CALCULUS I

Tarkan Öner

Contents

1 Review 31.1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Graph of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 Limit 182.1 Definition of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2 Limit Properties : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.4 Limits At Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5 More Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.6 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 Derivative 373.1 Tangent Line, Rate of Change and Instantaneous Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2 Definition of Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.3 The Differentiation Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.4 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.5 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.6 Derivatives of Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.7 Derivatives of Exponential and Logarithm Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.8 Logarithmic Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.9 Derivatives of Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.10 Derivatives of Inverse Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.11 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.12 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4 Applications of Derivatives 554.1 L’Hospital Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.2 Rolle’s Theorem and The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.3 Increasing and Decreasing Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.4 Minimum and Maximum Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.5 Concave up and Concave down . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.6 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.7 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.8 Linear Approximations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

5 Integral 805.1 Definite Integral. Area Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.2 Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.3 Fundamental Theorems of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.4 Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 925.5 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 965.6 Trigonometric Integrals and Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

5.6.1 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 995.6.2 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

5.7 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1045.8 More on Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

1

6 Applications of Integral 1126.1 Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6.1.1 Volumes by Cross Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126.1.2 Volumes by Cylindrical Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

6.2 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1206.2.1 First Type Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1206.2.2 Second Type Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

6.3 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1276.4 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1316.5 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

6.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1336.5.2 Tangents with Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1356.5.3 Area with Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1366.5.4 Arc Length with Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1376.5.5 Surface Area with Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

6.6 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1416.6.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1416.6.2 Tangents with Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1466.6.3 Area with Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1476.6.4 Arc Length with Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

6.7 Other Problems Related Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1526.7.1 Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1526.7.2 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1546.7.3 Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

2

1 Review

1.1 Functions

A function from A into B, denoted by f : A→ B, is a “rule”that assigns to each element of A exactly one element of B.Instead of saying “a function from A into B”, we simply say “a function”and instead of writing f : A→ B, we simply

write f .The sets A and B are called the domain and codomain of f respectively. The domain of f is denoted by dom(f).Let f be a function. For each x belonging to the domain of f , the corresponding element (in the codomain of f )

assigned by f is denoted by f(x) and is called the image of x under f .In this course, most of the functions we consider are functions whose domains and codomains are subsets of R.A variable that represents the “input numbers” for a function is called an independent variable. A variable that

represents the “output numbers”is called a dependent variable because its value depends on the value of the independentvariable.

Example 1.1.1 For the function f : R→ R given by

y = f (x) = x2

, x is independent variable and y is dependent variable.

In this course we will usually not be careful about specifying the domain of the function. By convention, if no domainis stated for a function, then the domain is assumed to be the largest set of inputs for which the function is defined. Forinstance, if we say that h is the function h(x) =

√x then the domain of h is understood to be the set of all nonnegative

real numbers since√x is well-defined for all x > 0 and undefined for x < 0.

Let f : A→ B be a function and let S ⊂ A. The image of S under f , denoted by f [S], is the subset of B given by

f [S] = {f (x) : x ∈ S}

The range of f , denoted by ran(f), is the image of A under f , that is, ran(f) = f [A].

Example 1.1.2 Let f (x) = 2x− 1.For S = {1, 2, 3}

f [S] = {f (x) : x ∈ S}= {f (1) , f (2) , f (3)}= {1, 3, 5}

For S = (−1, 1]

f [S] = {f (x) : x ∈ S}= {f (x) : −1 < x ≤ 1}= {f (x) : −2 < 2x ≤ 2}= {f (x) : −2− 1 < 2x− 1 ≤ 2− 1}= {f (x) : −3 < 2x− 1 ≤ 1}= {f (x) : −3 < f (x) ≤ 1}= (−3, 1]

Example 1.1.3 Find the domain and range of f (x) =√

1− x2.Solution : In order for the function to be defined, the expression under the square root cannot be negative.

1− x2 > 0

Hence the domain of f is [−1, 1].

−1 ≤ x ≤ 1⇒ 0 ≤ x2 ≤ 1

⇒ −1 ≤ −x2 ≤ 0

⇒ 0 ≤ 1− x2 ≤ 1

therefore, the range of f is [0, 1].

3

In general determining the range of a function can be somewhat diffi cult. We work with domains of functions muchmore than range of functions.

Example 1.1.4 Find the domain of f (x) = 2x− 1.Solution : dom (f) = −∞ < x <∞

Example 1.1.5 Find the domain of g (x) =x

x2 − 4.

Solution : dom (g) = R− {−2, 2}

Example 1.1.6 Find the domain of h (x) =

√x− 1

x2 − 9.

Solution : dom (h) = [1, 3) ∪ (3,∞)

The next topic is function composition. Let g : A→ B, f : C → D,B ⊂ C. The composition of f and g is a functiondefined such that

f ◦ g : A→ D,

(f ◦ g) (x) = f (g (x)) .

In other words, compositions are evaluated by plugging the second function listed into the first function listed. Notethat the order is important here. Interchanging the order will usually result in a different answer.

Example 1.1.7 Let f : R→ [2,∞), f (x) = x2 + 2 and g : [−1, 1]→ [0, 1] , g (x) =√

1− x2. Since [2,∞) 6⊂ [−1.1], g ◦ fcan’t be defined. But f ◦ g can be defined.

(f ◦ g) (x) = f (g (x)) = f(√

1− x2)

=(√

1− x2)2

+ 2

= 3− x2.

Example 1.1.8 Given f (x) = x2 + 1 and g (x) = 1− 2x, find each of the following.a) (f ◦ g) (x) = f (g (x)) = f (1− 2x) = (1− 2x)

2+ 1 = 4x2 − 4x+ 2

b) (g ◦ f) (x) = g (f (x)) = g(x2 + 1

)= 1− 2

(x2 + 1

)= −2x2 − 1

c)(f ◦ f) (x) = f (f (x)) = f(x2 + 1

)=(x2 + 1

)2+ 1 = x4 + 2x2 + 2

d) (g ◦ g) (x) = g (g (x)) = g (1− 2x) = 1− 2 (1− 2x) = 4x− 1

If f and g are functions, then

(f + g) (x) = f (x) + g (x)

(kf) (x) = kg (x) where k is a constant

(f − g) (x) = f (x)− g (x)

(fg) (x) = f (x) g (x)(f

g

)(x) =

f (x)

g (x)where g (x) 6= 0

for every x that belongs to both dom (f) and dom (g).

Example 1.1.9 Find f + g, 3f, f − g, fg, f/g when f (x) = x2 + 1 and g (x) = 1− 2x.Solution :

(x2 + 1

)(1− 2x) = − (2x− 1)

(x2 + 1

)=

(f + g) (x) = f (x) + g (x) = x2 + 1 + 1− 2x = x2 − 2x+ 2

(3f) (x) = 3f (x) = 3(x2 + 1

)= 3x2 + 3

(f − g) (x) = f (x)− g (x) = x2 + 1− (1− 2x) = x2 + 2x

(fg) (x) = f (x) g (x) =(x2 + 1

)(1− 2x) = −2x3 + x2 − 2x+ 1(

f

g

)(x) =

f (x)

g (x)=x2 + 1

1− 2x

4

1.2 Graph of Functions

Let f : A→ B be a function where A ⊂ R. The graph of f is the following subset of R2 :{(x, f (x)) ∈ R2 : x ∈ A

}Translations of Graphs :The graph of the function f (x+ h) is the graph of the function f (x) shifted h units to the left and the graph of the

function f (x− h) is the graph of the function f (x) shifted h units to the right where h > 0.

x

y

Graph of f (x)

x

y

h

Graph of f (x+ h)

x

y

h

Graph of f (x− h)

In contrast, the graph of function f (x) + h is the graph of the function f (x) shifted h units to the upward and thegraph of the function f (x)− h is the graph of the function f (x) shifted h units to the downward where h > 0.

x

y

Graph of f (x)

x

y

h

Graph of f (x) + h

x

y

h

Graph of f (x)− h

Example 1.2.1 By using translations, draw the graph of f(x) = x2 − 4x.

x2 − 4x = (x− 2)2 − 4

x

y

y = x2

­2 0 2 4 6

5

10

15

x

y

y = (x− 2)2

5

­8­6­4­2

2468

1012

x

y

y = (x− 2)2 − 4

Symmetry :

5

The graph of −f (x) is the the same as the graph of f (x) reflected through the x-axis.

x

y

y = f (x)

x

y

y = −f (x)

Example 1.2.2 Draw the graph of f (x) = −x2 + 4x.

−x2 + 4x = −(x2 − 4x

)= −

((x− 2)

2 − 4)

x

y

y = (x− 2)2 − 4

x

y

y = −(

(x− 2)2 − 4

)

The graph of f (−x) is the the same as the graph of f (x) reflected about the y-axis.

x

y

y = f (x)

x

y

y = f (−x)

Example 1.2.3 Draw the graph of f (x) = x2 + 4x.

x2 + 4x = ((−x)− 2)2 − 4

6

x

y

y = (x− 2)2 − 4

x

y

y = ((−x)− 2)2 − 4

Even Function : A function which satisfies the identity f (−x) = f (x) is said to be even function. Hence the graphof an even function is symmetric about the y-axis.

Odd Function : A function which satisfies the identity f (−x) = −f (x) is said to be odd function. Hence the graphof an even function is symmetric about the origin.

x

y

Graph of an even function

x

y

Graph of an odd function

Example 1.2.4

x

y

cos (−x) = cos (x)

x

y

sin (−x) = − sin (x)

7

1 Constant Functions : f(x) = c; where c is a constant (a real number).

x

y

c

Grap of f (x) = c

dom (f) = R, ran (f) = c

2 Linear Functions : f (x) = ax+ b, where a, b are constant and a 6= 0.

x

y

Graph of y = ax+ b, a > 0

x

y

Graph of y = ax+ b, a < 0

dom (f) = R, ran (f) = R

3 Quadratic Functions : f (x) = ax2 + bx+ c, where a, b, c are constant and a 6= 0. f can be also written as

f (x) = a (x± h)2 ± r

for some h, r ∈ R+.

x

y

r

h

Graph of y = a (x± h)2 ± r, a > 0

x

yr

h

Graph of y = a (x± h)2 ± r, a < 0

dom (f) = R, ran (f) = [r,∞) dom (f) = R, ran (f) = (−∞, r]

4 Polynomial Functions : f (x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0, where a0, . . . , an are constant and an 6= 0.

8

­4 ­2 2 4

­100

­50

50

x

y

y = x3 − x2 − 4x+ 10

­4 ­2 2 4

­5

5

10

x

y

y = x4 − 5x2 + 4

dom (f) = R, ran (f) =?

5 Rational Functions : f (x) =p (x)

q (x), where p and q polynomial functions.

­4 ­2 2 4

­2

­1

1

2

x

y

y = 1x

­4 ­2 0 2 4

0.2

0.4

0.6

0.8

1.0

x

y

y = 1x2

­4 ­2 2 4

­0.6

­0.4

­0.2

0.2

x

y

y = x−1x2+2

dom (f) = R− {0} dom (f) = R− {0} dom (f) = Rran (f) = R− {0} ran (f) = R+ ran (f) =?

6 Square Root Functions : f (x) =√x

0 2 40

1

2

x

y

y =√x

dom (f) = [0,∞), ran (f) = [0,∞)

9

7 Exponential Functions : f (x) = ax, where a > 0.

x

y

y = ax, a ∈ (0, 1)

x

y

y = ax, a > 1

dom (f) = R, ran (f) = (0,∞) dom (f) = R, ran (f) = (0,∞)

8 Logarithmic Functions : f (x) = log ax, where a > 0.

x

y

y = loga x, a ∈ (0, 1)

x

y

y = loga x, a > 1

dom (f) = (0,∞), ran (f) = R dom (f) = (0,∞), ran (f) = R

9 Trigonometric and Inverse Trigonometric Functions :

­6 ­4 ­2 2 4 6­1

1

x

y

y = sinx

­6 ­4 ­2 2 4 6­1

1

x

y

y = cosx

dom (f) = R, ran (f) = [−1, 1] dom (f) = R, ran (f) = [−1, 1]

10

­6 ­4 ­2 2 4 6­2

2

x

y

y = tanx

­6 ­4 ­2 2 4 6­2

2

x

y

y = cotx

dom (f) = R− {kπ, k ∈ Z}, ran (f) = R dom (f) = R− {kπ/2, k ∈ Z}, ran (f) = R

­6 ­4 ­2 2 4 6­2

2

x

y

y = secx

­6 ­4 ­2 2 4 6­2

2

x

y

y = cscx

dom (f) = R− {kπ, k ∈ Z} dom (f) = R− {kπ/2, k ∈ Z}ran (f) = (−∞,−1] ∪ [1,∞) ran (f) = (−∞,−1] ∪ [1,∞)

­1 1

­1.5

­1.0

­0.5

0.5

1.0

1.5

x

y

y = arcsinx

­1 0 1

1

2

3

x

y

y = arccosx

dom (f) = [−1, 1] , ran (f) = [−π/2, π/2] dom (f) = [−1, 1] , ran (f) = [0, π]

11

­6 ­4 ­2 2 4 6­1

1

x

y

y = arctanx

­3 ­2 ­1 1 2 3­1

1

x

y

y = arccotx

dom (f) = (−∞,∞) , ran (f) = (−π/2, π/2) dom (f) = (−∞,∞) , ran (f) = (0, π)

­5 ­4 ­3 ­2 ­1 0 1 2 3 4 5

123

x

y

y = arcsecx

­6 ­4 ­2 2 4 6­1

1

x

y

y = arccscx

dom (f) = (−∞,−1] ∪ [1,∞) dom (f) = (−∞,−1] ∪ [1,∞)ran (f) = [0, π/2) ∪ (π/2,∞) ran (f) = [−π/2, 0) ∪ (0, π/2]

