ > 1.
n=1
43. In this exercise, you will show that the root test is more general than the ratio test. To be

precise, consider the following. Let {xn }∞

n=1 ⊆ R be a sequence.
xn+1
(a) If lim = r ̸= 1, then lim |an |1/n = r.
n→∞ xn n→∞

(b) Interpret the result of part (a) in terms of how likely the Ratio Test or Root Test is to give a

definite conclusion.

(c) Show that the result of part (a) is not a biconditional statement.

(d) In spite of this conclusion, give a condition as to whether using the ratio test may be preferable

75
to using the root test.

Chapter 36: Power Series

Definition 241: Let a ∈ R, {cn }∞ n=1 ⊆ R be a sequence and x be a variable. The power series
∞
X
centered at x = a with coefficient set {cn }∞
n=1 is the series cn (x − a)n .
n=1
Note 242: In the definition above,

(i) If a ̸= 0, then the power series centered at a is is sometimes called the power series about a or

the power series in x − a.

(ii) If a = 0, then the power series centered at a is usually called the power series in x.
X∞
Theorem 243: Let cn (x − a)n be a power series in x − a with coefficient set {cn }∞n=1 . Then
n=1
one of the following is true.
X∞
(i) The power series cn (x − a)n converges if and only if x = a;
n=1
∞
X
(ii) The power series cn (x − a)n converges for every x ∈ R;
n=1
∞
X
(iii) There is an R > 0 such that the power series cn (x − a)n converges absolutely for every
n=1
x ∈ (a − R, a + R) and diverges for every x ∈ R \ [a − R, a + R].

Theorem 244 (Least Upper Bound Property of R): Every nonempty subset of R that is

bounded above has an supremum.

Proof: Proved in Analysis. ■

∞
X
Theorem 245: For any power series cn (x−a)n , there is an R ∈ R+ ∪{0} such that the following
n=1
are satisfied.
∞
X
(i) If the power series cn (x − a)n converges absolutely whenever |x − a| < R and diverges
n=1
whenever |x − a| > R, then R is finite.
∞
X
(ii) If the power series cn (x − a)n converges absolutely for every x ∈ R, then R is infinite.
n=1
∞
X
Proof: Let cn (x − a)n be a power series. Throughout this proof, we will assume a = 0 to
n=1
∞
X
simplify the notation. If the power series cn (x − a)n converges only at x = 0, then set R = 0.
n=1
∞
X
So suppose that the power series cn (x − a)n converges to B for some B ∈ R.
n=1
∞
X
n
Claim: The power series cn x converges absolutely for every x ∈ R with |x| < |B|.
n=1

76
 ∞
X
Proof (Claim): Since the series cn B n converges, lim cn B n = 0. So, there is an M ∈ R+ such
n→∞
n=1
n
that |cn B | < M for all n. Therefore, we obtain the following.
X∞
|cn xn |
n=0
∞ n
X x
= |cn B n |
n=1
B
∞ n
X x
<M .
n=1
B
∞
x X x n
If |x| < |B|, then < 1, and so, the series is a convergent geometric series. Hence, by
B n=1
B
∞
X X∞
the Comparison Test, the series |cn xn | converges, and so, the power series cn xn converges
n=0 n=1
absolutely(for every x ∈ R with |x| < |B|. ■Claim
∞
)
X
Let S = x ∈ R | the power series cn xn converges . Note that 0 ∈ S and whenever B ∈ S
n=1
with B ̸= 0, then (−|B|, |B|) ⊆ S. Consider two cases.

Case 1: S is bounded

By the Least Upper Bound Property, the set S has a least upper bound ℓ. This implies that

(−B, B) ⊆ S for all B ∈ (0, ℓ), and so, (−ℓ, ℓ) ⊆ S. Note that if |x| > ℓ, then x ∈
/ S. So, at least

one of ±ℓ ∈ S. Set R = ℓ.

Case 2: S is unbounded

In this case, (−B, B) ⊆ S for every arbitrarily large B. Hence, S = R, and set R = ∞. ■

Definition 246: In the theorem above, the real number R is called the radius of convergence.
X∞
Definition 247: Let cn (x − a)n be a power series in x − a with coefficient set {cn }∞
n=1 . The
n=1
∞
( )
X
n
interval of convergence is the set x ∈ R | cn (x − a) converges .
n=0
∞
X
Note 248: Let cn (x − a)n be a power series in x − a with coefficient set {cn }∞
n=1 .
n=1
(i) If the radius of convergence is zero, then the interval of convergence is [a, a] = {a}.

(ii) If the radius of convergence is ∞, then the interval convergence is (−∞, ∞) = R.

(iii) If the radius of convergence is R for some R ∈ R+ , then the interval of convergence is one of

(a − R, a + R), [a − R, a + R), (a − R, a + R], or [a − R, a + R].

Note 249: To determine the interval of convergence for a power series, there are two steps to

consider:

(i) Find the radius of convergence (likely using the ratio test).

(ii) Determine the convergence on the endpoints.

Example 250: For each of the following, find the radius and interval of convergence.

77
 ∞
X xn
(a)
n=0
n!
By a previous example,
xn+1 n!
lim ·
n→∞ (n + 1)! xn
x
= lim
n→∞ n + 1

= 0.
xn+1 n!
Note that lim · = 0 for all x ∈ R, and so, by part (ii) of the theorem above, the
n→∞ (n + 1)! xn

interval of convergence is R and the radius of convergence is ∞.

∞
X (x − 8)n
(b)
n=1
n
∞ ∞
X (x − 8)n X 1
If x = 9, then = , which diverges. So 9 is not in the interval of convergence.
n=1
n n=1
n
∞ ∞
X (x − 8)n X (−1)n
If x = 7, then = , which converges.
n=1
n n=1
n
Hence, by part (iii) of the theorem above, the interval of convergence is given by [7, 9) and the

radius of convergence is 1.
∞
X (−1)n (x + 4)n
(c)
n=1
n3n
Consider the following.
(−1)n+1 (x + 4)n+1 n3n
lim ·
n→∞ (n + 1)3n+1 (−1)n (x + 4)n
x+4
= lim .
n→∞ 3
∞
X (−1)n (x + 4)n x+4
This implies that the series n
is absolutely convergent if < 1, or |x + 4| < 3.
n=1
n3 3
Hence, the radius of convergence is 3.
∞ ∞
X (−1)n (x + 4)n X 1
Set x = −7. Then n
= , which diverges.
n=1
n3 n=1
n
∞ ∞
X (−1)n (x + 4)n X (−1)n
Set x = −1. Then n
= , which converges.
n=1
n3 n=1
n
Therefore, the interval of convergence is (−7, −1].

Example 251: For each of the following, find a power series representation for the given function.
1
(a) f (x) =
1 − x2
∞ ∞
X 1 X
Note that x2n = 2
if x 2
< 1. So, f (x) = x2n for x ∈ (−1, 1).
n=0
1 − x n=0
x2
(b) g(x) =
1−x
∞ ∞
X x2 X
Note that x2 · xn = for |x| < 1. So, g(x) = xn+2 for x ∈ (−1, 1).
n=0
1 − x n=0
∞
X
Theorem 252: Let a ∈ R and {cn }∞ n=0 ⊆ R be a sequence such that the power series cn (x − a)n
n=0
has positive or infinite radius of convergence R. Also, suppose that I is the interval of convergence

78
 ∞
X ∞
X
n
of the power series cn (x − a) . Define f : I → R by f (x) = cn (x − a)n .
n=0 n=0
(i) f is continuous on int(I);
∞
X
(ii) f is differentiable on int(I), f ′ (x) = ncn (x − a)n−1 , and f ′ (x) has radius of convergence R;
n=0
∞
(x − a)n+1
Z X Z
(iii) f is integrable on int(I), f (x) dx = cn + C, and f (x) dx has radius of
n=0
n+1
convergence R.

Proof: The proof of (i) is left as an exercise to the reader and the remaining proofs are technical

and are omitted from this text. ■

∞
X
Example 253: Find the function whose power series representation is given by nxn .
n=1
∞
X ∞
X ∞
X
Note that nxn = x nxn−1 . Set g(x) = nxn−1 . Then,
Z n=1 n=1 n=1

g(x) dx
∞
X
= xn + C
n=1
1
= −1+C
1−x
d 1
⇐⇒ g(x) = −1+C
dx 1 − x
1
= .
(1 − x)2
This implies the following.
∞
X X∞
nxn = x nxn−1
n=1 n=1
= xg(x)
x
= , for x ∈ (−1, 1).
(1 − x)2

Exercises

Section 11.8: 3-41 odd.

Section 11.9: 3-19 odd, 25, 27, 37a, 39, 41.

44. In this exercise, you will prove part (i) of theorem 252 by proving the intermediate claims.

Claim 1: |xn − y n | ≤ n|x − y|(|x|n−1 + |y|n−1 ).

Choose an R1 ∈ (0, R).

X∞
Claim 2: The series 2n|an |R1n converges.
n=0
Claim 3: If |x| < R1 and |y| < R1 , then |F (x) − F (y)| ≤ M |x − y|.

Claim 4: If |x| < R, then F is continuous at x.

79
 Chapter 37: Taylor Polynomials
Definition 254: Let f, g : I → R, where I is an open interval, and suppose all higher derivatives

f (k) (x) and g (k) exist on I. Then f and g agree if their derivatives are the same.

Definition 255: The nth -order Taylor polynomial centered at x = a of a function f is a

polynomial of degree n defined by the following.

n
X f (k) (a)
Tn (x) = (x − a)k .
k!
k=0
Note 256: When a = 0, the nth -order Taylor polynomial centered at x = a is called the nth -order

Maclaurin polynomial.

Theorem 257: The nth -order Taylor polynomial centered at x = a is the unique polynomial of

degree n which agrees with a function f (x) to order n.

Proof: Exercise. ■

Example 258: Find the nth -order Maclaurin polynomial for f (x) = ex .

Note that 1 = f (0) = f ′ (0) = f ′′ (0) = . . . = f (n) (0).

n
X 1 k
So, f (x) = ex = x .
k!
k=0
Example 259: Calculate the nth -order Taylor polynomial for f (x) = ln(x) centered at a = 1.
1 1
Note that f (1) = 0, and that f ′ (x) = , f ′′ (x) = − 2 , etc. Also, the general pattern is
x x
that f (k) (x) is a multiple of x−k with a coefficient of ±(k − 1)! that alternates in sign (i.e.,

f (k) (x) = (−1)k−1 (k − 1)!x−k ). Furthermore, the coefficient of (x − 1)k in Tn (x) is given by
n
f (k) (1) (−1)k−1 (k − 1)! (−1)k−1 X (−1)k−1
= = , for k ≥ 1. Thus, Tn (x) = (x − 1)k .
k! k! k k
k=0
Example 260: Calculate the nth -order Maclaurin polynomial for f (x) = cos(x).

Note that the derivatives form a repeating pattern of 4. So, the coefficients of x2k+1 are zero
(−1)k
and the coefficents of x2k are given by with alternating signs for every k ∈ ω. Thus,
(2k)!
n
X (−1)k 2k
cos(x) = x .
(2k)!
k=1
Definition 261: The nth -order remainder of a Taylor polynomial Tn (x) for a function f (x) is

given by Rn (x) = f (x) − Tn (x).

Note 262: The error of the Taylor polynomial approximation for a function is given by the absolute

value of the remainder.

Lemma 263 (Taylor’s Theorem): Let f ∈ R[a,b] be a function such that f (n+1) (x) exists and is
Zx
1
continuous for every x ∈ [a, b]. Then Rn (x) = (x − u)n f (n+1) (u) du, for u ∈ [a, x].
n!
a

80
 Zx
1
Proof: Set In (x) = (x − u)n f (n+1) (u) du, for u ∈ [a, x]. We must show that In (x) = Rn (x).
n!
a
If n = 0, then I0 (x) = f (x) − f (a) = R0 (x) by the Fundamental Theorem of Calculus. So suppose
1
n > 0. Integrating In by parts and setting h(u) = (x − u)n and g(u) = f (n) (u), which means
n!
g ′ (u) = f (n+1) (u), we obtain the following.

In (x)
Zx
= h(u)g ′ (u) du
a
x
Zx
= h(u)g(u) − h′ (u)g(u) du
a
a
1
= − (x − a)n f (n) (a) + In−1 (x)
n!
f (n) (a)
⇐⇒ In−1 (x) = (x − a)n + In (x).
n!
Continuing this process and noting that I0 (x) = f (x) − f (a), we obtain the following.

f (x)
n
X f (k) (a)
= + In (x)
k!
k=0
= Tn (x) + In (x)

⇐⇒ In (x) = Rn (x), as desired. ■

Theorem 264 (Error Bound Theorem): Let f ∈ R[a,b] such that f (n+1) (x) exists and is con-

tinuous for all x ∈ [a, b], and suppose K ∈ R such that |f (n+1) (u)| ≤ K for all u ∈ [a, x]. Then
|x − a|n+1
|Rn (x)| ≤ K .
(n + 1)!
Proof: Suppose x ≥ a. Then, note the following.

|Rn (x)|
Zx
1
= (x − u)n f (n+1) (u) du
n!
a
Zx
1
≤ |(x − u)n f (n+1) (u)| du
n!
a
Zx
K
≤ |(x − u)n | du
n!
a
K −(x − u)n+1 x

=
n! n+1 a
K|x − a|n+1
= .
(n + 1)!
If x ≤ a, the result follows by an argument similar to the one above. ■

81
 Exercises

45. In this exercise, you will prove theorem 257 by proving the intermediate claims.


dj

(x − a)k

k(k − 1)...(k − j + 1)(x − a)k−j 1 if k = j;

Claim 1: For all k ∈ N, = =
dxj k! k! 
0 if k ̸= j.

Claim 2: The polynomial Tn agrees with a function f at x = a to order n.

Let a ∈ R and p(x) = an xn + . . . + a1 x + a0 be any polynomial of degree n or less.

Claim 3: If p(j) (a) = 0 for j = 0, 1, . . . , n, then aj = 0 for all j ∈ N. (Hint: Use induction.)

Claim 4: Conclude that Tn is the only polynomial of degree n or less that agrees with f at x = a

to order n.

Chapter 38: Taylor and Maclaurin Series

Definition 265: Let f (x) be a function and c ∈ R. The Taylor series centered at c for f (x) is an
∞
X
approximation f (x) = Tn (x), where Tn (x) is the nth -order Taylor polynomial of f centered at c.
n=0
Theorem 266: Let f (x) be a function and c ∈ R. If f is represented by a power series centered at

c in an interval |x − c| < r for some r > 0, then the power series is the Taylor series centered at c.

Proof: Exercise. ■

Definition 267: The Taylor series of a function centered c = 0, is called the Maclaurin series.

Example 268: Find the Taylor series for f (x) = x−3 centered at c = 1.
1 1
Note that f (n) (x) = (−1)n ( )(n+2)!x−3−n . This means f (n) (1) = (−1)n ( )(n+2)!. By the theorem
2 2
f (n) (1) (−1)n (n + 2)!
above, the coefficients for the Taylor series for f (x) are given by an = = =
n! 2n!
(n + 2)(n + 1)
(−1)n .
2
∞
X (−1)n (n + 2)(n + 1)
Therefore, by the theorem above, the Taylor series is given by f (x) = (x −
n=0
2
1)n .

Proposition 269: A Taylor series of a function converges to the function if and only if lim Rk = 0.
k→∞
Proof: Exercise. ■

Theorem 270: Let c ∈ R so that |c| < r, for some r > 0, I = (c − r, c + r), and suppose there is a

K > 0 such that all derivatives of a differentiable function f are bounded by K on I. Then for all

x ∈ I, f (x) is represented by its Taylor series expansion in I.

|x − c|k+1
Proof: By the error bound theorem for Taylor polynomials, |Rk (x)| = |f (x)−Tk (x)| ≤ K .
(k + 1)!
rk+1 rk
If x ∈ I, then |x − c| < r and |Rk (x)| ≤ K . Recall that lim = 0. Therefore, lim Rk = 0
(k + 1)! k→∞ k! k→∞

82
for all x ∈ I, as desired. ■

Proposition 271: For all x ∈ R, the following Maclaurin expansions hold.

∞
X x2n+1
(i) sin(x) = (−1)n ;
n=0
(2n + 1)!
∞
X x2n
(ii) cos(x) = (−1)n ;
n=0
(2n)!
∞
X xn
(iii) ex = .
n=0
n!
Proof: For (i), recall that the even derivatives of sin(x) at x = 0 are zero and the odd derivatives
(−1)n
at zero alternate in sign. So, the Taylor coefficients are given by a2n+1 = , and apply a
(2n + 1)!
previous theorem.

For (ii), use an argument similar to the one above.

For (iii), note that if f (x) = ex , then f (n) (0) = e0 = 1. Thus, the Maclaurin series for ex is given
∞
X xn
by ex = , as desired. ■
n=0
n!
Example 272: Express e as an infinite series.
∞
X 1
By the proposition above, e = .
n=0
n!
Definition 273: Let p ∈ ω. The Bessel function of order p is the power series Jp (x) =
∞
X (−1)k x2k+p
2k+p
.
2 k!(k + p)!
k=0
Theorem 274: J0 (x) converges for every x ∈ R.

Proof: Exercise. ■

Proposition 275: Let p ∈ ω. The zeros of Jp (x) and Jp+1 (x) alternate.

Proof: Exercise. ■

Definition 276: Let k ∈ R with |k| < 1. The Elliptic function of the first kind is defined by
Zπ/2
1
E(k) = q dt.
1 − k 2 sin2 (t)
0
Zπ/2q
Definition 277: The function G(k) = 1 − k 2 sin2 (t) dt is called an elliptic function of the
0
second kind.

83
 Exercises

Section 11.10: 5-27 odd, 35-43 odd, 73-79 odd.

46. Prove theorem 266.

47. Prove proposition 269.

48. Prove theorem 274.

49. Prove proposition 275.

∞  2
π π X 1 · 3 · . . . · (2n − 1) k 2n
50. Show that for |k| < 1, G(k) = − · .
2 2 n=1 2 · 4 · . . . · (2n) 2n − 1
 x 2  y 2
51. Suppose that a < b and let L denote the arc length (circumference) of the ellipse + =
a b
1. It
r is known that L = 4bG(k), where G(k) is the Elliptic function of the second kind and
a2
k = 1− 2.
b
(a) Approximate L with a = 4 and b = 5.
a2
 
a π
(b) If a < b and is near 1 (i.e., the ellipse is nearly circular), show that L ∼
= · 3b + .
b 2 b
52. In this exercise, you will prove that e is irrational by proving the intermediate claims.