10 Hyperbolic and Inverse Hyperbolic Functions

­2 2

­10

­5

5

10

x

y

y = sinhx =ex − e−x

2

­3 ­2 ­1 0 1 2 3

1

2

3

4

5

x

y

y = coshx =ex + e−x

2

dom (f) = R, ran (f) = R dom (f) = R, ran (f) = [1,∞)

12

­3 ­2 ­1 1 2 3­1

1

x

y

y = tanhx

­2 ­1 1 2

­3

­2

­1

1

2

3

x

y

y = cothx

dom (f) = R, ran (f) = (−1, 1) dom (f) = R− {0}, ran (f) = (−∞,−1) ∪ (1,∞)

­3 ­2 ­1 0 1 2 3

1

x

y

y = sechx

x

y

y = cschx

dom (f) = R, ran (f) = (0, 1] dom (f) = R− {0}, ran (f) = R− {0}

x

y

y = arcsinhx = ln(x+√x2 + 1

)0 1 2 3 4 5 6 7

0

1

2

x

y

y = arccoshx = ln(x+√x2 − 1

)dom (f) = R, ran (f) = R dom (f) = [1,∞), ran (f) = [0,∞)

13

­1.0 ­0.5 0.5 1.0

­0.5

0.5

1.0

x

y

y = arctanhx = 12 ln

(1 + x

1− x

)

­5 ­4 ­3 ­2 ­1 1 2 3 4 5

­5

5

x

y

y = coth−1 x = 12 ln

(1 + x

−1 + x

)dom (f) = (−1, 1), ran (f) = R dom (f) = (−∞,−1) ∪ (1,∞), ran (f) = R− {0}

0 10

2

4

x

y

y = arc sechx = ln

(1 +√

1− x2x

)

­6 ­4 ­2 2 4 6

­2

2

x

y

y = arccschx = ln

(1x +

√1 + x2

|x|

)

dom (f) = (0, 1], ran (f) = [0,∞) dom (f) = R− {0}, ran (f) = R− {0}

9 Absolute Value Function : |x| =

x ,if x > 00 ,if x = 0−x ,if x < 0

x

y

y = |x|

14

1.3 Inverse Functions

Let f be a function. We say that f is injective (one to one) if the following condition is satisfied:

x1, x2 ∈ dom (f) and x1 6= x2 ⇒ f (x1) 6= f (x2)

Condition means that different elements of the domain are mapped to different elements of the range and it is equivalentto the following condition:

x1, x2 ∈ dom (f) and f (x1) = f (x2)⇒ x1 = x2

Example 1.3.1 Let f(x) = x2. The domain of f is R. The function f is not injective. This is because −1 6= 1, butf (−1) = f (1) = 1. But if we restrict the domain set to [0,∞), then f became an injective function.

Example 1.3.2 Let f(x) = 2x. The domain of f is R. The function f is injective. This is because if x1, x2 ∈ R andx1 6= x2, then 2x1 6= 2x2 .

Let f be an injective function. Then given any element y of ran(f), there is exactly one element x of dom(f) such thatf(x) = y. This means that if we use an element y of ran(f) as input, we get one and only one output x. The functionobtained in this way is called the inverse of f .

Definition 1.3.3 Let f : X → Y be an injective function and let Y1 be the range of f . The inverse function of f ,denoted by f−1, is the function from Y1 to X such that for every y ∈ Y1, f−1(y) is the unique element of X satisfyingf(f−1(y)

)= y.

f and f−1 are also satisty f−1 (f(x)) = x for x ∈ dom (f). To find the inverse function of f means to find the domainof f−1 as well as a formula for f−1(y). If the formula for f(x) is not complicated, dom(f−1) and f−1(y) can be found bysolving the equation y = f(x) for x.

Step 1 Put y = f(x).

Step 2 Solve x in terms of y. The result will be in the form x = an expression in y.

Step 3 Replace y in the right side of the equation with x and x in the left side of the equation with f−1 (x).

Step 4 From the expression obtained in Step 3, the range of f can be determined. This is the domain of f−1.

Remark 1.3.4 If the function is not injective, the expression in y obtained in Step 2 does not give a function.

Example 1.3.5 Given f (x) = 3x− 2 find f−1 (x).Solution : Since

x1 6= x2 ⇒ 3x1 6= 3x2 ⇒ 3x1 − 2 6= 3x2 − 2

⇒ f (x1) 6= f (x2)

, f is one to one.

y = 3x− 2⇒ x =y + 2

3

Hence f−1 (x) =x+ 2

3. At this point we can check the identity(

f−1 ◦ f)

(x) = x

(or(f ◦ f−1

)(x) = x) to verify our calculation.(

f−1 ◦ f)

(x) = f−1 (f (x)) = f−1 (3x− 2)

=(3x− 2) + 2

3= x.

Since f−1 (x) can be solved for all real numbers x, dom(f−1

)= R.

dom (f) = R, ran (f) = Rdom

(f−1

)= R, ran

(f−1

)= R

15

­10 ­5 5 10

­10

­5

5

10

x

y

f and f−1

Example 1.3.6 Given g (x) =x+ 4

2x− 5find g−1 (x).

Solution :

y =x+ 4

2x− 5⇒ y (2x− 5) = x+ 4

⇒ 2xy − 5y = x+ 4

⇒ x (2y − 1) = 5y + 4

⇒ x =5y + 4

2y − 1

Therefore g−1 (x) =5x+ 4

2x− 1and dom

(g−1

)= R−

{12

}.

dom (g) = R− {5

2}, ran (g) = R− {1

2}

dom(g−1

)= R− {1

2}, ran

(g−1

)= R− {5

2}

­4 ­2 2 4

­4

­2

2

4

x

y

g and g−1

Example 1.3.7 Given h (x) =√x− 1 find h−1 (x).

Solution :

y =√x− 1⇒ y2 = x− 1

⇒ x = y2 + 1

16

Therefore h−1 (x) = x2 + 1.

dom (h) = [1,∞), ran (h) = [0,∞)

dom(h−1

)= [0,∞), ran

(h−1

)= [1,∞)

0 1 2 3 4 50

1

2

3

4

5

x

y

h and h−1

17

2 Limit

2.1 Definition of Limit

If f is some function thenlimx→a

f (x) = L

is read " as x approaches a the limit of f(x) is L." It means that if you choose values of x which are close but not equalto a, then f(x) will be close to the value L; Moreover, as x gets closer and closer to a, f(x) gets closer and closer to L.

The following alternative notation is sometimes used

f(x)→ L as x→ a

(read " as x approaches a, f(x) approaches L ").When we are computing limits, the question that we are really asking is what y value is our graph approaching as we

move in towards x = a on our graph. We are not asking what y value the graph takes at the point in question. In otherwords, we are asking what the graph is doing around the point x = a.

Example 2.1.1 If f(x) = 3x− 2 thenlimx→2

f (x) = 4

, because if you substitute numbers x close to 2 in f(x), the result will be close to 5.

­2 2 4­2

2

4

6

8

x

y

y = 3x− 2

Example 2.1.2 Estimate the value of the following limit.

limx→2

x2 + 4x− 12

x2 − 2x.

Solution : We will choose values of x that get closer and closer to x = 2 and plug these values into the function.

x f (x) x f (x)

2.5 3.4 1.5 5.02.1 3.857142857 1.9 4.1578947372.01 3.985074627 1.99 4.0150753772.001 3.998500750 1.999 4.0015007502.0001 3.999850007 1.9999 4.0001500082.00001 3.999985000 1.99999 4.000015000

From this table it appears that the function is going to 4 as x approaches 2, so

limx→2

x2 + 4x− 12

x2 − 2x= 4.

Also notice that we can’t actually plug in x = 2 into the function as this would give us a division by zero error.

18

0 1 2 3 40

2

4

6

8

10

x

y

y = x2+4x−12x2−2x

Definition 2.1.3 (Formal Definition of Limit) Let f(x) be a function defined on an interval that contains x = a,except possibly at x = a. Then we say that,

limx→a

f (x) = L

if for every number ε > 0 there is some number δ > 0 such that

|f (x)− L| < ε whenever 0 < |x− a| < δ. (1)

The quantity ε is how close you would like f(x) to be to its limit L; the quantity δ is how close you have to choose xto a to achieve this. To prove that lim

x→af (x) = L you must assume that someone has given you an unknown ε > 0, and

then find a positive δ for which (1) holds. The δ you find will depend on ε.

Example 2.1.4 By using the formal definition, show that limx→2

(3x− 2) = 4.

Solution : According to the definition, if we want to prove

limx→2

(3x− 2) = 4

then for every number ε > 0, we need to produce a corresponding number δ > 0 such that

|(3x− 2)− 4| < ε whenever 0 < |x− 2| < δ.

Now we can rewrite the first inequality above :

|(3x− 2)− 4| < ε⇐⇒ |3x− 6| < ε

⇐⇒ 3 |× − 2| < ε

⇐⇒ |x− 2| < ε

3

So we see that if we put δ = ε3 , we should be fine. But let’s check it : Whenever 0 < |x− 2| < δ = ε

3 we get

|x− 2| <ε

3⇒ 3 |× − 2| < ε

⇒ |3x− 6| < ε

⇒ |(3x− 2)− 4| < ε

So for every ε > 0, if we choose δ = ε3 we should get what we want.

When we let " x approach a" we allow x to be both larger or smaller than a, as long as x gets close to a. Here are theinformal definitions for the two one sided limits.

Definition 2.1.5 (Right-handed limit) We say

limx→a+

f (x) = L

provided we can make f(x) as close to L as we want for all x suffi ciently close to a and x > a without actually letting xbe a.

19

Definition 2.1.6 (Left-handed limit) We say

limx→a−

f (x) = L

provided we can make f(x) as close to L as we want for all x suffi ciently close to a and x < a without actually letting xbe a.

Theorem 2.1.7 Given a function f(x) if,

limx→a+

f (x) = limx→a−

f (x) = L

thenlimx→a

f (x) = L.

Likewise, iflimx→a

f (x) = L.

then,limx→a+

f (x) = limx→a−

f (x) = L.

From this fact, we can also say that if the two one-sided limits have different values,i,e

limx→a+

f (x) 6= limx→a−

f (x)

then the normal limit will not exist.

Example 2.1.8 Let

f (x) =

x− 1 , if x < 00 ,if x = 0x+ 1 , if x > 0

­3 ­2 ­1 1 2 3

­3

­2

­1

1

2

3

x

y

y = f (x)

Here we havelimx→0−

f (x) = −1 6= limx→0+

f (x) = 1.

Hencelimx→0

f (x)

does exist.

Example 2.1.9 Let f (x) =x3

x.

20

­4 ­2 0 2 4

5

10

15

20

25

x

y

y = x3

x

Sincelimx→0+

f (x) = limx→0−

f (x) = 0

,we havelimx→0

f (x) = 0.

This example shows us that, we can have limits at a point even if the function itself does not defined at that point.

Example 2.1.10 Let

f (x) =

x2 , if x < 12 ,if x = 1−x+ 2 , if x > 1

.

­2 ­1 1 2 3 4

­2

­1

1

2

3

4

x

y

y = f (x)

In this caselimx→1+

f (x) = limx→1−

f (x) = 1 = limx→1

f (x)

butf (x) = 2.

Note that, even if a function exists at a point, there is no reason to think that the limit will have the same value asthe function at that point.

21

2.2 Limit Properties :

First we will assume that limx→a

f (x) and limx→a

g (x) exist and that c is any constant. Then,

1. limx→a

c = c

2. limx→a

x = a

3. limx→a

[cf (x)] = c limx→a

f (x)

4. limx→a

[f (x)± g (x)] = limx→a

f (x)± limx→a

g (x)

5. limx→a

[f (x) g (x)] = limx→a

f (x) limx→a

g (x)

6. limx→a

[f (x)

g (x)

]=

limx→a

f (x)

limx→a

g (x), provided lim

x→ag (x) 6= 0

7. limx→a

[f (x)]n

=[

limx→a

f (x)]n, where n is any real number.

Example 2.2.1 Compute the following limitlimx→2

(x2 − 2x+ 3

).

Solution :

limx→2

(x2 − 2x+ 3

)= lim

x→2x2 − lim

x→22x+ lim

x→23

=[

limx→2

x]2− 2 lim

x→2x+ lim

x→23

= 22 − 2.2 + 3

= 3

Note that when we plug x = 2 into the function we have.f (2) = 3.

Fact : If p(x) is a polynomial then,limx→a

p (x) = p (a) .

Example 2.2.2 Compute the following limit

limx→1

x3 − x+ 2

x+ 1.

Solution :

limx→1

x3 − x+ 2

x+ 1=

limx→1

(x3 − x+ 2

)limx→1

(x+ 1)

=13 − 1 + 2

1 + 1= 1

Example 2.2.3 Evaluate the following limitlimx→2

√x.

Solution :

limx→2

√x = lim

x→2(x)

12

=[

limx→2

x] 12

= (2)12

=√

2

22

It is possible that both f (x) and g (x) approach 0 as x approaches a. In that case, we say the limit

limx→a

f (x)

g (x)

is the form 00 . A limit of the form

00 may exist, or it may not. However it must be simplified before it can be analyzed.

Example 2.2.4 Compute the following limit

limx→−2

x2 − 4

x2 + x− 2.