Proof: Toward a contradiction, suppose that e is rational. Then there are m, n ∈ Z \ {0} such that
m
e= .
n
Claim 1: The number m!e−1 ∈ ω.
 
−1 m+1 1 1
Claim 2: There is a b ∈ Z such that m!e = b + (−1) − + ... .
m + 1 (m + 1)(m + 2)
Claim 3: |m!e−1 − b| ∈ (0, 1).

Note that claim 3 contradicts claim 1. Therefore, e is irrational. ■

∞  2
π π X 1 · 3 · . . . · (2n − 1) k 2n
53. Prove that for |k| < 1, G(k) = − · .
2 2 n=1 2 · 4 · . . . · 2n 2n − 1

84
 Chapter 39: Fourier Series
Theorem 278: Let f : [−L, L] → R be discontinuous on at most finitely many values in [−L, L].

Then f is bounded on [−L, L].

Definition 279: Let f be a function and p ∈ R. Then p is the period of f if it is the smallest

positive integer such that f (x + p) = f (x) for every x ∈ dom(f ).

Definition 280: A function is periodic if it has a period.

Definition 281: Let f : [−L, L] → R be periodic. The Fourier series for f is the series
∞
a0 X  nπx   nπx 
+ an cos + bn sin , where
2 n=1
L L
ZL
1
(i) a0 = f (x) dx;
L
−L
ZL
1  nπx 
(ii) ak = f (x) cos dx;
L L
−L
ZL
1  nπx 
(iii) bk = f (x) sin dx.
L L
−L
∞
a0 X  nπx   nπx 
Note 282: Suppose that f (x) = + an cos + bn sin . Consider the following.
2 n=1
L L
∞
a0 X  nπx   nπx 
(i) f (x) = + an cos + bn sin
2 n=1
L L
ZL ZL ∞
a0 X  nπx   nπx 
⇐⇒ f (x) dx = + an cos + bn sin dx
2 n=1
L L
−L −L
ZL ZL
 
X ∞  nπx   nπx 
= a0 L + an cos dx + bn sin dx
n=1
L L
−L −L
ZL
1
⇐⇒ a0 = f (x) dx.
L
−L
∞
a0 X  nπx   nπx 
(ii) f (x) = + an cos + bn sin
2 n=1
L L
ZL  
kπx
⇐⇒ f (x) cos dx
L
−L
ZL   ZL " X
∞    #
a0 kπx  nπx  kπx  nπx  kπx
= cos dx + an cos cos + bn sin cos dx
2 L n=1
L L L L
−L −L
 L
ZL

∞    
nπx kπx  nπx  kπx
X Z  
=  an cos cos dx + bn sin cos dx.
n=1
L L L L
−L −L
By a previous note and a previous theorem, we obtain the following.

85
ZL  
kπx
f (x) cos dx = an L
L
−L
ZL  
1 kπx
⇐⇒ an = f (x) cos dx.
L L
−L
∞
a0 X  nπx   nπx 
(iii) f (x) = + an cos + bn sin
2 n=1
L L
L  
kπx
Z
⇐⇒ f (x) sin dx
L
−L
ZL   ZL " X
∞    #
a0 kπx  nπx  kπx  nπx  kπx
= sin dx + an cos sin + bn sin sin dx
2 L L L L L
−L −L n=1
 L
ZL

∞  nπx     
kπx nπx kπx
X Z  
=  an cos sin dx + bn sin sin dx.
n=1
L L L L
−L −L
By a previous note and a previous theorem, we obtain the following.
ZL  
kπx
f (x) sin dx = bn L
L
−L
ZL  
1 kπx
⇐⇒ bn = f (x) sin dx.
L L
−L 

0
 if x ∈ (−π, 0];
Example 283: Consider the square-wave function f (x) = where f is as-

1
 if x ∈ (0, π],
sumed to be periodic outside of the interval [−π, π]. Find the Fourier series expansion for the

function f .

Consider the following.

Zπ Z0 Zπ
1 0
(i) a0 = f (x) dx = 0 dx + 1 dx = 1;
π π
−π −π 0
Zπ Zπ π
1 0 1
(ii) If k ≥ 1, then ak = f (x) cos(kx) dx = cos(kx) dx = sin(kx) = 0;
π π πk 0
−π 0
Zπ
1
(iii) bk = f (x) sin(kx) dx
π
−π
Zπ
1
= sin(kx) dx
π
0
1
=− [cos(kπ) − cos(0)]
πk
1 
(−1)k − 1

=−
πk

0
 if k is even;
=
 2 if k is odd.


πk

86
Notice that we can write the even- and odd-indexed coefficients separately as b2k = 0 and b2k−1 =
2
for any k ∈ N. Hence, the Fourier series is given by
(2k − 1)π
∞
1 X
+ bk sin(kx)
2 n=1
∞
1 X
= + b2k−1 sin[(2k − 1)x]
2 n=1
∞
1 X 2
= + sin[(2k − 1)x].
2 n=1 (2k − 1)π

Theorem 284 (Fourier Convergence Theorem): Let f be a differentiable function of period 2l

such that f and f ′ are continuous on [−l, l] except for a finite amount of jump discontinuities. Then

f has a Fourier series expansion.

(i) When f is continuous at x, the series converges to f (x).

 
1
(ii) Whenever f is discontinuous at a point x, then the series converges to lim+ f (t) + lim− f (t) .
2 t→x t→x
Proof: The proof is technical and is omitted from this text. ■

Note 285: The Fourier Convergence Theorem says that a Fourier series may converge to a discon-

tinuous function, even though every term in the series is continuous (and differentiable).

Exercises

In exercises #14-16, find the Fourier series expansion of the given function on the given interval.

Assume that the functions are periodic outside the interval.

54. f (x) = −x; [−1, 1]

2
55. g(x) = x
 ; [−1, 1]

0
 if x ∈ (−1, 0);
56. h(x) =

1 − x if x ∈ (0, 1).

∞
X 1 π2
57. Conclude that = .
k! 6
k=1
∞
X sin(2k − 1) π
58. Show that = .
2k − 1 4
k=1
∞
X 1 π2
59. Show that = .
(2k − 1)2 8
k=1
60.

(a) If a function f is even, show that the Fourier series expansion of f consists only of a constant

and cosine terms.

(b) If a function f is odd, show that the Fourier series expansion of f consists only of a constant

and sine terms.

87
 Chapter 40: Separable Differential Equations
Definition 286: A differential equation y ′ = f (x, y) is separable if it can be expressed as g(y)y ′ =

h(x).

Example 287: Determine whether y ′ = xy 2 − 2xy is separable.

To see that the equation is separable, note the following.

y′
y ′ = xy 2 − 2xy = x(y 2 − 2y) ⇐⇒ 2 = x.
y − 2y
Therefore, the equation is separable.

Example 288: The equation y ′ = xy 2 − 2x2 y is not separable.

Proposition 289: The differential equation y ′ = f (x, y) = g(y)h(x) is separable if and only if
y′
Z Z
dy = h(x) dx.
g(y)
x2 + 7x + 3
Example 290: Solve the differential equation y ′ = .
y2
Note the following.
x2 + 7x + 3
y′ =
y2
2 ′
⇐⇒ y y = x2 + 7x + 3
Z Z
2 ′
⇐⇒ y y dx = x2 + 7x + 3 dx
Z Z
⇐⇒ y dy = x2 + 7x + 3 dx
2

1 3 1 7
⇐⇒ y = x2 + x2 + 3x + C
3 3 2
21
⇐⇒ y 3 = x3 + x2 + 9x + 3C
r 2
21
⇐⇒ y = x3 + x2 + 9x + 3C.
3

2
x2 + 7x + 3
Example 291: Find the solution of the differential equation y ′ = that satisfies the
y2
given condition y(0) = 3. r
21 2
By the previous example, we get y = 3
x3 +
x + 9x + 3C. From the initial condition, we get
2 r
√ 21
3 = y(0) = 3 3C, which means C = 9. Thus, y = 3 x3 + x2 + 9x + 27.
2
Example 292: A tank contains 100 gallons of brine which contains .5 pounds of salt per gallon.

Brine that contains .25 pounds of salt per gallon flows into the tank at a rate of 4 gallons per minute.

The solution is kept thoroughly mixed and drains from the tank at a rate of 4 gallons per minute.

Express the amount of salt in the tank as a function of time.

dQ 1 Q Q
Let Q(t) denote the amount of salt in the tank at time t. Then = ·4− ·4=1− and
dt 4 100 25
Q(0) = 50. This implies the following.
dQ Q 25 − Q
=1− =
dt 25 25

88
 dQ 1
⇐⇒ = dt
Z25 − Q 25 Z
dQ 1
⇐⇒ = dt
25 − Q 25
1
⇐⇒ − ln |25 − Q| = t + C.
25
Since Q(0) = 50, we obtain the following.

Q(0) = 50

⇐⇒ − ln | − 25| = C

⇐⇒ C = − ln 25
1
⇐⇒ − ln |25 − Q| = t − ln 25
25
1
⇐⇒ ln |Q − 25| = ln 25 − t
25
1
ln 25− t
⇐⇒ Q − 25 = e 25 = eln 25 e−t/25 = 25e−t/25

⇐⇒ Q = 25e−t/25 + 25.

Exercises

Section 9.3: 11-17 odd, 45, 47.

Section 11.9: 37.

89
 Chapter 41: Arc Length
Theorem 293: Let a, b ∈ R with a ≤ b and suppose f : [a, b] → R such that f ′ is continuous. Then
Zb p
the length of the curve y = f (x) from a to b is 1 + [f ′ (x)]2 dx.
a
Example 294: For each of the following, find the length of the curve.

(b) y = 3x2/3 − 10; x ∈ [8, 27]

x2
(a) y = − ln x; x ∈ [2, 4]
8 Consider the following.
Consider the following.
y = 3x2/3 − 10
x2
y= − ln x
8 ⇐⇒ y ′ = 2x−1/3
x 1
⇐⇒ y ′ = −
4 x ⇐⇒ [y ′ ]2 = 4x−2/3
′ 2 x2 1 1
⇐⇒ [y ] = − + 2 ⇐⇒ 1 + [y ′ ]2 = 1 + 4x−2/3
16 2 x
′ 2 x2 1 1 √
⇐⇒ 1 + [y ] =
p
+ + 2 ⇐⇒ [y ′ ]2 = 1 + 4x−2/3
2 16 2 x s 1+ √
3
4 + x2

1 1
= x+ = √ .
4 x 3
x2
p 1 1 By the theorem above, the length of the curve
⇐⇒ 1 + [y ′ ]2 = x + .
4 x
By the theorem above, the length of the curve is s
Z27 √
3
4 + x2
is √3
Z4 x2
1 1 8
x + dx Z13
4 x 3√
2 = u du
1 2 4 2
= x + ln x 8
8 2 13
1 = u3/2
= 2 + ln 4 − − ln 2 √8 √
2
3 = 13 13 − 8 8
= + ln 2. √ √
2
= 13 13 − 16 2.

Exercises

Section 8.1: 9-19 odd, 37, 39.

90
 Chapter 42: Solids of Revolution and Surface Area
Theorem 295: Suppose f : [a, b] → R such that f ′ is continuous on [a, b]. Also, let R denote the

region bounded by the curves y = f (x), x = a, x = b, and the x-axis. Then the surface area of the
Zb
solid obtained by revolving R about the x-axis is
p
2πf (x) 1 + [f ′ (x)]2 dx.
a
Note 296: In the theorem above, let y = f (x) and suppose that s denotes the arc length of the

curve. Then by a previous theorem and the Fundamental Theorem of Calculus,

Zb p
s= 1 + [f ′ (x)]2 dx
a
ds p
⇐⇒ = 1 + [f ′ (x)]2
dx p
⇐⇒ ds = 1 + [f ′ (x)]2 dx
Zb p Zb
′ 2
⇐⇒ 2πf (x) 1 + [f (x)] dx = 2πy ds.
a a √
Example 297: Find the surface area of the solid obtained by rotating the curve f (x) = x−1

about the x-axis for x ∈ [1, 13].

Consider the following.

√
y = x−1
1
⇐⇒ y ′ = √
p2 x − 1 √
⇐⇒ 2πy [y ′ ]2 + 1 = π 4x − 3
Z13 Z13
p √
′ 2
⇐⇒ 2πy [y ] + 1 dx = π 4x − 3 dx
1 1
πp 13
= (4x − 3)3
6 1
π
= (343 − 1)
6
= 57π.
√
Example 298: Find the surface area of the solid obtained by rotating the curve y = 1 + ex about

the x-axis for x ∈ [0, 1].

Consider the following.

√
y = 1 + ex
ex
⇐⇒ y ′ = √
2 1 + ex s
p √ e2x
⇐⇒ 2πy [y ′ ]2 + 1 = 2π 1 + ex · +1
4(1 + ex )
= π(ex + 2)
Z1 p Z1
⇐⇒ 2πy [y ] + 1 = π(ex + 2) dx
′ 2

0 0

91
 1
= ex + 2x
0
= π(e + 1).
  
1
Example 299 (Gabriel’s Horn): Consider the region R = (x, y) ∈ R | x ≥ 1 and y ∈ 0,
2
.
x
The solid obtained after revolving R about the x-axis is called Gabriel’s Horn.

(a) Find the surface area of Gabriel’s Horn.

To find the surface area of R, consider the following.

1
y′ = − 2
x r
p
′ 2
1 1
⇐⇒ 2π [y ] + 1 = 2π 1+ 4
x x
1
≥ .
x
Z∞
1
Recall that dx diverges. Therefore, by the Comparison Theorem for Improper Integrals,
x
1
Z∞ r
1 1
2π 1 + 4 dx diverges, and so, the surface area of R is infinite.
x x
1
(b) Find the volume of Gabriel’s Horn.

To find the volume of R, consider the following.

Z∞  2
1
π dx
x
1
11−2
=π·
2−1
= π.

Note that the volume of R is finite, but it has an infinite surface area.

Exercises

Section 8.2: 7-17 odd.

61. Find the surface area of the solid obtained by rotating the circle x2 + (y − b)2 = r2 about the

axis.

Definition: The solid described above is called a torus.

√
62. Show that the surface area of a right circular cone of radius r and height h is πr r2 + h2 .

92
 Chapter 43: Centers of Mass
Theorem 300: Suppose that f : R → R is continuous with f (x) ≥ 0 for every x ∈ [a, b] and let R

denote the region bounded by the curves x = a, x = b, y = 0, and y = f (x). Also, let A denote the

area of R. Then the center of mass of R is the point (x, y), where
Zb
1
(i) x = xf (x) dx;
A
a
Zb
1
(ii) y = [f (x)]2 dx.
2A
a
Theorem 301: Suppose that f, g : R → R are continuous and f (x) ≤ g(x) for every x ∈ [a, b] and

let R denote the region bounded by the curves x = a, x = b, y = f (x), and y = g(x). Also, let A

denote the area of R. Then the center of mass of R is the point (x, y), where
Zb
1
(i) x = x[g(x) − f (x)] dx;
A
a
Zb
1
(ii) y = [g(x)]2 − [f (x)]2 dx.
2A
a
Example 302: Suppose that masses m1 = 5, m2 = 4, m3 = 3, and m4 = 6 lie on the points

P1 = (−4, 2), P2 = (0, 5), P3 = (3, 2), and P4 = (1, −2), respectively. Find the moments and center

of mass of the system.

4
X
The mass of the system is 5 + 4 + 3 + 6 = 18. Also, the x-moment is Mx = mi yi = 5(2) +
i=1
X 4
4(5) + 3(2) + 6(−2) = 24, and the y-moment is My = mi xi = 5(−4) + 4(0) + 3(3) + 6(1) = −5.
   i=1   
My M x 5 24 5 4
Therefore, the center of mass is , = − , = − , .
m m 18 18 18 3
√
Example 303: Find the center of mass of the region bounded by y = x, y = 0, x = 0, and x = 4.
Z4
√ 16
Note that the area of the region is x dx = . So, by the theorems above,
3
0
Z4
3 12
x= x3/2 dx = , and
16 5
0
Z4
3 3
y= x dx = .
32 4
0  
12 3
Therefore, the center of mass is the point , .
5 4

93
 Exercises

Section 8.3: 23-33 odd.

63. In this exercise, you will determine how far a stack of identical books (of mass m and unit

length) can extend without tipping over. The stack will not tip over if the (n + 1)st book is placed

at the bottom of the stack with its right edge located at the center of mass of the first n books. Let

cn be the center of mass of the first n books, measured along the x-axis, where we take the positive

x-axis to the left of the origin. Note that if an object of mass m1 has center of mass at x1 and a

second object of m2 has center of mass x2 , then the center of mass of the system has x-coordinate
m1 x1 + m2 x2
.
m1 + m2
(a) Show that if the (n + 1)st book is placed with its right edge at cn , then its center of mass is
1
located at cn + .
2
(b) Consider the first n books as a single object of mass nm with center of mass at cn and the

(n + 1)st book as a second object of mass m. Show that if the (n + 1)st book is placed with its right
1
edge at cn , then cn+1 = cn + .
2(n + 1)
(c) Prove that lim cn = ∞.
n→∞

Note: Part (c) implies that if we use enough books, the stack can be extended as far as desired

without tipping over.

Chapter 44: Hydrostatic Pressure

Notation 304: Consider the table below.

Symbol Measurement Metric U.S. Customary Equations

m mass kg
g acceleration due to gravity 9.8 m/s2 32 ft/s2
A area of horizontal plate (lamina) m2 ft2
ρ mass density kg/m3
δ weight density lb/ft3 δ = ρg
2
P pressure N/m = 1 Pa P = ρgd = δd
F force N lb F = mg = ρgdA = P a

Example 305: A tank is 8 meters long, 4 meters wide, 2 meters high, and contains kerosene with

density 820 kilograms per cubic meter to a depth of 1.5 meters.

(a) Find the hydrostatic pressure on the bottom of the tank.

P = ρgd = 820 · 9.8 · 1.5 = 12054 Pascals.

(b) Find the hydrostatic force on the bottom of the tank.

94
F = P A = 12054 · 8 · 4 = 385628 Newtons.

(c) Find the hydrostatic force on one end of the tank.

The area of the ith strip is 4 · ∆x and the pressure is ρgd = ρgxi . Thus,
Z1.5 Z1.5
F = ρgx · 4 dx = 820 · 9.8 · 4 x dx = 36162 Newtons.
0 0
Example 306: In the figure below, a square vertical plate of side length a is submerged in water.