Solution : At x = −2 the limit is the form 00 and it must be reduced to a non

00 form.

limx→−2

x2 − 4

x2 + x− 2= lim

x→−2

(x+ 2) (x− 2)

(x− 1) (x+ 2)

= limx→−2

x− 2

x− 1

=4

3

Theorem 2.2.5 (Squeeze (Sandwich)Theorem) Suppose that for all x on [a, c] (except possibly at x = b) we have,

f (x) ≤ h (x) ≤ g (x) .

Also suppose that,limx→b

f (x) = limx→b

g (x) = L

for some a ≤ b ≤ c. Thenlimx→b

h (x) = L

Example 2.2.6 Evaluate the following limit.

limx→0

x2 cos

(1

x

)Solution : Previous examples cannot help us to find this limit. We know the following fact.

−1 ≤ cos (x) ≤ 1

We can also have

−1 ≤ cos

(1

x

)≤ 1

for x 6= 0. Since we are taking a limit, we can ignore x = 0. By multiply everything by x2, we get following.

−x2 ≤ x2 cos

(1

x

)≤ x2

Since we have alsolimx→0−x2 = lim

x→0x2 = 0

by the Squeeze theorem we must also have

limx→0

x2 cos

(1

x

)= 0.

23

2.3 Infinite Limits

Definition 2.3.1 Let f(x) be a function defined on an interval that contains x = a, except possibly at x = a. Then wesay that,

limx→a

f (x) =∞

if for every number M > 0 there is some number δ > 0 such that

f (x) > M whenever 0 < |x− a| < δ.

Definition 2.3.2 Let f(x) be a function defined on an interval that contains x = a, except possibly at x = a. Then wesay that,

limx→a

f (x) = −∞

if for every number N < 0 there is some number δ > 0 such that

f (x) < N whenever 0 < |x− a| < δ.

The notationlimx→a

f (x) =∞

means that the values of f(x) can be made arbitrary large (as large as we like) by taking x suffi ciently close to a (on eitherside of a) but not equal to a.

Similarly,limx→a

f (x) = −∞

means that the values of f(x) can be made as large negative as we like by taking x suffi ciently close to a (on either sideof a) but not equal to a.Similar definitions can be given for the one-sided infinite limits

limx→a+

f (x) = ∞

limx→a−

f (x) = ∞

limx→a+

f (x) = −∞

limx→a−

f (x) = −∞

Example 2.3.3

x

y

y = 1x

limx→0+

1

x=∞, lim

x→0−1

x= −∞, hence lim

x→0

1

xdoes not exist

24

Example 2.3.4

2 4 6 8 10

­8

­6

­4

­2

0

2

4

6

8

x

y

y = 15−x

limx→5−

1

5− x =∞, limx→5+

1

5− x = −∞, hence limx→5

1

5− x does not exist

Example 2.3.5

x

y

y = 1x2

limx→0+

1

x2=∞, lim

x→0−1

x2=∞, hence lim

x→0

1

x2=∞

Example 2.3.6

1 2 3 4 5

­0.5

0.0

0.5

1.0

1.5

x

y

y = lnx

25

limx→0+

lnx = −∞

Notice that we can only evaluate the right-handed limit here.

Definition 2.3.7 (Vertical Asymptote) The line x = a is called a vertical asymptote of the curve y = f(x) if at leastone of the following statements is true :

limx→a

f (x) =∞ limx→a+

f (x) =∞ limx→a−

f (x) =∞limx→a

f (x) = −∞ limx→a+

f (x) = −∞ limx→a−

f (x) = −∞

Example 2.3.8 Find all vertical asymptotes of f(x) =−2x

x+ 3.

Solution : There are potential vertical asymptotes where x+ 3 = 0, that is x = −3.

limx→−3+

−2x

x+ 3=∞, lim

x→−3−−2x

x+ 3= −∞

This shows that the lines x = −3 is all vertical asymptotes of f :

­4 ­2 2 4

­40

­20

20

40

x

y

y =−2x

x+ 3

Example 2.3.9 Find all vertical asymptotes of f(x) = − 8

x2 − 4.

Solution : There are potential vertical asymptotes where x2 − 4 = 0, that is x = ±2.

limx→−2−

(− 8

x2 − 4

)= −∞, lim

x→2−

(− 8

x2 − 4

)=∞

26

This shows that the lines x = ±2 are all vertical asymptotes of f :

­4 ­2 2 4

­10

­5

5

x

y

y = − 8

x2 − 4

27

2.4 Limits At Infinity

Definition 2.4.1 Let f(x) be a function defined on x > K for some K. Then we say that,

limx→∞

f (x) = L

if for every number ε > 0 there is some number M > 0 such that

|f (x)− L| < ε whenever x > M.

Definition 2.4.2 Let f(x) be a function defined on x < K for some K. Then we say that,

limx→−∞

f (x) = L

if for every number ε > 0 there is some number N < 0 such that

|f (x)− L| < ε whenever x < N.

The notationlimx→∞

f (x) = L

means that the values of f(x) can be made as close to L as we like by taking x suffi ciently large and similarly, the notation

limx→−∞

f (x) = L

means that the values of f(x) can be made arbitrarily close to L by taking x suffi ciently large negative.

Example 2.4.3 Show that limx→∞

1x = 0.

Solution : For given ε > 0 we need to choose right N such that∣∣∣∣ 1x − 0

∣∣∣∣ < ε whenever x > N.∣∣∣∣ 1x − 0

∣∣∣∣ < ε⇒ x >1

ε

Hence for a given any positive ε, we will simply choose N = 1ε .

­4 ­2 2 4

­2

­1

1

2

x

y

y = 1x

Example 2.4.4

limx→∞

1

xn= 0, lim

x→−∞

1

xn= 0 where n is a positive integer

Example 2.4.5 Compute

limx→∞

5x3 + 2x− 1

−x5 + x+ 2.

Solution :

limx→∞

5x3 + 2x− 1

−x5 + x+ 2= lim

x→∞

x3(5 + 2

x2 −1x3

)x3(−x2 + 1

x2 + 2x3

)= lim

x→∞

5

−x2= 0.

28

Notice that here we have another indeterminate form ∞∞ .

Example 2.4.6 Compute

limx→∞

2x2 + x

3x2 + 2x− 7.

Solution :

limx→∞

2x2 + x

3x2 + 2x− 7= lim

x→∞

x2(2 + 1

x

)x2(3 + 2

x −7x2

)= lim

x→∞

2

3

=2

3.

Definition 2.4.7 (Horizontal Asymptote) The line y = L is called a horizontal asymptote of the curve y = f(x) ifeither

limx→−∞

f (x) = L or limx→∞

f (x) = L

Example 2.4.8 Find all horizontal asymptotes of f(x) =−2x

x+ 3.

Solution : We have

limx→∞

−2x

x+ 3= lim

x→∞

−2x

x(1 + 3

x

) = −2

limx→−∞

−2x

x+ 3= lim

x→−∞

−2x

x(1 + 3

x

) = −2

It follows that y = −2 is the horizontal asymptote of f(x).

­10 ­5 5 10

­8

­6

­4

­2

2

4

6

x

y

y =−2x

x+ 3

Example 2.4.9 Find all horizontal asymptotes of f(x) =x√

x2 + 1.

Solution : We have

limx→∞

x√x2 + 1

= limx→∞

x

|x|√

1 + 1x2

= 1

limx→−∞

x√x2 + 1

= limx→−∞

x

|x|√

1 + 1x2

= −1

29

It follows that y = ±1 are the horizontal asymptote of f(x).

­2 ­1 1 2

­1

1

x

y

y =x√

x2 + 1

Definition 2.4.10 Let f(x) be a function defined on x > K for some K. Then we say that,

limx→∞

f (x) =∞

if for every number N > 0 there is some number M > 0 such that

f (x) > N whenever x > M.

Definition 2.4.11 Let f(x) be a function defined on x > K for some K. Then we say that,

limx→∞

f (x) = −∞

if for every number N < 0 there is some number M > 0 such that

f (x) < N whenever x > M.

The other two definitions are almost identical. The only differences are the signs ofM and/or N and the correspondinginequality directions.

Example 2.4.12lim

x→−∞x = −∞ lim

x→∞x =∞

limx→−∞

x2 =∞ limx→∞

x2 =∞lim

x→−∞x3 = −∞ lim

x→∞x3 =∞

......

Example 2.4.13

limx→∞

(2x3 − x2 − 8

)= lim

x→∞x3(

2− 1

x− 8

x3

)=∞.2 =∞

limx→−∞

(2x3 − x2 − 8

)= lim

x→−∞x3(

2− 1

x− 8

x3

)= −∞.2 = −∞

Example 2.4.14 Compute

limx→∞

x3 + 1

2x2 + x− 1.

Solution :

limx→∞

x3 + 1

2x2 + x− 1= lim

x→∞

x2(x+ 1

x3

)x2(2 + 1

x −1x2

)= lim

x→∞

x

2=∞

30

2.5 More Example

Example 2.5.1

x

y

y = ax, a > 1

limx→−∞

ax = 0

limx→∞

ax =∞

x

y

y = bx, b ∈ (0, 1)

limx→−∞

bx =∞limx→∞

bx = 0

Example 2.5.2lim

x→−∞ex = 0, lim

x→∞ex =∞, lim

x→−∞e−x =∞, lim

x→∞e−x = 0

Example 2.5.3

limx→∞

ex2−x+1 =∞

limx→∞

e−x+1 = 0

limx→∞

e2x+1 = 1

limx→0+

e1x =∞

limx→0−

e1x = 0

Example 2.5.4limx→∞

(e2x

2

+ ex)

=∞+∞ =∞

Example 2.5.5

limx→∞

(e2x

2

− ex)

= ∞−∞

= limx→∞

e2x2

(1− ex

e2x2

)= lim

x→∞e2x

2(

1− ex−2x2)

= ∞

31

Example 2.5.6

limx→∞

(√x2 + x− x

)= ∞−∞

= limx→∞

(√x2 + x− x

) (√x2 + x+ x)(√

x2 + x+ x)

= limx→∞

(x2 + x− x2

)√x2 + x+ x

= limx→∞

x

|x|(√

1 + 1x2 + 1

)=

1

2

Here y = 12 is a horizontal asymptote.

Example 2.5.7

x

y

y = lnx

limx→0+

lnx = −∞limx→∞

lnx =∞

Example 2.5.8

limx→∞

ln(3x3 − 2x− 1

)= ∞

limx→∞

ln

(2

x2 − 7x

)= −∞

Example 2.5.9 limx→0

sin xx = 1

Example 2.5.10 limx→0

1−cos xx = 0

Example 2.5.11

limx→0

tanx

x= lim

x→0

sin xcos x

x= limx→0

sinx

x cosx

= limx→0

sinx

x

1

cosx= 1.1 = 1

Example 2.5.12

limx→0

sin (3x)

x= limx→0

3sin (3x)

3x= 3.1 = 3

Example 2.5.13

limx→0

1− cosx

sinx= lim

x→0

1−cos xx

sin xx

=0

1= 0

32

Example 2.5.14

x

y

y = arctanx

limx→−∞

arctanx = −π2limx→∞

arctanx = π2

Example 2.5.15

limx→∞

arctan(x2 − 7

)=

π

2

limx→−∞

arctan

(1

x+ 1

)= 0

limx→−1−

arctan

(1

x+ 1

)= −π

2

Example 2.5.16

limx→∞

x

√1

x2 + 3x= ∞.0

= limx→∞

√x2

x2 + 3x

=

√limx→∞

x2

x2 + 3x

=√

1

= 1

Here y = 1 is a horizontal asymptote.

Example 2.5.17

limx→0

(1− cosx) cotx = 0.∞

= limx→0

(1− cosx)cosx

sinx

= limx→0

(1− cosx

x

)x

sinxcosx

= 0.1.1

= 0

33

2.6 Continuity

A function f is continuous at a iflimx→a

f (x) = f (a) .

A function is continuous if it is continuous at every a in its domain.

Example 2.6.1 For the following function, determine if f(x) is continuous at x = −2, x = 0 and x = 1.

­4 ­2 2 4

­2

­1

1

2

3

4

x

y

y = f (x)

Solution :x = −2 : Since

limx→−2−

f (x) = 2, 6= limx→−2+

f (x) = 4,

function has no limit at x = −2.x = 0 :

limx→0−

f (x) = limx→0+

f (x) = 0⇒ limx→0

f (x) = 0 = f (x)

Hence the function is continuous at x = 0 since the function and limit have the same value.x = 1

limx→1−

f (x) = limx→1+

f (x) = 1⇒ limx→1

f (x) = 1 6= 2 = f (1)

f (x) is not continuous at x = 1. This kind of discontinuity is called a removable discontinuity.

Here is the formal definition of continuity.

Definition 2.6.2 (Continuous Function) Let f(x) be a function defined on an interval that contains x = a. Then wesay that f(x) is continuous at x = a if for every number ε > 0 there is some number δ > 0 such that

|f (x)− f (a)| < ε whenever 0 < |x− a| < δ.

Example 2.6.3 Show that f (x) = x continuous at x = 0.Solution : For any ε > 0, we need to find δ > 0 such that

|f (x)− f (a)| < ε whenever 0 < |x− a| < δ

|x| < ε whenever 0 < |x| < δ

If we choose ε = δ, the condition is satisfied.It can be shown that f (x) = x is continuous for any x ∈ R.

Theorem 2.6.4 If f and g are continuous at x = a and k is a number, then f ± g, kf and fg are also continuous atx = a. Moreover, if g (a) 6= 0, then f

g is also continuous at x = a.

Theorem 2.6.5 Every polynomial function is continuous on R.