Find the hydrostatic force against one side of the plate.

a a

a a

95
Consider the figure below.
y

√a
2

a a

∆y
yi∗
x

a a

− √a2

We will consider the top and bottom halves separately. For the top half, the length of the ith strip is
 2
2a 2a 2a
√ and the area is √ ∆y. The pressure on the strip is approximately 2δ √ ∆y.
2 − yi∗ 2 − yi∗ 2 − yi∗
So the total force on the top half is
n  2
X 2a ∗
F1 = lim 2δ √ − yi ∆y
n→∞
i=1
2
√
a/
Z 2 2
2a
= 2δ √ −y dy
2
0 √
"  3 a/ 2
1 2a
= 2δ − √ −y
3 2
√ 3 0
2a δ
= .
6
Z0 √
a2 2 2a3 δ
Similarly, the force on the bottom half is F2 = − y 2 dy = . Therefore, the total
√
2 6
−a/ 2
√
2a3 δ
force is F = F1 + F2 = Newtons.
2
Example 307: A milk truck carries milk with density 64.6 pounds per cubic foot in a horizontal

cylindrical tank with diameter 6 feet.

(a) Find the force exerted by the milk on one end of the tank when the tank is full.

The pressure on a strip is approximately δdi = 64.6(3 − yi∗ ) and the total force is
Xn q
lim 64.6(3 − yi∗ ) · 2 9 − (yi∗ )2 ∆y
n→∞
i=1
Z3 p
= 129.2 (3 − y) 9 − y 2 dy
−3
Z3 p Z3 p
= 129.2 · 3 9 − y dy − 129.2 y 9 − y 2 dy
2

3 −3

96
 387.7π
= ·9
2
= 1744.2π pounds.

(b) What if the tank is half full?

If the tank is half-full, then the surface area of the milk is 0, so the pressure on a strip is approximately

δdi = 64.6(0 − yi∗ ) and the upper limit becomes 0. Hence, the force is
Z0  0
p 1
129.2 −y 9 − y dy = 129.2 · (9 − y 2 )3/2
2 = 1162.8 pounds.
3 −3
−3
Example 308: A vertical dam has a semicircular gate as shown in the figure below. Find the

hydrostatic force against the gate.

Z2 p
The force is given by F = ρg(10 − x) · 2 4 − x2 dx
0
Z2 p Z2 p
= 20ρg 4− x2 dx − ρg 4 − x2 · 2x dx
0 0
16
= 20πρg − ρg
 3 
16
= ρg 20π −
 3 
16
= 9800 20π − Newtons.
3

Exercises

Section 8.3: 1-17 odd.

97
 Chapter 45: Parametric Equations
Definition 309: Let I ⊆ R be a (possibly infinite) interval and suppose f, g : I → R are continuous.

(i) A set C = {(x, y) ∈ R2 | x = f (t), y = g(t), t ∈ I} is called a plane curve;

(ii) x = f (t) and y = g(t) are called parametric equations for C ;

(iii) c(t) = (f (t), g(t)) is called a parametrization;

(iv) t is called a parameter.

Example 310: For each of the following, eliminate the parameter to find a Cartesian equation of

the curve and sketch the curve.

√ t
(a) x = t − 2 and y = − 7, t ≥ 0
2
Note the following.
√
x= t−2

⇐⇒ t = (x + 2)2
1
⇐⇒ y = (x + 2)2 − 7
2
1 2
= x + 2x − 5.
2
The curve is sketched below.

√
(b) x = t, y = 2 ln t, t > 0

Note the following.

√
x= t

⇐⇒ x2 = t

⇐⇒ y = 2 ln(x2 ) = 4 ln(x).

98
The curve is sketched below.

π
(c) x = sin(t), y = cos2 (t), 0 ≤ t ≤
2
Note the following.

1 = cos2 (t) + sin(t) = y + x2

⇐⇒ y = 1 − x2 .

Start at (0, 1) and end at (1, 0). The curve is sketched below.

Exercises

Section 10.1: 1-21 odd, 24, 28.

99
 Chapter 46: Parametric Equations and Tangents
Definition 311: A plane curve with parametric equations x = f (t) and g = y(t) defined on an

interval I is smooth if it satisfies the two properties below.

(i) f ′ , g ′ : I → R) are continuous;

(ii) If f ′ (x) = 0 = g ′ (x), then x is an endpoint of I.

Theorem 312: Suppose that C is a smooth curve in R2 with x = f (t) and y = g(t). Then the
g ′ (t0 )
slope of the tangent line to C at a point (x0 , y0 ) = (f (t0 ), g(t0 )) is given by ′ , where f ′ (t0 ) ̸= 0.
f (t0 )
dy
Notation 313: The slope of the tangent line in the situation above is usually denoted .
dx
Note 314: Recall the following.
dy dy dx
= ·
dt dx dt
dy
dy g ′ (t)
⇐⇒ = dt = ′ .
dx dx f (t)
dt
Example 315: Let C denote the curve defined by x = t sin2 t and y = t cos t. Find the equation of
π
the line that is tangent to the curve at the point corresponding to t = .
2
dy cos t − t sin t π
By the note above, = . If t = , then
dx 2t cos t sin t + sin2 t 2
π
(i) x = ;
2
(ii) y = 0;
π
dy − π
(iii) = 2 =− .
dx t=π/2 1 2
π
This implies that the tangent line has slope - . Hence, the tangent line is given by y − 0 =
2
π π π π2
− x− , which means y = − x − .
2 2 2 4
Example 316: Let C denote the curve defined by x = et and y = ln t. Find the equation of the

line that is tangent to the curve at the point corresponding to t = 2.

dy et
By the note above, = = tet . If t = 2, then
dx 1
t
(i) x = e2 ;

(ii) y = ln 2;
dy
(iii) = 2e2 .
dx t=2
This implies that the tangent line has slope 2e2 . Hence, the tangent line is given by y − ln 2 =

2e2 (x − e2 ), or y = 2e2 x − 2e4 + ln 2.

Example 317: Set x = t2 + 1 and y = t ln t.

dy
(a) Find .
dx

100
 dy 1 + ln t
By the note above, = .
dx 2t
d2 y
(b) Find .
dx2
Consider the following.
d2 y
dx2  
d dy
=
dt dx
2 − 2(1 + ln t)
=
4t2
ln t
= − 2.
2t

Exercises

Section 10.2: 1-19 odd.

101
 Chapter 47: Parametric Equations and Area
Example 318: Consider the area A of the circle centered at the point (0, 0) with radius 5.
Z5 p
A=2 25 − x2 dx.
−5 x √
Set x = 5 sin θ. Then θ = sin−1 , 25 − x2 = 5 cos θ, and dx = 5 cos θ dθ. Hence,
5
A
Z5
=2 25 cos2 θ dθ
−5
Z5
= 50 cos2 θ dθ
−5
 
1 1
= 50 cos θ sin θ + θ
2 2
= 25 [cos θ sin θ + θ]
h h  x i x x 5
= 25 cos sin−1 · + sin−1
5 5 5 −5
"r 5
2
x x  x 
= 25 1− · + sin−1
25 5 5
hπ πi −5
= 25 +
2 2
= 25.

Theorem 319: Let C be a plane curve with parametric equations x = f (t) and y = g(t) such

that f, g : [a, b] → R are differentiable. Also, suppose that for all t1 , t2 ∈ [a, b] with t1 < t2 ,

(f (t1 ), g(t1 )) = (f (t2 ), g(t2 )) if and only if t1 = a and t2 = b. Then the area of the region enclosed
Zb Zb
′
by C is f (t)g (t) dt = g(t)f ′ (t) dt .
a a
Example 320: Consider the example above.

The circle is parametrized by x = 5 cos θ and y = 5 sin θ for θ ∈ [0, 2π]. Note that y ′ = 5 cos θ. By

the theorem above,

Z2π
A = 25 cos2 θ dθ
0
 2π
1 1
= 25 sin θ cos θ + θ
2 2 0
25 2π
= [sin θ cos θ + θ|0
2
= 25π.

102
 Exercises

Section 10.2: 31-34.

Chapter 48: Parametric Equations and Arc Length

Theorem 321: Let C be a smooth plane curve with parametric equations x = f (t) and y = g(t)

such that f, g : [a, b] → R are differentiable. Also, suppose that for all t1 , t2 ∈ [a, b] with t1 <

t2 , (f (t1 ), g(t1 )) = (f (t2 ), g(t2 )) if and only if t1 = a and t2 = b. Then the arc length of C is
Zb p
[f ′ (t)]2 + [g ′ (t)]2 dt.
a
Example 322: Find the length of the plane curve given by x = et and y = 4 where t ∈ [0, ln 3].

Note that x′ = et and y ′ = 0. By the theorem above, the arc length is given by
Zln 3p
(et )2 + 02 dt
0
Zln 3
= et dt
0
ln 3
= et
0
=3−1

= 2.

Exercises

Section 10.2: 37-43 odd.

103
 Chapter 49: Parametric Equations and Surface Area
Theorem 323: Let C be a smooth plane curve with parametric equations x = f (t) and y = g(t)

such that f : [a, b] → R and g : [a, b] → [0, ∞) are differentiable. Also, suppose that for all

t1 , t2 ∈ [a, b] with t1 < t2 , (f (t1 ), g(t1 )) = (f (t2 ), g(t2 )) if and only if t1 = a and t2 = b. Then the
Zb
surface area of the solid formed by rotating C about the x-axis is
p
2πg(t) [f ′ (t)]2 + [g ′ (t)]2 dt.
a
Example 324: Let C denote the parametric curve defined by x = x2 + 1 and y = 6t, where
√ √
t ∈ [ 7, 55]. Find the surface area of the solid formed by rotating C about the x-axis.

Note
√
that x′ = 2x and y ′ = 6. So by the theorem above, the surface area is given by
Z 55p
4t2 + 36 dt
√
7
√
Z 55p
=2 t2 + 9 dt.
√
7
√
 
−1 t
Set t = 3 tan θ. Then θ = tan , t2 + 9 = 3 sec θ, and dt = 3 sec2 θ dθ. Now,
3
√
Z 55p
2 t2 + 9 dt
√
7
√
Z 55
= 18 sec3 θ dθ
√
7
√
 55
tan θ sec θ 1
= 18 + ln | sec θ + tan θ √
2 2
     7   
t −1 t −1 t t
=9 · sec tan + ln sec tan +
3 3 √
3 3
 p  55
p t
= 3t 1 + t2 + 9 ln 1 + t2 +
3 √7
√ √ !
√ √ 55 √ √ 7
= 3 55 · 56 + 9 ln 56 + − 3 7 · 50 + 9 ln 50 +
3 3
√ √ !
√ √ 6 14 + 55
= 12 770 − 15 14 + 9 ln √ √ .
15 2 + 7

Exercises

Section 10.2: 57-63 odd.

104
 Chapter 50: Polar Coordinates
Note 325:

In the figure above, note the following.

(i) r is the distance from the origin to the point (x, y);

(ii) θ is the angle made by the x-axis and the line segment of length r.

Definition 326: In the figure above,

(i) The ordered pair (r, θ) in the figure above is a polar coordinate representation of a point

(x, y).

(ii) r is usually called the radial coordinate;

(iii) θ is usually called the angular coordinate.

Note 327:

(i) Polar coordinate representations of a point (x, y) are not unique.

(ii) We may choose to specify unique polar coordinates for points other than the origin by placing

restrictions on r and θ;

(iii) The origin does not have a well-defined polar coordinate representation.

(iv) When plotting points in polar coordinates, the x-axis is called the pole or the polar axis;

(v) We usually do not include a y-axis when plotting points in polar coordinates.

Proposition 328: Let (x, y) ∈ R2 and suppose (r, θ) is a corresponding polar coordinate represen-

tation for (x, y).

(i) x = r cos(θ);

(ii) y = r sin(θ);

105
(iii) x2 + y 2 = r2 ;
y
(iv) tan(θ) = , provided x ̸= 0.
x
Proof:

(iii) Use the Pythagorean theorem.

y r sin(θ)
(iv) Note that = = tan(θ). ■
x r cos(θ)
Example 329: Plot the given points.
 π
(a) 2,
4

 π
(b) 3,
3

 π
(c) 2, −
2

106
  π
(d) 0,
6

(e) (0, 4)

107
Note 330: If r < 0, then (r, θ) and (−r, θ + π) give the same polar coordinate representation of a

given point.

Example 331: Plot the following points.

 π
(a) −3,
6

 
5π
(b) −2, −
6

108
Example 332:

(a) Give a polar coordinate representation for the point (1, 1).
√ π  π
Note that r2 = 2, which means r = 2 and that θ = tan−1 (1) = . Thus, the point 2, is a
4 4
polar coordinate representation for (1, 1).
 
5π
(b) Find a rectangular coordinate representation of the point 3, .
6
  √ √ !
5π 3 π 3 3 3 3
Note that x = 3 cos = −3 and y = 3 sin = . Thus, the point − , is a
6 2 6 2 2 2
 
5π
rectangular coordinate representation of the point 3, .
6

Exercises

Section 10.3: 1-6.

109
 Chapter 51: Polar Curves
Definition 333:

(i) A polar equation is an equation that describes a curve using polar coordinates.

(ii) The graph of a polar equation is {(x, y) ∈ R2 | x = r cos(θ), y = r sin(θ), and r = f (θ)}.

(iii) The graph of a polar equation is called a polar curve.

Example 334: Find a polar equation for the curve described by the given Cartesian equation.

(a) y = 5

Consider the following.

y=5

⇐⇒ r sin θ = 5
5
⇐⇒ r = = 5 csc θ.
sin θ
x2
(b) y = .
4
Consider the following.
x2
y=
4
r2 cos2 θ
⇐⇒ r sin θ =
4
1 cos2 θ
⇐⇒ =
r 4 sin θ
4 sin θ
⇐⇒ r = = 4 tan θ sec θ.
cos2 θ
Example 335: Find a Cartesian equation for the curve described by the given polar equation and

sketch a graph.
√
(a) r = 5 2
√
This is a circle of radius 5 2 centered at the point (0, 0). The graph is sketched below.

110
(b) r2 cos(2θ) = 1.

Note the following.

r2 cos(2θ) = 1

⇐⇒ r2 (cos2 θ − sin2 θ) = 1

⇐⇒ r2 cos2 θ − r2 sin2 θ = 1

⇐⇒ x2 − y 2 = 1.

This is a hyperbola. The curve is sketched below.

(c) r = θ for θ ≥ 0.

As θ increases, r also increases. Thus, the polar curve is a spiral, called an Archimedian spiral,

111
and is sketched below.
105 90 75
120 60

135 45

150 30

165 15

0 200 400 600

180 0

195 345

210 330

225 315

240 300
255 270 285

Exercises

Section 10.3: 7-47 odd, 65.

112
 Chapter 52: Polar Curves and Tangents
Example 336: For each of the following, find the slope of the line that is tangent to the given curve

at the indicated point.

π
(a) r = 7; θ =
6
dy 7 cos θ π
Note that x = 7 cos θ and y = 7 sin θ. Now, = = − cot θ. If θ = , then
√ dx −7 sin θ 6
7 3
(i) x = ;
2
7
(ii) y = ;
2
dy √
(iii) = − 3.
dx
Hence, the equation of√ the
! tangent line is given by
7 √ 7 3
y− =− 3 x−
2 2
√ 21
= − 3x +
√2
⇐⇒ y = − 3x + 14.
π
(b) r = θ; θ =
2
dy sin θ + θ cos θ π
Note that x = θ cos θ and y = θ sin θ. Now, = . If θ = , then
dx cos θ − θ sin θ 2
(i) x = 0;
π
(ii) y = ;
2
dy 2
(iii) =− .
dx π
2 π
Thus, the line is given by y = − x + .
π 2

Exercises

Section 10.3: 55-63 odd.

113
 Chapter 53: Polar Curves and Area
Theorem 337: Let a, b ∈ [0, 2π] with a < b. Also, suppose that f : [a, b] → R is continuous and

f (θ) ≥ 0 for all θ ∈ [a, b]. Then the area of the region bounded by the polar curves θ = a, θ = b,
Zb
1
and r = f (θ) is [f (θ)]2 dθ.
2
a
Example 338: Consider the graph of r = 4 sin θ.
105 90 75
120 60

135 45

150 30

165 15

0 1 2 3 4
180 0

195 345

210 330

225 315

240 300
255 270 285

The area under the curve is given by

Zπ
1
16 sin2 θ dθ
2
0
Zπ
=8 sin2 θ dθ
0
π
= 4θ − 2 sin 2θ
0
= 4π.

114
(b) Consider the graph of r = 3 cos 2θ.
105 90 75
120 60

135 45

150 30

165 15

0 1 2 3
180 0

195 345

210 330

225 315

240 300
255 270 285

We will find the area of one loop. To do this, consider the following.
Zπ/4
1
2 · 9 cos2 (2θ) dθ
2
0
Zπ/4
=9 cos2 (2θ) dθ
0
 π/4
9 1
= θ + sin 4θ
2 4 0
9π
= .
8

Exercises

Section 10.4: 9-41 odd.

115
 Chapter 54: Polar Curves and Arc Length
Theorem 339: Let f : [a, b] → R is differentiable such that f ′ is continuous and suppose that C is
Zb p
the graph of the curve with polar equation r = f (θ). Then the length of C is [f (θ)]2 + [f ′ (θ)]2 dθ.
a
Example 340: Consider the polar curve r = eθ for θ ∈ [0, 2π]. The length of the curve is given by
Z2πp
e2θ + e2θ dθ
0
2π
√ Z θ
= 2 e dθ
√ 0
= 2(e2π − 1).

Exercises

Section 10.4: 45-48.

116
 Chapter 55: Parabolas
Definition 341: A parabola is the set of points in R2 that are equidistant from a fixed point F

(called the focus) and a fixed line l (called the directrix).

Definition 342:

(i) The midpoint between the focus and the directrix of a parabola is called the vertex of the

parabola;

(ii) The line that passes through the vertex and the focus of a parabola is called the axis (of

symmetry) of the parabola.

Theorem 343: The graph of the equation (x − h)2 = 4p(y − k) with p ̸= 0 is a parabola with vertex

(h, k), a vertical axis, focus (h, k + p), and directrix y = k − p.

Proof: If a point (x, y) ∈ R2 lies on the parabola, then it is equidistant to the focus (h, k + p) and

the directrix y = k − p. So,

p
(x − h)2 + [y − (k + p)]2 = y − (k − p)

⇐⇒ (x − h)2 + [y − (k + p)]2 = [y − (k − p)]2

⇐⇒ (x − h)2 + y 2 − 2y(k + p) + (k + p)2 = y 2 − 2y(k − p) + (k − p)2

⇐⇒ (x − h)2 + y 2 − 2ky − 2py + k 2 + 2pk + p2 = y 2 − 2ky + 2py + k 2 − 2pk + p2

⇐⇒ (x − h)2 − 2py + 2pk = 2py − 2pk

⇐⇒ (x − h)2 = 4py − 4pk = 4p(y − k). ■

Corollary 344: The graph of the equation x2 = 4py is a parabola with vertex (0, 0), focus (0, p),

and directrix y = −p.