34

Theorem 2.6.6 Rational functions are continuous everywhere except where we have division by zero

Example 2.6.7 Determine where the function below is not continuous

f (x) =x+ 3

x2 − 5x− 6.

Solution : We need to is determine where the denominator is zero.

x2 − 5x− 6 = (x+ 1) (x− 6) = 0

⇒ x = −1 and x = 6

Therefore, the function will not be continuous at x = −1 and x = 6.

Let f (x) be a continuous at a. Since limx→a

x = a, condition

limx→a

f (x) = f (a)

means thatlimx→a

f (x) = f(

limx→a

x)

So this means we can interchange lim and f .This fact can be extent as follows:

Theorem 2.6.8 Let f (x) is continuous at x = b and limx→a

g (x) = b then,

limx→a

f (g (x)) = f(

limx→a

g (x)).

Moreover, if g (a) = b, f ◦ g is also continuous.

Exercise 2.6.9

limx→0

ln(x2 − 5x+ 1

)= ln lim

x→0

(x2 − 5x+ 1

)= ln 1

= 0

Example 2.6.10limx→0

ecos x = elimx→0

cos x= e1 = e

Theorem 2.6.11 (Intermediate Value Theorem) If f(x) is a continuous function on [a, b], then for every d betweenf(a) and f(b), there exists a value c in between a and b such that f(c) = d.

Intermediate Value Theorem says that a continuous function will take on all values between f(a) and f(b).

Example 2.6.12 Show that the function f (x) = lnx− 1 has a solution between 2 and 3.Solution : When we plug in 2 and 3 into f(x), we see that f(2) = ln(2)−1 ≈ −0.307 and f(3) = ln(3)−1 ≈ 0.099. Since

35

the first number is negative and the second number is positive and f(x) is a continuous function on the interval [2, 3], bythe Intermediate Value Theorem, f(x) must have a solution between 2 and 3.lnx− 1

1 2 3 4 5

­3

­2

­1

0

x

y

y = lnx− 1

36

3 Derivative

3.1 Tangent Line, Rate of Change and Instantaneous Velocity

Consider two points P (x1, y1) and Q(x2, y2) on the graph of y = f(x). The line joining these two points is called a secantline and has a slope given by

msecant =y2 − y1x2 − x1

.

If we let h = x2 − x1, then x2 = x1 + h and y2 = f (x2) = f (x1 + h).The slope of the secant line joining P and Q is then

msecant =f (x1 + h)− f (x1)

h

(or msecant =f (x2)− f (x1)

x2 − x1).

and the equation of the secant line that joining P and Q is

y = msecant (x− x1) + f (x1)

ory = msecant (x− x2) + f (x2) .

Let’s now imagine that point Q slides along the curve towards point P . As it does so, the slope of the secant linejoining P and Q will more closely approximate the slope of a tangent line to the curve at P . We can in fact define theslope of the tangent line at point P as the limiting value of msecant as point Q approaches P .As point Q approaches P , the value of h = x2 − x1 approaches zero (x2 approaches x1). The slope of the tangent line

at P is then

mtangent = limh→0

f (x1 + h)− f (x1)

h

(or mtangent = limx2→x1

f (x2)− f (x1)

x2 − x1).

The equation of the tangent line that goes through (x1, f (x1)) is given by

y = mtangent (x− x1) + f (x1) .

Example 3.1.1 Let y = x2.a) Find the slope of the secant line joining (0, f(0)) to (3, f(3)).b) Find the secant line joining (0, f(0)) to (3, f(3)).c) Find the slope of the tangent line to f(x) at x = 3.d) Find the tangent line to f(x) at x = 3.Solution : a)

msecant =f (3)− f (0)

3− 0=

9− 0

3− 0= 3.

b)

y = msecant (x− x1) + f (x1)

= 3 (x− 0) + f (0)

= 3x

37

­5 ­4 ­3 ­2 ­1 1 2 3 4 5

­10

10

20

x

y

y = x2 and y = 3x

c)

mtangent = limh→0

f (x1 + h)− f (x1)

h

= limh→0

f (3 + h)− f (3)

h= limh→0

(3 + h)2 − 32

h

= limh→0

6h+ h2

h= limh→0

(6 + h) = 6

d)

y = mtangent (x− x1) + f (x1)

= 6 (x− 3) + 9

= 6x− 9

­4 ­2 2 4

­10

10

20

x

y

y = x2 and y = 6x− 9

Example 3.1.2 Let y = 1x .

a) Find the slope of the tangent line to f(x) at x = 2.b) Find the tangent line to f(x) at x = 2.Solution : a)

mtangent = limx→a

f (x)− f (a)

x− a

= limx→2

f (x)− f (2)

x− 2= limx→2

1x −

12

x− 2

= limx→2

2−x2x

x− 2= limx→2

(2− x)

2x (x− 2)

= limx→2

−1

2x= −1

4.

38

b)

y = mtangent (x− x1) + f (x1)

= −1

4(x− 2) +

1

2

­4 ­2 2 4

­3

­2

­1

1

2

3

x

y

y = 1x and y = − 14 (x− 2) + 1

2

There is another interpretation of the slopes of the secant line and the tangent line. The slope of the secant line joining(a, f(a))to(b, f(b)) is

f (b)− f (a)

b− a .

This is the change in f divided by the change in x, so it represents the average rate of change of f as x goes from ato b (i.e. on the interval a ≤ x ≤ b).

Similarly the slope of the tangent line at a represents the instantaneous rate of change at x = a.

Example 3.1.3 Let f (x) =√x.

a) Find the average rate of change of f(x) on the interval 1 ≤ x ≤ 4.b) Find the instantaneous rate of change of f(x) at x = 4.Solution : a)

msecant =f (b)− f (a)

b− a

=f (4)− f (1)

4− 1=

√4−√

1

4− 1=

1

3.

b) The instantaneous rate of change of f(x) at x = 4 is mtangent at a = 4.

mtangent = limx→a

f (x)− f (a)

x− a

= limx→4

f (x)− f (4)

x− 4= limx→4

√x−√

4

x− 4

= limx→4

√x− 2

(√x− 2) (

√x+ 2)

= limx→4

1√x+ 2

=1

4.

Thus, the instantaneous rate of change of f(x) at x = 4 is 14 . This means that if f continued to change at the same rate,

then for every 4 units that x increased, the function would increase by 1 unit.

Suppose that the function under investigation gives the position of an object moving in one dimension. For instance,suppose that s(t) is the position of the object at time t.The average velocity of the object from t = a to t = b is the change in position divided by the time elapsed :

vaverage =s (b)− s (a)

b− a .

39

Notice that this is the same as the slope of the secant line to the curve, or the average rate of change.The instantaneous velocity at t = a is

v (a) = limh→0

s (a+ h)− s (a)

h.

This is the slope of the tangent line to the curve, or the instantaneous rate of change. You can also use the second formula

v (a) = limt→a

s (t)− s (a)

t− a

Roughly speaking, the instantaneous velocity measures how fast the object is travelling at a particular instant.

Example 3.1.4 An object moves along a straight line so that its position in meters is given by

s (t) = t3 − 6t2 + 9t

for all time t in seconds.a) Find the average velocity of the object between t = 2 and t = 5 seconds.b) Find the instantaneous velocity at t = 2 seconds.Solution : a)

vaverage =s (b)− s (a)

b− a

=s (5)− s (2)

5− 2=

(53 − 6.52 + 9.5

)−(23 − 6.22 + 9.2

)3

= 6

b)

v (a) = limt→a

s (t)− s (a)

t− a

= limt→2

s (t)− s (2)

t− 2= v (a) = lim

t→2

(t3 − 6t2 + 9t

)−((

23 − 6.22 + 9.2))

t− 2

= limt→2

(3t2 − 12t+ 9

)= −3.

40

3.2 Definition of Derivative

Definition 3.2.1 Let f be a function which is defined on some interval (c, d) and let a be some number in this interval.The derivative of the function f at a is the value of the limit

f ′ (a) = limx→a

f (x)− f (a)

x− a .

f is said to be differentiable at a if this limit exists.f is called differentiable on the interval (c, d) if it is diffrentiable every point a in (c, d).

We often “read”f ′ (a) as “f prime of a”.When we substitute x = a+ h in the limit and let h→ 0 instead of x→ a, we obtain the formula

f ′ (a) = limh→0

f (a+ h)− f (a)

h

Often you will writte x instead of a in this equation.

f ′ (x) = limh→0

f (x+ h)− f (x)

h

Let y = f (x). All of the following are equivalent and represent the derivative of f (x) with respect to x :

f ′ (x) = y′ =df

dx=dy

dx=

d

dx(f (x)) =

d

dx(y)

f ′ (a) = y′ (a) =df

dx

∣∣∣∣x=a

=dy

dx

∣∣∣∣x=a

Example 3.2.2 (Derivative of xn for n = 1, 2, . . .) Let f (x) = xn.

f ′ (x) = limh→0

f (x+ h)− f (x)

h

= limh→0

(x+ h)n − xnh

= limh→0

xn +(n1

)xn−1h+ · · ·+

(nn−1)xhn−1 + hn − xn

h

= limh→0

h(nxn−1 + · · ·+ nxhn−2 + hn−1

)h

= limh→0

(nxn−1 + · · ·+ nxhn−2 + hn−1

)= nxn−1

Example 3.2.3 (Derivative of constant function) Let f (x) = c be a constant function.

f ′ (x) = limh→0

f (x+ h)− f (x)

h

= limh→0

c− ch

= limh→0

0

h= lim

h→00

= 0

Theorem 3.2.4 If a function f is differentiable at some a in its domain, then f is also continuous at a.

41

Proof. Since f is dfferentiable at a, by definition, lima→a

f (x)− f (a)

x− a exists (a real number). Hence we have

limx→a

(f (x)− f (a)) = limx→a

f (x)− f (a)

x− a (x− a)

= limx→a

f (x)− f (a)

x− a limx→a

(x− a)

= limx→a

f (x)− f (a)

x− a (a− a)

= limx→a

f (x)− f (a)

x− a .0

= 0

Therefore, we get

limx→a

f (x) = limx→a

(f (x)− f (a) + f (a))

= limx→a

(f (x)− f (a)) + limx→a

f (a)

= 0 + f (a)

= f (a) .

That means f is continuous at a.

Example 3.2.5 Let f (x) = |x|. The domain of f is R and the function f is continuous at a since

limx→0

f (x) = limx→0|x| = 0 = f (0) .

However, f is not differentiable at 0. This is because

f ′ (0) = limh→0

f (0 + h)− f (0)

h

does not exist.

limh→0+

f (0 + h)− f (0)

h= lim

h→0+|0 + h| − |0|

h= limh→0+

h

h= limh→0+

1 = 1

limh→0−

f (0 + h)− f (0)

h= lim

h→0−|0 + h| − |0|

h= limh→0−

−hh

= limh→0+

−1 = −1

x

y

y = |x|

42

3.3 The Differentiation Rules

1. (f ± g)′

= f ′ ± g′ (or d

dx(f ± g) =

df

dx± dg

dx)

2. (fg)′

= f ′g + fg′ (ord

dx(fg) =

df

dxg + f

dg

dx)

3. (cf)′

= cf ′ (or ddx (cf) = c ddx (f))

4.(f

g

)′=f ′g − fg′

g2(or

d

dx

(f

g

)=

dfdxg − f

dgdx

g2)

5. If f(x) and f−1(x) are inverses of each other then,(f−1

)′(x) =

1

f ′ (f−1 (x))

Proof. (1)

(f + g)′(x) = lim

h→0

(f + g) (x+ h)− (f + g) (x)

h

= limh→0

f (x+ h) + g (x+ h)− (f (x) + g (x))

h

= limh→0

f (x+ h)− f (x) + g (x+ h)− g (x)

h

= limh→0

(f (x+ h)− f (x)

h+g (x+ h)− g (x)

h

)= lim

h→0

f (x+ h)− f (x)

h+ limh→0

g (x+ h)− g (x)

h

= f ′ (x) + g′ (x)

Example 3.3.1 f (x) = x3 − 2x2 − x+ 1

f ′ (x) =(x3 − 2x2 − x+ 21

)′=

(x3)′ − (2x2)′ − (x)

′+ (21)

′

= 3x2 − 2(x2)′

+−1 + 0

= 3x2 − 2.2x− 1

= 3x2 − 4x− 1

Example 3.3.2 f (x) =(3x5 + 4x

) (2x7 − 5x3

)f ′ (x) =

(3x5 + 4x

)′ (2x7 − 5x3

)+(3x5 + 4x

) (2x7 − 5x3

)′=

(15x4 + 4

) (2x7 − 5x3

)+(3x5 + 4x

) (14x6 − 15x2

)= 30x11 − 67x7 − 20x3 + 42x11 + 11x7 − 60x3

= 72x11 − 56x7 − 80x3

Example 3.3.3 f (x) =x3

x2 − 1

f ′ (x) =

(x3)′ (

x2 − 1)−(x3) (x2 − 1

)′(x2 − 1)

2

=

(3x2) (x2 − 1

)−(x3)

(2x)

(x2 − 1)2

=x4 − 3x2

(x2 − 1)2

43

Example 3.3.4 Let f (x) =x+ 4

2x− 5and f−1 (x) =

5x+ 4

2x− 1. Since

f ′ (x) =13

(2x− 5)2

by using 5-th property we have (f−1

)′(x) =

1

f ′ (f−1 (x))

=113(

2

(5x+ 4

2x− 1

)−5)2

=13

(2x− 1)2

Example 3.3.5 (The Power rule) Let f is the nth power of some function,

f (x) = (u (x))n,

thenf ′ (x) = n (u (x))

n−1u′ (x) .

Example 3.3.6 f (x) =(x2 + 2x− 7

)5f ′ (x) = 5

(x2 + 2x− 7

)4(2x+ 2)

Example 3.3.7 (The Power Rule for Negative Integer Exponents) Let f (x) = (u (x))−n,then

f ′ (x) = −n (u (x))−n−1

u′ (x) .

Example 3.3.8 f (x) =(x2 + 2x− 7

)−3f ′ (x) = −3

(x2 + 2x− 7

)−4(2x+ 2)

Example 3.3.9 (The Power Rule for Rational Exponents) Let f (x) = (u (x))mn , then

f ′ (x) =m

n(u (x))

mn −1 u′ (x) .

Example 3.3.10 f (x) =√x

f ′ (x) =(√x)′

=(x12

)′=

1

2(x)

12−1 (x)

′

=1

2x−

12

=1

2√x

Example 3.3.11 f (x) =(x2 + 2x− 7

) 53

f ′ (x) =((x2 + 2x− 7

) 53

)′=

5

3

(x2 + 2x− 7

) 53−1 (x2 + 2x− 7

)′=

5

3

(x2 + 2x− 7

) 23 (2x+ 2)

44

3.4 Chain Rule

Theorem 3.4.1 (Chain Rule) If f and g are differentiable, so is the composition f g.

The derivative of f ◦ g is given by(f ◦ g)

′(x) = f ′ (g (x)) g′ (x) .

The chain rule tells you how to find the derivative of the composition f ◦ g of two functions f and g provided you nowhow to differentiate the two functions f and g.Following is the other form of the Chain Rule :If we write u = g (x) and y = f (u), then the derivative of y with rexpect to x is

dy

dx=dy

du

du

dx.

Example 3.4.2 Use the Chain Rule to differentiate f (x) =(2x3 + 5x

)40.

Solution :Let u = g (x) = 2x3 + 5x and y = f (u) = u40. Then

dy

dx=

dy

du

du

dx= 40u39.

(6x2 + 5

)= 40

(2x3 + 5x

) (6x2 + 5

)Example 3.4.3 Use the Chain Rule to differentiate f (x) =

√1− x2.

Solution :Let y = f (u) = u12 and u = g (x) = 1− x2. Then

dy

dx=

dy

du

du

dx=

1

2√u. (−2x)

=1

2√

1− x2. (−2x)

45

3.5 Derivatives of Trigonometric Functions

(sinx)′

= cosx

(cosx)′

= − sinx

(tanx)′

= sec2 x

(cotx)′

= − csc2 x

(secx)′

= secx tanx

(cscx)′

= − cscx cotx

•

d

dx(sinx) = lim

h→0

sin (x+ h)− sinx

h

= limh→0

sinx cosh+ cosx sinh− sinx

h

= limh→0

sinx (cosh− 1) + cosx sinh

h

= limh→0

sinx (cosh− 1)

h+ limh→0

cosx sinh

h

= sinx limh→0

(cosh− 1)

h+ cosx lim

h→0

sinh

h= sinx.0 + cosx.1

= cosx

Example 3.5.1 f (x) = secx− 5 tanx

f ′ (x) = (secx− 5 tanx)′

= secx tanx− 5 sec2 x

Example 3.5.2 f (x) =cosx

2 sinx− 1

f (x) =

(cosx

2 sinx− 1

)′=

(cosx)′(2 sinx− 1)− cosx (2 sinx− 1)

′

(2 sinx− 1)2

=− sinx (2 sinx− 1)− cosx2 cosx

(2 sinx− 1)2

=−2 + sinx

(2 sinx− 1)2

Example 3.5.3 f (x) = sin3(x2 + 1

)f ′ (x) =

(sin3

(x2 + 1

))′= 3 sin2

(x2 + 1

) (sin(x2 + 1

))′= 3 sin2

(x2 + 1

)cos(x2 + 1

) (x2 + 1

)′= 3 sin2

(x2 + 1

)cos(x2 + 1

)(2x)

46

3.6 Derivatives of Inverse Trigonometric Functions

(arcsinx)′

=1√

1− x2

(arccosx)′

= − 1√1− x2

(arctanx)′

=1

1 + x2

(arccotx)′

= − 1

1 + x2

(arcsecx)′

=1

|x|√x2 − 1

(arccscx)′

= − 1

|x|√x2 − 1

• (arcsinx)′

= 1√1−x2

For the rule of derivative of inverse function (f−1

)′(x) =

1

f ′ (f−1 (x))

, if we choose f (x) = sinx and f−1 (x) = arcsinx then we have

(arcsinx)′

=1

cos (arcsinx).

Let arcsinx = y. Then we obtainsin y = x⇒ cos y =

√1− x2

andcos (arcsinx) = cos y =

√1− x2

which gives

(arcsinx)′

=1√

1− x2.

Example 3.6.1 f (x) = arcsin(√

1− x4)

f ′ (x) =1√

1−(√

1− x4)2 (√1− x4

)′

=1√

1− 1x4

(1− x4

)′2√

1− x4

=1

x2−4x3

2√

1− x4

=−2x√1− x4

Example 3.6.2 f (x) = arccot

(x

1− x2

)

f ′ (x) = − 1

1 +

(x

1− x2

)2 ( x

1− x2

)′

= − 1

1 + x2

(1−x2)2

1− x2 − x (−2x)

(1− x2)2

= − 1 + x2

(1− x2)2 + x2

= − 1 + x2

1− x2 + x4

47

3.7 Derivatives of Exponential and Logarithm Functions

(ax)′

= ax ln a

(ex)′

= ex

(loga x)′

=1

x ln a

(lnx)′

=1

x

Example 3.7.1 f (x) = log5(x2 + 1

)f ′ (x) =

1

(x2 + 1) ln 5

(x2 + 1

)′=

2x

(x2 + 1) ln 5

Example 3.7.2 f (x) = 2sin x

f ′ (x) = 2sin x ln 2 (sinx)′

= 2sin x ln 2 cosx

48

3.8 Logarithmic Differentiation

Example 3.8.1 Let f (x) = xx. Find f ′ (x).

y = xx ⇒ ln y = ln (xx)⇒ ln y = x lnx

⇒ (ln y)′

= (x lnx)′

⇒ y′

y= x′ lnx+ x (lnx)

′= lnx+ x

1

x= lnx+ 1

⇒ y′ = y. (lnx+ 1)

⇒ y′ = xx (lnx+ 1)

Example 3.8.2 Let f (x) = (sin 2x)x3 . Find f ′ (x).

y = (sin 2x)x3 ⇒ ln y = ln

((sin 2x)

x3)⇒ ln y = x3 ln (sin 2x)

⇒ (ln y)′

=(x3 ln (sin 2x)

)′⇒ y′

y=(x3)′

ln (sin 2x) + x3 (ln (sin 2x))′

= 3x2 ln (sin 2x) + x3cos 2x.2

sin 2x

⇒ y′ = y.

(3x2 ln (sin 2x) + x3

cos 2x.2

sin 2x

)⇒ y′ = (sin 2x)

x3(

3x2 ln (sin 2x) + x3cos 2x.2

sin 2x

)Example 3.8.3 Find the derivative of

y =x2 3√

7x− 14

(1 + x2)4

Solution :

y =x2 3√

7x− 14

(1 + x2)4 ⇒ ln y = ln

(x2 3√

7x− 14

(1 + x2)4

)= 2 lnx+

1

3ln (7x− 14)− 4 ln

(1 + x2

)

(ln y)′

=

(2 lnx+

1

3ln (7x− 14)− 4 ln

(1 + x2

))′y′

y= 2

1

x+

1

3

7

7x− 14− 4

2x

1 + x2

y′ = y

(2

1

x+

1

3

7

7x− 14− 4

2x

1 + x2

)y′ =

x2 3√

7x− 14

(1 + x2)4

(2

1

x+

1

3

7

7x− 14− 4

2x

1 + x2

)

49

3.9 Derivatives of Hyperbolic Functions

(sinhx)′

= coshx

(coshx)′

= sinhx

(tanhx)′

= sech2 x

(cothx)′

= − csch2 x

(sechx)′

= − sechx tanhx

(cschx)′

= − cschx cothx

•(sinhx)

′=

(ex − e−x

2

)′=ex + e−x

2= coshx

Example 3.9.1 f (x) = sech2 (lnx)

f ′ (x) =(sech2 (lnx)

)′= 2 sech (lnx) (sech (lnx))

′

= 2 sech (lnx) (− sech (lnx) tanh (lnx)) (lnx)′

= − 2

xsech2 (lnx) tanh (lnx)

Example 3.9.2 f (x) = coth

(1

x

)

f ′ (x) = − csch2(

1

x

)(1

x

)′= − csch2

(1

x

)(− 1

x2

)

50

3.10 Derivatives of Inverse Hyperbolic Functions

(arcsinhx)′

=1√

1 + x2

(arccoshx)′

=1√

x2 − 1(x > 1)

(arctanhx)′

=1

1− x2 (|x| < 1)

(arccothx)′

=1

1− x2 (|x| > 1)

• To show (arcsinhx)′

= 1√1+x2

, we use following rule

(f−1

)′(x) =

1

f ′ (f−1 (x)).

f (x) = sinhx, f−1 (x) = arcsinhx

(arcsinhx)′

=1

cosh (arcsinhx).

Let arcsinhx = y. Then we obtainsinh y = x⇒ cosh y =

√1 + x2

sincecosh2 x− sinh2 x = 1

andcosh (arcsinhx) = cosh y =

√1 + x2

which gives

(arcsinhx)′

=1√

1 + x2.

y = arcsinhx⇒ x = sinh y

⇒ dx

dy= (sinh y)

′= cosh y =

√x2 + 1

⇒ dy

dx=

1√x2 + 1

(cosh2 y = sinh2 +1 = x2 + 1)

Example 3.10.1 f (x) = arccosh (2x+ 1)

f ′ (x) =1√

(2x+ 1)2 − 1

(2x+ 1)′

=2√

(2x+ 1)2 − 1

=1√

x2 + x

Example 3.10.2 f (x) = arcsinh(1x

)f ′ (x) =

1√1 +

(1x

)2(

1

x

)′

=1√

1 + 1x2

(− 1

x2

)

=−1

x (1 + x2)

51

3.11 Implicit Differentiation

Some functions are defined implicitly by a relation between x and y. In some cases it is possible to solve such an equationfor x or for y; but sometimes it is impossible. In this section we to show how to find derivatives of implicitly definedfunctions.

Example 3.11.1 Find the equation of the tangent line to x = y2 at the point (4, 2).Solution : To find equations of the tangent we first find dy

dx by differentiating both sides of x = y2.

x = y2 ⇒ x′ =(y2)′

⇒ 1 = 2y.y′

⇒ y′ =1

2y

Hencemtangent = y′ (2) =

1

4

and the equation is

y = mtangent (x− x0) + f (x0)

y =1

4(x− 4) + 2.

­5 5 10

­3

­2

­1

1

2

3

x

y

Example 3.11.2 Find the equations of the tangent lines to x2 + y2 = 13 at x = 2.Solution : x = 2⇒ y = ±3. So we need to find the tanget line equations at (2,−3) and (2, 3). To find dy

dx we differentiateboth sides : (

x2 + y2)′

= (13)′

2x+ 2yy′ = 0

y′ = −2x

2y= −x

y

Therefore for (2,−3)

mtangent = y′ (2) =2

3

and for (2, 3)

mtangent = y′ (2) = −2

3.

The equations are

y =2

3(x− 2) + (−3)

andy = −2

3(x− 2) + 3.

52

­4 ­2 2 4

­4

­2

2

4

x

y

Example 3.11.3 Finddy

dxif sin (x+ y) = y2 cos 2x.

Solution :sin (x+ y) = y2 cos 2x⇒ (sin (x+ y))

′=(y2 cos 2x

)′cos (x+ y) . (x+ y)

′=

(y2)′

cos 2x+ y2 (cos 2x)′

cos (x+ y) (1 + y′) = 2yy′ cos 2x+ y2 (− sin 2x) .2

y′ (cos (x+ y)− 2y cos 2x) = −2y2 sin 2x− cos (x+ y)

y′ =−2y2 sin 2x− cos (x+ y)

cos (x+ y)− 2y cos 2y

53

3.12 Higher Order Derivatives

Let f be a function that is differentiable at some points belonging to dom(f). Then f ′ is a function. If, in addition, f ′ isdifferentiable at some points belonging to dom(f ′), then the derivative of f ′ exists and is denoted by f ′′ and is called thesecond derivative of f . In general, the n-th derivative of f (where n is a positive integer), denoted by f (n), is defined tobe the derivative of the (n− 1)-th derivative of f (where the 0-th derivative of f means f ). For n = 1, the first derivativeof f is simply the derivative f ′ of f . For n > 1, f (n) is called a higher-order derivative of f .

Similar to first order derivative, we have different notations for second order derivative of f.

f ′′ (x) = y′′ =d2f

dx2=d2y

dx2=

d2

dx2(f (x)) =

d2

dx2(y)

Example 3.12.1 If f (x) = x3 − 2x2 + 3x+ 1 then

f (x) = x3 − 2x2 + 3x+ 1

f ′ (x) = 3x2 − 4x+ 3

f ′′ (x) = 6x− 4

f ′′′ (x) = 6

f ′′′′ (x) = 0

...