Theorem 345: The graph of the equation (y − k)2 = 4p(x − h) with p ̸= 0 is a parabola with vertex

(h, k), a horizontal axis, focus (h + p, k), and directrix x = h − p.

Proof: If a point (x, y) ∈ R2 lies on the parabola, then it is equidistant to the point (h + p, k) and

the directrix x = h − p. So,

p
[x − (h + p)]2 + (y − k)2 = x − (h − p)

⇐⇒ [x − (h + p)]2 + (y − k)2 = [x − (h − p)]2

⇐⇒ x2 − 2x(h + p) + (h + p)2 + (y − k)2 = x2 − 2x(h − p) + (h − p)2

⇐⇒ x2 − 2x(h + p) + h2 + 2ph + p2 + (y − k)2 = x2 − 2x(h − p) + h2 − 2hp + p2

⇐⇒ −2px + 2ph + (y − k)2 = 2px − 2ph

⇐⇒ (y − k)2 = 4px − 4ph = 4p(x − h). ■

Corollary 346: The graph of the equation y 2 = 4px is a parabola with vertex (0, 0), focus (p, 0),

and directrix x = −p.

117
Example 347: Determine the equation of the parabola with vertex (0, 0) and focus (2, 0).

Note that the axis is horizontal. So the equation of the parabola is y 2 = 4px with h = 0, k = 0, and

p = 2. In other words, the equation of the parabola is y 2 = 8x.

1 1
Example 348: Find the focus of the parabola y = − x2 − x + .
2 2
Consider the following.
1 1
y = − x2 − x +
2 2
⇐⇒ −2y = x2 + 2x − 1

⇐⇒ 1 − 2y = x2 + 2x = (x + 1)2 − 1

⇐⇒ 2 − 2y = (x + 1)2

⇐⇒ −2(y − 1) = (x + 1)2 .
 
1 1
Hence, h = −1, k = 1, and p = − . Therefore, the focus is the point (h, k + p) = −1, .
2 2

118
 Chapter 56: Ellipses
Definition 349: An ellipse is the set of all points P such that the sum of the distances to two

fixed points F1 and F2 , called the foci is a constant K > 0. In other words, P F1 + P F2 = K.

Definition 350:

(i) The intersection of an ellipse and the line passing through its foci is called the set of vertices;

(ii) The chord joining the vertices of an ellipse is called the major axis;

(iii) The midpoint of the major axis of an ellipse is called the center;

(iv) The chord perpendicular to the major axis of an ellipse at the center is called the minor axis.
(x − h)2 (y − k)2
Theorem 351: The graph of the equation + = 1 is an ellipse centered at the
a2 b2 √
point (h, k) with vertices (h ± a, k), foci (h ± c, k) with c = a2 − b2 , a horizontal major axis of

length 2a, and a vertical minor axis of length 2b.

Example 352: Find the foci of the ellipse 16x2 + 9y 2 = 144.

x2 y2
Dividing by 144, we obtain + = 1. Since 16 > 9, this is an ellipse with its foci on the y-axis
9 16 √ √
and with a = 4 and b = 3. Hence, c = 7, and so the foci are the points (0, ± 7).
(x − h)2 (y − k)2
Theorem 353: The graph of the equation 2
+ 2
= 1 is an ellipse centered at the
b √a
point (h, k) with vertices (h, k ± a), foci (h, k ± c) with c = a2 − b2 , a vertical major axis of length

2a, and a horizontal minor axis of length 2b.

Example 354: Find the center, vertices, and foci of the ellipse 4x2 + y 2 − 8x + 4y − 8 = 0.

Consider the following.

0 = 4x2 + y 2 − 8x + 4y − 8

= 4x2 − 8x + y 2 + 4y − 8

= 4(x − 1)2 + (y + 2)2 − 16

⇐⇒ 16 = 4(x − 1)2 + (y + 2)2

(x − 1)2 (y + 2)2
⇐⇒ + = 1.
4 16 √
√
Since 16 > 4, the major axis is vertical, h = 1, k = −2, a = 4, b = 2, and c = 16 − 4 = 2 3.

Therefore, the center is the point (1, −2), the vertices are the points (1, −6) and (1, 2), and the foci
√ √
are the points (1, −2 − 2 3) and (1, −2 + 2 3).

119
 Chapter 57: Hyperbolas
Definition 355: A hyperbola is the set of all points P in a plane, the difference of whose distances

from two fixed points F1 and F2 , called the foci, is a constant K > 0. In other words, P F1 − P F2 =

±K.

Definition 356:

(i) The intercepts of a hyperbola are called the vertices of the hyperbola.

(ii) The line segment containing the vertices of a hyperbola is called the transverse axis.

(iii) The line segment perpendicular to the transverse axis passing through the center of a hyperbola

is called the conjugate axis.

(x − h)2 (y − k)2
Theorem 357: The graph of the equation − = 1 is a hyperbola centered at the
a2 b2
point (h, k) with vertices (h ± a, k), a horizontal transverse axis of length 2a, a vertical conjugate
√
axis of length 2b, and foci (h ± c, k), where a, b > 0 and c = a2 + b2 .
(y − k)2 (x − h)2
Theorem 358: The graph of the equation 2
− = 1 is a hyperbola centered at the
a b2
point (h, k) with vertices (h, k ± a), a vertical transverse axis of length 2a, a horizontal conjugate
√
axis of length 2b, and foci (h, k ± c), where a, b > 0 and c = a2 + b2 .

Example 359: Find the equation of the hyperbola with foci (−1, 2) and (5, 2) and vertices (0, 2)

and (4, 2).

By the Midpoint Formula, the center of the hyperbola is the point (2, 2). Furthermore, c = 5 − 2 = 3
√ √
and a = 4 − 2 = 2, and so b = c2 − a2 = 5. So the transverse axis is horizontal and the equation
(x − 2)2 (y − 2)2
of the hyperbola is − = 1.
4 5
(x − h)2 (y − k)2
Theorem 360: The asymptotes of the hyperbola − = 1 with horizontal transverse
a2 b2
b
axis are the lines y = k ± (x − h).
a
(y − k)2 (x − h)2
Theorem 361: The asymptotes of the hyperbola − = 1 with vertical transverse
a2 b2
a
axis are the lines y = k ± (x − h).
b
Example 362: Determine the foci and asymptotes of the hyperbola 4x2 − 3y 2 + 8x + 16 = 0.

Consider the following.

0 = 4x2 − 3y 2 + 8x + 16

= 4x2 + 8x − 3y 2 + 16

= 4(x + 1)2 − 3y 2 + 12

⇐⇒ −12 = 4(x + 1)2 − 3y 2

y2 (x + 1)2
⇐⇒ − = 1.
4 3

120
 √ √ √
Hence, k = 0, a = 2, h = −1, b = 3, and c = a2 + b2 = 7. So, the asymptotes are the lines
2 √ √
y = ± √ (x + 1) and the foci are the points (−1, −2 − 7) and (−1, −2 + 7).
3

Exercises

Section 10.5: 1-48.

121
 Chapter 58: Rotations of Axes
Theorem 363: Suppose that the x- and y-axes in R2 are rotated by an acute angle ϕ to produces

axes u and v. Then the coordinates of the point (x, y) ∈ R2 in the xy-plane and (u, v) ∈ R2 in the

uv-plane are related as follows.

(i) x = u cos ϕ − v sin ϕ; (iii) u = x cos ϕ + y sin ϕ;

(ii) y = u sin ϕ + v cos ϕ; (iv) v = −x sin ϕ + y cos ϕ.

π
Example 364: If the x- and y-axes in R2 are rotated by rad, find the new coordinates of the
6
point (2, −4).
π
Set x = 2, y = −4, and ϕ = . Then,
6
π π π π
u = 2 cos − 4 sin v = −2 sin − 4 cos
√ 6 6 6√ 6
3 1 1 3
=2· −4· = −2 · − 4 ·
√ 2 2 2√ 2
= 3−2 = −1 − 2 3.
√ √
Hence, the new point is ( 3 − 2, −1 − 2 3).

Theorem 365: Consider a conic Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0, where A, B, C, D, E, F ∈ R.

A−C
To eliminate the xy-term, it suffices to rotate the axes by an acute angle ϕ such that cot 2ϕ = .
B
Proof: We apply the theorem above to rotate the axes by an acute angle ϕ to create the uv-plane

as follows.

(i) x = u cos ϕ − v sin ϕ;

(ii) y = u sin ϕ + v cos ϕ.

Now,

0 = Ax2 + Bxy + Cy 2 + Dx + Ey + F

= A(u cos ϕ − v sin ϕ)2 + B(u cos ϕ − v sin ϕ)(u sin ϕ + v cos ϕ) + C(u sin ϕ + v cos ϕ)2 + D(u cos ϕ −

v sin ϕ) + E(u sin ϕ + v cos ϕ) + F

= (A cos2 ϕ + B sin ϕ cos ϕ + C sin2 ϕ)u2 + [2(C − A) sin ϕ cos ϕ + B(cos2 ϕ − sin2 ϕ)]uv

+ (A sin2 ϕ − B sin ϕ cos ϕ + C cos2 ϕ)v 2 + (D cos ϕ + E sin ϕ)u + (−D sin ϕ + E cos ϕ)v + F .