Example 3.12.2 Find the first four derivatives of f (x) = sin (2x) + e−2x + lnx.Solution :

f (x) = sin (2x) + e−2x + lnx

f ′ (x) = cos (2x) 2− 2e−2x + x−1

f ′′ (x) = − sin (2x) 4 + 4e−2x − x−2

f ′′′ (x) = − cos (2x) 8− 8e−2x + 2x−3

f ′′′′ (x) = sin (2x) 16 + 16e−2x − 6x−4

Example 3.12.3 Findd2y

dx2if x2 + y3 = 5.

Solution : Here we need to do implicit differentiation.(x2 + y3

)′= (5)

′

2x+ 3y2y′ = 0

y′ =−2x

3y2.

This is the first derivative. We get the second derivative by differentiating this.

y′′ =

(−2x

3y2.

)′=−2.3y2 − (−2x) 6yy′

(3y2)2

=−6y2 + 12x6yy′

9y4

=−6y2 + 12x6y−2x3y2

9y4

=−18y3 + 24x2y

27y5.

54

4 Applications of Derivatives

4.1 L’Hospital Rule

Theorem 4.1.1 (L’Hospital Rules) If f and g are differentiable in a neighborhood of x = c, and

limx→c

f (x)

g (x)=

0

0or lim

x→c

f (x)

g (x)=±∞±∞

, then

limx→c

f (x)

g (x)= limx→c

f ′ (x)

g′ (x)

provided the limit on the right exists.The same result holds for one-sided limits and c can be ±∞.

Example 4.1.2

limx→1

x2 − 1

x− 1=

00L′Hos.

limx→1

(x2 − 1

)′(x− 1)

′ = limx→1

2x

1= 2

Example 4.1.3

limx→0

sinx

x=

00L′Hos.

limx→0

(sinx)′

(x)′ = lim

x→0

cosx

1= 1

limx→0

1− cosx

x=

00L′Hos.

limx→0

(1− cosx)′

(x)′ = lim

x→0

sinx

1= 0

Example 4.1.4

limx→0+

x lnx = 0.∞

= limx→0

lnx1x

=00L′Hos.

limx→0

(lnx)′(

1x

)′= lim

x→0

1x

− 1x2

= limx→0

(−x) = 0.

Example 4.1.5

limx→∞

ex

x2=

∞∞L′Hos.

limx→∞

(ex)′

(x2)′ = lim

x→∞

ex

2x=

∞∞L′Hos.

limx→∞

(ex)′

(2x)′

= limx→∞

ex

2=∞

Example 4.1.6 Show that

limx→∞

(1 +

k

ax+ b

)cx+d= e

ck

a .

Solution :

55

n = limx→∞

(1 +

k

ax+ b

)cx+dlnn = ln lim

x→∞

(1 +

k

ax+ b

)cx+d= limx→∞

ln

[(1 +

k

ax+ b

)cx+d]

= limx→∞

(cx+ d) . ln

(1 +

e

ax+ b

)=∞.0

= limx→∞

ln(

1 + kax+b

)1

cx+d

=00L′Hos.

limx→∞

− ka(ax+b)2

1 + kax+b

− c(cx+d)2

= limx→∞

(cx+ d)2

c

ka

(ax+ b)

1

(ax+ b+ k)

=ck

a

lnn =kc

a⇒ n = lim

x→∞

(1 +

k

ax+ b

)cx+d= e

ck

a

Example 4.1.7

limx→∞

(1 +

3

x

)5x= e15

Example 4.1.8

limx→∞

(x+ 7

x+ 3

)2x+3= lim

x→∞

(1 +

4

x+ 3

)2x+3

= e

4.2

1

= e8

Example 4.1.9 Show that

limx→∞

(2 +

1

x

)x=∞.

Solution :

n = limx→∞

(2 +

1

x

)xlnn = ln lim

x→∞

(2 +

1

x

)x= limx→∞

ln

[(2 +

1

x

)x]= lim

x→∞x. ln

(2 +

1

x

)=∞.0 =∞

lnn =∞⇒ n = limx→∞

(2 +

1

x

)x= e∞ =∞

Example 4.1.10limx→∞

x1x =∞0

y = limx→∞

x1x

ln y = ln limx→∞

(x1x

)= limx→∞

(ln (x)

1x

)= limx→∞

(1

xlnx

)= lim

x→∞

lnx

x=

∞∞L′Hos.

limx→∞

(lnx)′

(x)′ = lim

x→∞

1x

1= 0

ln y = 0⇒ y = limx→∞

x1x = e0 = 1

56

4.2 Rolle’s Theorem and The Mean Value Theorem

In this section we want to take a look at the Mean Value Theorem. Before we get to the Mean Value Theorem we needto cover the following theorem

Theorem 4.2.1 (Rolle’s Theorem) If f is continuous on the closed interval [a, b], differentiable on the open interval(a, b) and f (a) = f (b) then there is some number c such that a < c < b and f ′ (c) = 0.

x

y

x

y

x

y

Example 4.2.2 Show that f (x) = x3 + 2x+ 10 has exactly one real root.Solution : Since f is a 3th degree polynomial, it has tree roots. We’re being asked to prove here is that only one of those3 is a real number. First, we should show that it does have at least one real root. To do this note that f (−3) = −23 andthat f (−1) = 7 and so we can see that f (−3) < 0 < f (1) Now, because f is a polynomial we know that it is continuouseverywhere and so by the Intermediate Value Theorem there is a number c such that −3 < c < −1 and f (c) = 0. In otherwords f has at least one real root. We now need to show that this is in fact the only real root. To do this we’ll use anargument that is called contradiction proof. Assume that f has at least two real roots.This means there are real numbersa and b such that f (a) = f (b) = 0. From Rolle’s Theorem there must be another number c such that f ′ (c) = 0. But thederivative of this function is

f ′ (x) = 3x2 + 2

which is always greater than zero. This contradicts the statement above. So we can only have a single real root.

­4 ­2 2 4

­100

100

x

y

y = x3 + 2x+ 10

Theorem 4.2.3 (The Mean Value Theorem) If f is continuous on the closed interval [a, b] and differentiable on theopen interval (a, b), then there is some number c such that a < c < b and

f ′ (c) =f (b)− f (a)

b− a .

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c thatwill satisfy the conclusion of the theorem.

57

Mean Value Theorem tells that the secant line connecting (a, f (a)) and (b, f (b)) and the tangent line at x = c mustbe parallel.

x

y

Example 4.2.4 Determine all the numbers c which satisfy the conclusions of the Mean Value Theorem for the followingfunction.

f (x) = 4x5 − x+ 2

on [0, 1] .Solution : Since f is a polynomial, it satisfies hypotheses of the Mean Value Theorem. Therefore there is a number c in(0, 1) such that

f ′ (c) =f (1)− f (0)

1− 0.

20c4 − 1 = 5− 2

20c4 − 1 = 3

c = ± 14√

5

But c must be lie in (0, 1), so c = 14√5 .

x

y

Example 4.2.5 Use the Mean Value theorem to show that

limx→∞

(√x+ 1−

√x)

= 0.

Solution : Define the function f(x) =√x. Then f is continuous on [0,∞) and differentiable on (0,∞). By the Mean

Value theorem there exists a number c such that

f (x+ 1)− f (x)

(x+ 1)− x = f ′ (c)

for c between x and x+ 1. But then

0 <√x+ 1−

√x =

1

2√c.

As x goes to infinity, so does c (it is always bigger than x). The left side of this equation goes to 0 as c goes to infinity.Therefore, the right side must also go to zero.

58

4.3 Increasing and Decreasing Functions

Definition 4.3.1 Let f be a function.A function is called increasing if a < b implies f(a) < f(b) for all numbers a, b ∈ dom (f).A function is called decreasing a < b implies f(a) > f(b) for all numbers a, b ∈ dom (f).A function is called non-increasing if a < b implies f(a) ≥ f(b) for all numbers a, b ∈ dom (f).A function is called non-decreasing a < b implies f(a) ≤ f(b) for all numbers a, b ∈ dom (f).

x

y

Increasing

x

y

Decreasing

x

y

Non-decreasing

x

y

Non-increasing

The sign of the derivative of f tells you if f is increasing or not.

Theorem 4.3.2 (Increasing/Decreasing Test) Suppose f is a differentiable function on an interval (a, b).1) f ′ (x) > 0 for all a < x < b, then f is increasing.2) f ′ (x) < 0 for all a < x < b, then f is decreasing.

Example 4.3.3 Find where the function f (x) = x4 − 4x3 + 5 is increasing or decreasing.Solution :

f ′ (x) =(x4 − 4x3 + 5

)′= 4x3 − 12x2

= 4x2 (x− 3) = 0

⇒ x = 0, x = 3

(−∞, 0) (0, 3) (3,∞)

4 + + +x2 + + +x− 3 − − +

f ′ − − +f ↘ ↘ ↗

From the table,1) on the interval (3,∞), f is increasing2) on the interval (−∞, 3), f is decreasing.

­4 ­2 2 4 6

­50

50

100

x

y

y = x4 − 4x3 + 5

59

Example 4.3.4 Find where the function f (x) =x

x2 + 4is increasing or decreasing.

Solution :

f ′ (x) =

(x

x2 + 4

)′=

(x)′ (x2 + 4

)− (x)

(x2 + 4

)′(x2 + 4)

2

=(1)(x2 + 4

)− (x) (2x)

(x2 + 4)2

=−x2 + 4

(x2 + 4)2 = 0

⇒ x = −2, x = 2

(−∞,−2) (−2, 2) (2,∞)

−x2 + 4 − + −(x2 + 4

)2+ + +

f ′ − + −f ↘ ↗ ↘

­4 ­2 2 4

­0.2

­0.1

0.1

0.2

x

y

y =x

x2 + 4

60

4.4 Minimum and Maximum Values

Definition 4.4.1 A function f has an absolute maximum (or global maximum) at c if f(c) ≥ f(x) for all x in dom (f).The number f(c) is called the maximum value of f on dom (f). Similarly, f has an absolute minimum (or global minimum)at c if f(c) ≤ f(x) for all x in dom (f) and the number f(c) is called the minimum value of f on dom (f).

The maximum and minimum values of f are called the extreme values of f .

Definition 4.4.2 A function f has a local maximum (or relative maximum) at c if f(c) ≥ f(x) for all x in some openinterval containing c. Similarly, f has a local minimum (or relative minimum) at c if f(c) ≤ f(x) for all x in some openinterval containing c.

Example 4.4.3 Let f (x) = sinx where dom (f) = [0, 2π]

1 2 3 4 5 6­1

0

1

x

y

f has an absolute (local) maximum value 1 at x=π

2.

f has an absolute (local) minimum value 1 at=3π

2.

Let f (x) = sinx where dom (f) = (−∞,∞)

­10 ­8 ­6 ­4 ­2 2 4 6 8 10­1

1

x

y

f has an absolute (local) maximum value 1 at x=π

2+ k2π, k ∈ Z.

f has an absolute (local) minimum value 1 at=3π

2+ k2π, k ∈ Z.

Example 4.4.4 Let f (x) = x2.

­4 ­2 0 2 4

10

20

x

y

f has no absolute (local) maximum value.

f has an absolute (local) minimum value 0 at=0.

61

Let f (x) = x3 + 3x2 − 1.

­4 ­3 ­2 ­1 1 2

­10

10

x

y

f has no absolute maximum value and absolute maximum value.

f has an local maximum value 3 at=− 2.

f has an local minimum value 0 at=0

Example 4.4.5 Let f (x) = arctanx.

x

y

f has no absolute maximum value and absolute maximum value.

f has no local maximum value and local minimum value.

Let f (x) = 1x .

­4 ­2 2 4

­2

­1

1

2

x

y

f has no absolute maximum value and absolute maximum value.

f has no local maximum value and local minimum value.

The following theorem is also useful, but in a difierent way. It doesn’t say how to find maxima or minima, but it tellsyou that they do exist.

Theorem 4.4.6 (The Extreme Value Theorem) If f is continuous on a closed interval [a, b], then f attains an ab-solute maximum value f(c) and an absolute minimum value f(d) at some numbers c and d in [a, b].

62

Example 4.4.7 Let f (x) = tanx, dom (f) =(−π2 ,

π2

).

­1.5 ­1.0 ­0.5 0.5 1.0 1.5

­6

­4

­2

2

4

6

x

y

f has no local maximum value and local minimum value.

Let f (x) = x2, dom (f) = [−1, 2] By the Extreme Value Theorem,

­1 0 1 2

1

2

3

4

x

y

f has an absolute maximum value.4 at x=2.

f has an absolute minimum value 0 at=0.

Let f (x) = x3, dom (f) = [−1, 1] By the Extreme Value Theorem,

­1.0 ­0.5 0.5 1.0

­1

1

x

y

f has an absolute maximum value.1 at x=1.

f has an absolute minimum value − 1 at=− 1.

Definition 4.4.8 Let c be in the domain of a function f . If f ′ (c) = 0 or f ′ (c) doesn’t exist, then c is called a criticalpoint of f .

Theorem 4.4.9 (Fermat’s Theorem) If f has a local maximum or minimum at c, and if f ′ (c) exists, then f ′ (c) = 0( c is a critical point of f ).

63

The converse of this theorem is not true. In other words, when f ′ (c) = 0, f does not necessarily have a local maximumor minimum. For example, if f(x) = −x3, then f ′ (x) = −3x2 equals 0 at x = 0, but x = 0 is not a local minimum orlocal maximum point.Sometimes f ′ (x) does not exist, but x = c is a local maximum or local minimum. For example, if f(x) = |x|, then

f ′ (x) does not exist. But f(x) has its local (and absolute) minimum at x = 0.

­4 ­2 2 4

­100

­50

50

100

x

y

y = −x3

­4 ­2 0 2 4

2

4

x

y

y = |x|

Example 4.4.10 Determine all the critical points for the function.

f (x) =x

x2 + 1.

Solution :

f ′ (x) =(x)′ (x2 + 1

)− x

(x2 + 1

)′(x2 + 1)

2

=(1)(x2 + 1

)− x (2x)

(x2 + 1)2

=−x2 + 1

(x2 + 1)2 = 0

=(−x+ 1) (x+ 1)

(x2 + 1)2 = 0

⇒ x = −1, x = 1 are critical points.

Example 4.4.11 Determine all the critical points for the function.

f (x) = (x− 1)23 .

Solution :

f ′ (x) =(

(x− 1)23

)′=

2

3(x− 1)

− 13 (x− 1)

′

=2

3 (x− 1)13

= 0

Notice that f ′ (x) 6= 0 but at x = 1, f ′ (x) is not defined, although f (1) is defined. Hence x = 1 is a critical point of f ,

64

and indeed, a minimum occurs at x = 1.

­2 0 2 4

1

2

x

y

y = (x− 1)23

Example 4.4.12 Determine all the critical points for the function.

f (x) =1

x− 1.

Solution :

f ′ (x) =

(1

x− 1

)′=

−1

(x− 1)2 = 0

There is no points that f ′ (c) = 0. For x = 1, f ′ (1) does not exist but 1 /∈ dom (f). Hence f has no critical points andhas no local extrema.

Fermat’s theorem says that the set of the local extreme points of f is a subset of the critical points of f . Thereforeto find the local exremes at first we need to find the critical points, then determine which one is local extreme. Followingtheorem help us for this.

Theorem 4.4.13 (First Derivative Test) Suppose c is a critical point of a continuous function f .(a) If f ′ (x) > 0 to the left of x = c and f ′ (x) < 0 to the right of x = c; then f has a local maximum at x = c.(b) If f ′ (x) < 0 to the left of x = c and f ′ (x) > 0 to the right of x = c; then f has a local minimum at x = c.(c) If f ′ does not change sign at x = c (that is, f ′ is positive on both sides of c or negative on both sides), then f has nolocal maximum or minimum at x = c.

Note that the first derivative test will only classify critical points as local extrema and not as absolute extrema.

Example 4.4.14 Find and classify all the critical points of the following function

f (x) = 27x− x3.

Solution :

f ′ (x) = 27− 3x2

= 3 (3 + x) (3− x) = 0

The critical points are x = −3 and x = 3.

← −3 −3→ ← 3 3→f ′ − + + −

local minima local maxima

THE CLOSED INTERVAL METHOD :To find the absolute maximum and minimum values of a continuous function f on a closed interval [a, b]:

1) Find the values of f at the critical points of f in (a, b).2) Find the values of f at the endpoints of the interval.3) The largest value of Step 1 and 2 is the absolute maximum value; the smallest value of Step 1 and Step 2 is the absoluteminimum value.

65

Example 4.4.15 Find the absolute maximum and minimum values of

f (x) = 2x3 − 3x2 − 12x+ 1

the interval [−2, 3] and determine where these values occur.Solution : 1) Critical points:

f ′ (x) = 6x2 − 6x− 12

= 6 (x− 2) (x+ 1) = 0

⇒ x = −1, x = 2

Notice that both in [−2, 3]. The values of f at the critical points of f in (−2, 3) are f (−1) = 8, f (2) = −192) The values of f at the endpoints of the interval [−2, 3] are f (−2) = −3, f (3) = −83) In interval [−2, 3], f attends its absolute maximum at x = −1 and its absolute minimum at x = 2. Its maximum andminimum values are 8 and −19 respectively.

­2 ­1 1 2 3

­20

­15

­10

­5

5

10

x

y

y = 2x3 − 3x2 − 12x+ 1

66

4.5 Concave up and Concave down

Definition 4.5.1 Given the function f then1) f is concave up (convex)on an interval I if all of the tangent lines to the curve on I are below the graph of f .2) f is concave down (concave) on an interval I if all of the tangent lines to the curve on I are above the graph of f .

Concave up,Decreasing

Concave up,Increasing

Concave down,Increasing

Concave down,Decreasing

Example 4.5.2

­4 ­2 0 2 4

10

20

x

y

y = x2

­4 ­2 02 4

­20

­10

xy

y = −x2

­4 ­2 2 4

­100

100

x

y

y = x3

­4 ­2 2 4

­2­1

12

x

y

y = 1x

Definition 4.5.3 A point x = c is called an inflection point if the function is continuous at the point and the concavityof the graph changes at that point.

Theorem 4.5.4 (Concavity Test) Given the function f (x) then,1. If f ′′ (x) > 0 for all x in some interval I then f (x) is concave up on I.2. If f ′′ (x) < 0 for all x in some interval I then f (x) is concave down on I.

Example 4.5.5 Find intervals of concavity and the infection points of

f (x) = x3 − x.

Solution : We need to solve f ′′ (x) = 0

f ′ (x) = 3x2 − 1

f ′′ (x) = 6x

67

6x = 0⇒ x = 0

Hence we have(−∞, 0) (3,∞)

6 + +x − +

f ′′ − +f _ ^

Therefore f is concave down on (−∞, 0); ; it is concave up (0,∞). The point 0 is the inflection point since the curvechanges from concave down to concave up.

­3 ­2 ­1 1 2 3

­4

­2

2

4

x

y

y = x3 − x

Example 4.5.6 Find intervals of concavity and the infection points of f (x) = xe−x.Solution :

f ′ (x) = e−x − xe−x

f ′′ (x) = −e−x − e−x + xe−x = (x− 2) ex

(x− 2) ex = 0⇒ x = 2

(−∞, 2) (2,∞)

x− 2 − +ex + +

f ′′ − +f _ ^

As a result, the inflection point is at x = 2.

2 4

­2.5

­2.0

­1.5

­1.0

­0.5

xy

y = xe−x

Theorem 4.5.7 Let f be a function and let c be a real number such that f is differentiable on an open interval containingc and that f ′′ (c) exists. Suppose that c is an inflection point of f . Then we have f ′′ (c) = 0.

68

The converse of this Theorem is not true: if f ′′ (c) = 0, c may not be an inflection number of f . The functionf (x) = −x4 have a point 0 which f ′′ (0) = 0 but 0 is not an inflection point.

f (x) = −x4

f ′ (x) = −4x3

f ′′ (x) = −12x2 = 0

⇒ x = 0

(−∞, 0) (0,∞)

f ′′ − −f _ _

xy

y = −x4

Theorem 4.5.8 (Second Derivative Test.) Suppose that c is a critical point of f such that f ′′ (x) is continuous in aregion around c Then1) If f ′′ (c) < 0 then c is a local maximum.2) If f ′′ (c) > 0 then c is a local minimım.3) If f ′′ (c) = 0 then c can be a local maximum, local minimum or neither.

Usually the second derivative test will work, but sometimes a critical point c has f ′′ (c) = 0. In this case the secondderivative test gives no information at all. As you can see, for the following two functions Second Derivative Test doesn’twork.

x

y

y = x3

x

y

y = x4

Example 4.5.9 Use the Second Derivative Test to classify the critical points of the function f (x) = 3x5 − 5x3 + 3.Solution :

f ′ (x) = 15x4 − 15x2

= 15x2(x2 − 1

)= 0

⇒ x = −1, x = 0, x = 1

f ′′ (x) = 60x3 − 30x

f ′′ (−1) = −30 < 0⇒ x = −1 is local maximum.

f ′′ (1) = 30 > 0⇒ x = 1 is local minimum.

f ′′ (0) = 0⇒ for x = 0, test gives no information

69

Hence we use the First Derivative Test for x = 0.

← 0 0→f ′ − −

not local extrem

70

4.6 Curve Sketching

GUIDELINES FOR SKETCHING A CURVE:A) DomainB) Intercepts: x− and y−intercepts.C) Symmetry: even (f(−x) = f(x)) or odd (f(−x) = −f(x)) function.D) Asymptotes: horizontal ( lim

x→±∞f (x) = L) and vertical ( lim

x→a±f (x) = ±∞) asymptotes.

E) Intervals of Increase or Decrease: Use the Increasing/Decreasing TestF) Local Maximum and Minimum Values: Use critical numbers.G) Concavity and Points of Inflection: Compute f ′′(x) and use the Concavity TestH) Sketch the Curve: Using the information in items A−G, draw the graph.

Example 4.6.1 Sketch the graph of f (x) = 14x

4 − 2x2 + 1Solution :A) Domain : dom (f) = RB) Intercepts :i-) x−intercept (y = 0) : f (x) = 0⇒ x = ±

√4± 2

√3

ii-) y−intercept (x = 0) : f (0) = 1C) Symmetryi) This function is even ( the function is symmetric with recpect to y−axis) since

f (−x) =1

4(−x)

2 − 2 (x)2

+ 1 =1

4x4 − 2x2 + 1 = f (x)

ii) This function is not odd since

f (−1) = −3

46= 3

4= −f (1) .

D) Asymptotes :i) Horizontal asymptotes :

limx→∞

f (x) =∞, limx→−∞

f (x) =∞⇒ No horizontal asymptotes

i) Vertical asymptotes :No vertical asymptotes

E) Intervals of Increase or Decrease :f ′ (x) = x3 − 4x = 0⇒ x = −2, 0, 2

(−∞,−2) (−2, 0) (0, 2) (2,∞)

f ′ − + − +f ↘ ↗ ↘ ↗

F) Local Maximum and Minimum Values :

(−2,−3) and (2,−3) are local minimum and (0, 1) is local maximum

G) Concavity and Points of Inflection :

f ′′ (x) = 3x2 − 4 = 0⇒ x = ± 2√3(

−∞,− 2√3

) (− 2√

3, 2√

3

) (2√3,∞)

f ′′ + − +f ^ _ ^

Since the concavity changes at(− 2√

3, −119

)and

(2√3, −119

), these are inflection points.

H) Sketch the Curve :

71

­4 ­3 ­2 ­1 1 2 3 4

­4

­3

­2

­1

1

2

3

4

x

y

y = 14x

4 − 2x2 + 1

Example 4.6.2 Sketch the graph of f (x) = x2

1−x2Solution :A) Domain : dom (f) = R− {−1, 1}B) Intercepts :i-) x−intecept (y = 0) : f (x) = 0⇒ x = 0ii-) y−intercept (x = 0) : f (0) = 0C) Symmetryi) This function is even ( the function is symmetric with recpect to y−axis) since

f (−x) =(−x)

2

1− (−x)2 =

x2

1− x2 = f (x)

ii) This function is not odd since

f (−2) = −4

36= 4

3= −f (2) .

D) Asymptotes :i) Horizontal asymptotes :

limx→∞

f (x) = −1, limx→−∞

f (x) = −1⇒ y = −1 is the horizontal asymptote

i) Vertical asymptotes :

limx→−1−

f (x) = −∞, limx→−1+

f (x) =∞, limx→1−

f (x) =∞, limx→1+

f (x) = −∞

⇒ x = ±1 are the vertical asymptotes

E) Intervals of Increase or Decrease :

f ′ (x) =2x

(1− x2)2= 0⇒ x = 0

(−∞, 0) (0,∞)

f ′ − +f ↘ ↗

F) Local Maximum and Minimum Values :

(0, 0) local minimum and no local maximum

72

G) Concavity and Points of Inflection :

f ′′ (x) =2 + 6x2

(1− x2)36= 0

f ′′ (x) changes sign at x = ±1. But f is continuous at these points. Hence f has no inflection point.

(−∞,−1) (−1, 1) (1,∞)

f ′′ − + −f _ ^ _

H) Sketch the Curve :

­4 ­3 ­2 ­1 1 2 3 4

­4

­3

­2

­1

1

2

3

4

x

y

y = x2

1−x2

Example 4.6.3 Sketch the graph of f (x) = xe−x

Solution :A) Domain : dom (f) = RB) Intercepts :i-) x−intecept (y = 0) : f (x) = 0⇒ x = 0ii-) y−intercept (x = 0) : f (0) = 0C) Symmetryi) This function is not even since

f (−1) = −e 6= 1

e= f (1)

ii) This function is not odd since

f (−1) = −e 6= −1

e= −f (1)

D) Asymptotes :i) Horizontal asymptotes :

limx→∞

f (x) = 0, limx→−∞

f (x) = −∞⇒ y = 0 is the horizontal asymptote

i) Vertical asymptotes :No vertical asymptotes

E) Intervals of Increase or Decrease :

f ′ (x) = e−x − xe−x = e−x (1− x) = 0⇒ x = 1

73

(−∞, 1) (1,∞)

f ′ − +f ↘ ↗

F) Local Maximum and Minimum Values : (1, e−1

)is local maximum

G) Concavity and Points of Inflection :

f ′′ (x) = −e−x − (1− x) e−x = e−x (−2 + x) = 0⇒ x = 2

(−∞, 2) (2,∞)

f ′′ − +f _ ^(

2, 2e−2)is inflection point.

H) Sketch the Curve :

­4 ­3 ­2 ­1 1 2 3 4

­4

­3

­2

­1

1

2

3

4

x

y

y = xe−x

74

4.7 Optimization

Optimization problems are applications in which the desired answer is maximum or minimum of a function.

Example 4.7.1 Find the points on the curve y = x2 + 1 which is closest to (0, 2).Solution :

­1 0 1

1

2

3

x

y

y = x2 + 1

In this case we need to minimize the distance between the point (0, 2) and any point that is one the graph (x, y).