To eliminate the xy-term, we choose ϕ such that [2(C − A) sin ϕ cos ϕ + B(cos2 ϕ − sin2 ϕ)] = 0. In

other words,

0 = [2(C − A) sin ϕ cos ϕ + B(cos2 ϕ − sin2 ϕ)]

= (C − A) sin 2ϕ + B cos 2ϕ

⇐⇒ B cos 2ϕ = (A − C) sin 2ϕ

122
 A−C
⇐⇒ cot 2ϕ = . ■
B √ √ √
Example 366: Eliminate the xy-term in the equation 6 3x2 + 6xy + 4 3y 2 = 21 3.

By the theorem above, we define the uv-axes by rotating the xy-axes

√ √as follows.
√
6 3−4 3 3
Rotate the xy-axes by an acute angle ϕ such that cot 2ϕ = = . This implies that
6 3
π
ϕ = . Now,
6 √ !  
3 1
(i) x = u −v ;
2 2
  √ !
1 3
(ii) y = u +v .
2 2
This implies the following.
√
21 3
√ √
= 6 3x2 + 6xy + 4 3y 2
" √ !  #2 " √ !  # "   √ !# "   √ !#2
√ 3 1 3 1 1 3 √ 1 3
=6 3 u −v +6 u −v u +v +4 3 u +v
2 2 2 2 2 2 2 2
√ √
= 7 3u2 + 3 3v 2
u2 v2
⇐⇒ + = 1.
3 7
Note 367: The graph of the equation Ax2 +Bxy+Cy 2 +Dx+Ey+F = 0, where A, B, C, D, E, F ∈ R

with A, C not being both zero, is

(i) a parabola if one of A or C is zero.

(ii) a circle if A = C.

(iii) an ellipse if A and C have the same sign.

(iv) a hyperbola if A and C have opposite signs.

Definition 368: A conic section is

(i) degenerate if the graph of the equation is either a pair of lines, a single line, or a single point.

(ii) non-degenerate if it is not degenerate.

Theorem 369: The graph of the equation Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 is a conic. In the

non-degenerate cases, the graph is

(i) a parabola if B 2 − 4AC = 0;

(ii) an ellipse if B 2 − 4AC < 0;

(iii) a hyperbola if B 2 − 4AC > 0.

Proof: We proceed as in the proof of the theorem above to rotate the axes by an acute angle ϕ. A

straightforward calculation, which is left to the reader, shows that

[2(C − A) sin ϕ cos ϕ + B(cos2 ϕ − sin2 ϕ)]2 − 4(A cos2 ϕ + B sin ϕ cos ϕ + C sin2 ϕ)

(A sin2 ϕ − B sin ϕ cos ϕ + C cos2 ϕ) = B 2 − 4AC.

123
This implies that the quantity B 2 − 4AC remains unchanged for any rotation. So, without loss of

generality, rotate the axes so that the xy term of the conic is eliminated. In other words, choose an

angle ϕ such that the equation below is satisfied.

(A cos2 ϕ + B sin ϕ cos ϕ + C sin2 ϕ)u2 + (A sin2 ϕ − B sin ϕ cos ϕ + C cos2 ϕ)v 2

+ (D cos ϕ + E sin ϕ)u + (−D sin ϕ + E cos ϕ)v + F = 0.

This implies that B 2 −4AC = −4(A cos2 ϕ+B sin ϕ cos ϕ+C sin2 ϕ)(A sin2 ϕ−B sin ϕ cos ϕ+C cos2 ϕ).

To complete the proof, we consider three cases.

Case 1: Either A cos2 ϕ + B sin ϕ cos ϕ + C sin2 ϕ = 0 or A sin2 ϕ − B sin ϕ cos ϕ + C cos2 ϕ = 0.

In this case, B 2 − 4AC = 0 and the graph of the equation is a parabola.

Case 2: The quantities A cos2 ϕ + B sin ϕ cos ϕ + C sin2 ϕ and A sin2 ϕ − B sin ϕ cos ϕ + C cos2 ϕ have

the same sign.

In this case, B 2 − 4AC < 0 and the graph of the equation is an ellipse.

Case 3: The quantities A cos2 ϕ + B sin ϕ cos ϕ + C sin2 ϕ and A sin2 ϕ − B sin ϕ cos ϕ + C cos2 ϕ have

opposite signs.

In this case, B 2 − 4AC > 0 and the graph of the equation is a hyperbola. ■

Chapter 59: Conic Sections and Polar Coordinates

 
2 2 + 2 d(P, F )
Theorem 370: Let F ∈ R , ℓ ⊆ R be a line, and e ∈ R . Then the set P ∈R | =e
d(P, ℓ)
is a conic.

Proof: If e = 1, then d(P, F ) = d(P, ℓ), which defines a parabola. So suppose that e ̸= 1. Set

F = (0, 0) and suppose ℓ is a line parallel to the y-axis d units to the right. In other words, set ℓ to

be the vertical line x = d, where d > 0. If the point P has polar coordinates (r, θ), then d(P, F ) = r

and d(P, ℓ) = d − r cos θ. Now,

d(P, F )
=e ⇐⇒ x2 + y 2 = e2 (d − x)2
d(P, ℓ)
⇐⇒ d(P, F ) = e · d(P, ℓ) ⇐⇒ (1 − e2 )x2 + 2de2 x + y 2 = e2 d2
2
e2 d y2 e2 d2

⇐⇒ r = e(d − r cos θ) ⇐⇒ x + + = .
1 − e2 1 − e2 (1 − e2 )2
Consider two cases.

Case 1: e < 1

Note the following.

2
e2 d y2 e2 d2

x+ 2
+ 2
=
1−e 1−e (1 − e2 )2

124
 e2 d
 
x+
1 − e2 y2
⇐⇒ 2 2 + 2 2 = 1.
e d e d
(1 − e ) 2 2 1 − e2
2
e d ed ed
Set h = − ,a= , and b = √ .
1 − e2 1 − e2 1 − e2
Then,
e2 d
 
x+
1 − e2 y2
+ =1
e2 d2 e2 d2
(1 − e2 )2 1 − e2
2 2
(x − h) y
⇐⇒ + 2 = 1.
a2 b
This is an equation for an ellipse with center (h, 0). Now,
e4 d2
c2 = a2 − b2 = = −h2
(1 − e2 )2
⇐⇒ c = −h.
c
This implies that the point c, the origin, is a focus of a conic section. It also follows that e = .
a
Case 2: e > 1

An argument similar to the one in case 1 shows that the graph of the equation is a hyperbola with
c
e = , where c2 = a2 + b2 . ■
a
Definition 371: In the theorem above, the constant e is called the eccentricity of the conic.
ed ed
Theorem 372: A polar equation of the form r = or r = represents a conic
1 ± e cos θ 1 ± sin θ
with one focus at the origin and with eccentricity e.

(i) If e = 1, then the conic section is a parabola.

(ii) If e < 1, then the conic section is an ellipse.

(iii) If e > 1, then the conic section is a hyperbola.

Example 373:
10
(a) Show that the conic given by the equation r = is an ellipse.
3 − 2 cos θ
Proof: Consider the following.
10 10/3
r= = .
3 − 2 cos θ 1 − 2/3 cos θ
2
Since < 1, the equation represents an ellipse. ■
3
The curve is sketched below.

125
 105 90 75
120 60

135 45

150 30

165 15

0 5 10
180 0

195 345

210 330

225 315

240 300
255 270 285

126
 π
(b) Rotate the ellipse by an angle rad about the origin. Find a polar equation for the resulting
4
ellipse.
π 10
Replace θ with θ − . So the new equation is r =  π.
4 3 − 2 cos θ −
4
105 90 75
120 60

135 45

150 30

165 15

0 5 10
180 0

195 345

210 330

225 315

240 300
255 270 285

Exercises

Section 10.6: 9-15 odd, 21, 23, 24.

127
 Chapter 60: Planetary Motion
Theorem 374 (Kepler’s Laws):

(i) A planet revolves around the sun in an elliptical orbit with the sun at one focus.

(ii) The line joining the sun to a planet sweeps out equal areas in equal times.

(iii) The square of the period of revolution of a planet is proportional to the cube of the length of

the major axis of its orbit.

Definition 375: The position of a planet that is the closest to the sun is called the perihelion of

the planet.

Definition 376: The position of a planet that is the furthest from the sun is called the aphelion

of the planet.

Lemma 377: The polar equation of an ellipse with focus at the origin, semimajor axis a, eccentricity
a(1 − e2 )
e, and directrix x = d can be written as r = .
1 + e cos θ
Theorem 378: The perihelion distance from a planet to the sun is a(1 − e) and the aphelion

distance is a(1 + e).

Example 379: Find a polar equation for the elliptical orbit of the earth around the sun (at one

focus) given that the eccentricity is about 0.017 and the length of the major axis is about 2.99 × 108

kilometers.

Note that 2a = 2.99 × 108 , which means that a = 1.495 × 108 . Hence, the earth’s orbit is described
(1.495 × 108 )[1 − (0.017)2 ] 43205.5
by the equation r = = .
1 + 0.017 cos θ 1 + 0.017 cos θ
Example 380: Find the perihelion and aphelion distances of the earth.

In this case, a = 1.495 × 108 and e = 0.017. So, the perihelion distance from the earth to the

sun is a(1 − e) = (1.495 × 108 )(1 − 0.017) = 146, 958, 500 kilometers, and the aphelion distance is

a(1 + e) = (1.495 × 108 )(1 + 0.017) = 152, 041, 500 kilometers.

Exercises

Section 10.6: 25-31.

128