D =

√(x− 0)

2+ (y − 2)

2=

√x2 + (y − 2)

2

D (x) =

√x2 + (x3 + 1)

2

We can now substitute y = x2 + 1 into the first equation to express D as a function of one variable.

D (x) =

√x2 + (x2 + 1− 2)

2

=

√x2 + (x2 − 1)

2

=√x4 − x2 + 1

The critical points accur when D′ (x) = 0.

D′ (x) =2x3 − x√x4 − x2 + 1

= 0

x(2x2 − 1

)= 0

⇒ x = 0, x = ± 1√2

We can use the First Derivative Test . (−∞,− 1√

2

) (− 1√

2, 0) (

0, 1√2

) (1√2,∞)

x − − + +2x2 − 1 + − − +√x4 − x2 + 1 + + + +

f ′ − + − +f ↘ ↗ ↘ ↗

Therefore the minimum value of the distance must occur at x = ± 1√2and the distance is D

(± 1√

2

)=√32 . The corre-

sponding y-coordinates are

y = x2 + 1 =

(1√2

)2+ 1 =

3

2

y = x2 + 1 =

(− 1√

2

)2+ 1 =

3

2

Thus, the points are(− 1√

2, 32

)and

(1√2, 32

).

75

Example 4.7.2 A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equalsquares from each corner and then folding up the sides. Find the length of the side of the square that must be cut o if thevolume of the box is to be maximized. What is the maximum volume?Solution : Let the length of the side of the square to be cut be x cm. Then the base of the box is a square with each sideequals to (18− 2x) cm. Hence we have 0 < x < 9. The volume V , in cm3, of the open box is

V (x) = x (18− 2x)2, 0 < x < 9

V ′ (x) = 12x2 − 144x+ 324

Solving V ′(x) = 0, that is

12x2 − 144x+ 324 = 0

12 (x− 9) (x− 3) = 0

⇒ x = 3, x = 9

we get the critical number of V, x = 3 (note that 0 < x < 9).

V ′′ (x) = 24x− 144

V ′′ (3) = −72

From the Second Derivative Test, at x = 3 V has a maximum value and the maximum volume is V (3) = 432cm3.

Example 4.7.3 We have a rectangle which one corner is on the parabole 3− x2 and other corners are on origin, x−axesand y−axes. What is the maximum area of this rectangle.Solution :

x

y

AO

C B

Let |OA| = x. Then |AB| = 3− x2 and area of the rectangle OABC

A (x) = x(3− x2

)= 3x− x3

is a function of x. Now find its critical points.

A′ (x) = 3− 3x2

⇒ x = ±1

Since the area is a positive number, x = 1 and A (1) = 2.

Example 4.7.4 Find the height of the cylinder that can be put into a sphere with radius 6 cm and has a maximum volume.Solution : h2 + r2 = 62 ⇒ r2 = 36− h2

V = π.r2.2h

V (h) = π(36− h2

)2h

= π(72h− 2h3

)V ′ (h) = π

(72− 6h2

)= 0

⇒ h = ±2√

3

Hence h = 2√

3 and 2h = 4√

3.

76

Example 4.7.5 If the polynomial p (x) = 3x4 + ax3 + bx2 + c is divided by (x− 1)3, c =?

Solution : Since p (x) is divided by (x− 1)3, we can write p (x) = (x− 1)

3.q (x).

p (x) = (x− 1)3.q (x)

p′ (x) = 3 (x− 1)2.q (x) + (x− 1)

3.q′ (x)

p′′ (x) = 6 (x− 1) .q (x) + 3 (x− 1)2.q′ (x) + 3 (x− 1)

2.q′ (x) + (x− 1)

3.q′′ (x)

Here we observe thatp (1) = p′ (1) = p′′ (1) = 0.

which give us a linear system we need to solve

p (x) = 3x4 + ax3 + bx2 + c⇒ p (1) = 0⇒ 3 + a+ b+ c = 0

p′ (x) = 12x3 + 3ax2 + 2xb⇒ p′ (1) = 0⇒ 12 + 3a+ 2b = 0

p′′ (x) = 36x2 + 6ax+ 2b⇒ p′′ (1) = 0⇒ 36 + 6a+ 2b = 0

3 + a+ b+ c = 012 + 3a+ 2b = 036 + 6a+ 2b = 0

⇒ a = −8, b = 6, c = −1

77

4.8 Linear Approximations

Given a function, f (x), we can find its tangent at x = a. The equation of the tangent line, which we’ll call L (x), is

L (x) = f (a) + f ′ (a) (x− a) .

x

y

From this graph we can see that near x = a the tangent line and the function have nearly the same graph. On occasionwe will use the tangent line, L (x), as an approximation to the function, f (x), near x = a. In these cases we call thetangent line the linear approximation to the function at x = a.

Example 4.8.1 Find the linearization of the function f(x) =√x at 4 and use it to approximate the numbers

√3, 98 and√

4, 05.Solution :

f(x) =√x⇒ f ′(x) =

1

2√x

L (x) = f (a) + f ′ (a) (x− a)

= 2 +1

4(x− 4)

= 1 +x

4

The corresponding linear approximation is √x ≈ 1 +

x

4

when x is near 4. In particular, we have√

3.98 ≈ 1 +3.98

4= 1.995

and √4.05 ≈ 1 +

4.05

4= 2.0125

­4 ­2 2 4 6 8

1

2

3

x

y

Example 4.8.2 Find the linearization of the function f(x) = 3√x+ 1 at 7 and use it to approximate the value of

√7.98

and√

8.05.Solution :

f(x) = 3√x+ 1⇒ f ′(x) =

1

33

√(x+ 1)

2

L (x) = f (a) + f ′ (a) (x− a)

= 2 +1

12(x− 7)

78

The corresponding linear approximation is3√x+ 1 ≈ 2 +

1

12(x− 7)

when x is near 8. In particular, we have

3√

7.98 = 3√

6.98 + 1 ≈ 2 +1

12(6.98− 7) = 1.9983

and3√

8.05 = 3√

7.05 + 1 ≈ 2 +1

12(7.05− 7) = 2.0041

­20 20

­2

2

4

x

y

Example 4.8.3 Find the linearization of the function f(x) = cosx at 0 and use it to approximate the numbers cos (−0, 1)and cos (0, 1) .Solution :

f(x) = cosx⇒ f ′(x) = − sinx

L (x) = f (a) + f ′ (a) (x− a)

= 1 + 0 (x− 0)

= 1

The corresponding linear approximation iscosx ≈ 1

when x is near 0. In particular, we havecos (−0.1) ≈ 1

andcos (0.1) ≈ 1

­2 2

­1

1

x

y

79

5 Integral

There are two important integral symbols in integral calculus.

∫f (x) dx and

b∫a

f (x) dx

Although they look like each other, their definition and meaning totally different.

Symbol Definition Meaning Result Callb∫a

f (x) dx limn→∞

n∑i=1

f (x∗i ) ∆xiArea under the graph of ffrom a to b

Scalar Definite integral∫f (x) dx F (x) + c

Antiderivative of fi.e., F ′ (x) = f (x)

Function Indefinite integral

In this section, we will investigate these two concepts and their relation.

80

5.1 Definite Integral. Area Problem

Motivation :

Example 5.1.1 Find the area under the graph of y = f (x) = 3 from 0 to 2.

A=6

0 1 20

1

2

3

x

y

y = 3

Example 5.1.2 Find the area under the graph of y = f (x) = 12x from 0 to 2.

A=10 1 2

0.0

0.5

1.0

x

y

y = 12x

Example 5.1.3 Find the area under the graph of y = f (x) = x2 from 0 to 1.

A=?

0.0 0.5 1.00.0

0.5

1.0

x

y

y = x2

This is not easy as others. At first, we will find the area approximately. General idea for approximation is that divide[0, 1] to n equal subintervals to obtain rectangular strips. The sum of the rectangular areas is an approximation to area A.Let start with n = 2 :

x0 = 0 x1 = 12 x2 = 1

We have different choices to obtain rectangular strips. If we use the right side values of subintervals to find the height of

81

rectangles, then the method is called Right Boxes.

0.0 0.5 1.00.0

0.5

1.0

x

y

Right Boxes Method

0.0 0.5 1.00.0

0.5

1.0

x

y

Left Boxes Method

0.0 0.5 1.00.0

0.5

1.0

x

y

Midpoint Method

Area≈2∑i=1

f (xi)(12

)= 5

8 Area≈2∑i=1

f (xi−1)(12

)= 1

8 Area≈2∑i=1

f(xi−1+xi

2

) (12

)= 5

16

For n = 5 :x0 = 0 x1 = 1

5 x2 = 25 x3 = 3

5 x4 = 45 x5 = 1

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

x

y

Right Boxes Method

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

x

y

Left Boxes Method

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

x

y

Midpoint Method

Area≈5∑i=1

f (xi)(15

)= 11

25 Area≈5∑i=1

f (xi−1)(15

)= 6

25 Area≈5∑i=1

f(xi−1+xi

2

) (15

)= 33

100

For n = 10 :x0=0 x1=

110

x2=210

x3=310

x4=410

x5=510

x6=610

x7=710

x8=810

x9=910

x10=1

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

x

y

Right Boxes Method

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

x

y

Left Boxes Method

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

x

y

Midpoint Method

Area≈10∑i=1

f (xi)(110

)= 77

200 Area≈10∑i=1

f (xi−1)(110

)= 57

200 Area≈10∑i=1

f(xi−1+xi

2

) (110

)= 133

400

As it seen from the graphs, with larger n, we get skinner rectangles and it gives better approximations.For arbitrary n :

x0 = 0 x1 = 1n · · · xi = i

n · · · xn−1 = n−1n xn = 1

82

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

x

y

Right Boxes Method

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

x

y

Left Boxes Method

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

x

y

Midpoint Method

Area ≈n∑i=1

f (xi)(1n

)=

n∑i=1

(in

)2 ( 1n

)= 1

n3

n∑i=1

i2

= 2+3/n+2/n2

6

Area ≈n∑i=1

f (xi−1)(1n

)=

n∑i=1

(i−1n

)2 ( 1n

)= 1

n3

n∑i=1

(i− 1)2

= 2−3/n+2/n26

Area ≈n∑i=1

f(xi−1+xi

2

) (1n

)=

n∑i=1

(2i−1n

)2 ( 1n

)= 1

4n3

n∑i=1

(2i− 1)2

= 4n3−n212n3

Limit of these approximations as n → ∞ (that means as the number of rectangles → ∞ and their widths → 0) givesprecisely the area we want.

Right Boxes Method Left Boxes Method Midpoint Method

limn→∞

2+3/n+2/n2

6 limn→∞

2−3/n+2/n26 lim

n→∞4n3−n212n3

1/3 1/3 1/3

The area under the graph of f (x) = x2 from 0 to 1 is precisely 1/3 and as seen the result does not depend on the samplepoint we choose in the subintervals.General Construction :Let f (x) be continuous and non-negative ( f (x) ≥ 0 ) on [a, b].

a b

A

x

y

To compute the area A under the graph of f from a to b, we proceed as follows :

1. Subdivide the interval [a, b] into n subintervals with endpoints

a = x0 < x1 < x2 < . . . < i < . . . < xn−2 < xn−1 < xn = b.

For each i, let

∆xi = xi − xi−1= length of [xi−1, xi] .

2. Select a point x∗i ∈ [xi−1, xi] for each i. Then evaluate

f (x∗1) ∆x1, f (x∗2) ∆x2, . . . , f (x∗n) ∆xn.

These values give us the area of the rectangles.

83

3. The area A is approximated by the sum of the areas of rectangles:

A ≈ f (x∗1) ∆x1 + f (x∗2) ∆x2 + . . .+ f (x∗n) ∆xn

≈n∑i=1

f (x∗i ) ∆xi

This sum is called Riemann sum.

4. Take the limit

A = limn→∞

n∑i=1

f (x∗i ) ∆xi.

As long as f is continuous the value of the limits is independent of the sample points x∗i we used.

That limit is represented byb∫a

f (x) dx and called definite integral from a to b.

Evaluating an integral directly from the definition is a hard work as seen in the example. We will soon study simpleand effi cient methods to evaluate integrals.Properties of the Definite Integral

1. Integral of a constant :b∫a

cdx = c (b− a)

2. Linearity :

b∫a

[f (x) + g (x)] dx =

b∫a

f (x) dx+

b∫a

g (x) dx

b∫a

kf (x) dx = k

b∫a

f (x) dx

3. Interval Additivity :

b∫a

f (x) dx+

c∫b

f (x) dx =

c∫a

f (x) dx

b∫a

f (x) dx = −a∫b

f (x) dx (Def.)

a∫a

f (x) dx = 0

4. Comparison :

f (x) ≥ 0⇒b∫a

f (x) dx ≥ 0

f (x) ≥ g (x)⇒b∫a

f (x) dx ≥b∫a

g (x) dx

m ≤ f (x) ≤M ⇒ m (b− a) ≤b∫a

f (x) dx ≤M (b− a)

84

5.2 Indefinite Integrals

Given a function f (x) on some interval I. An antiderivative of f (x) on I is another function F (x) satisfying

F ′ (x) = f (x)

for all x ∈ I.For example, if f (x) = 3x2, then all of the followings are antiderivatives for f (x) on (−∞,∞) :

x3, x3 + 3, x3 +π

3.

Let F (x) and G (x) be two antiderivatives of f (x) on I ( G (x)′

= f (x) , F (x)′

= f (x) ). Since

(G (x)− F (x))′

= G (x)′ − F (x)

′

= f (x)− f (x)

= 0

, G (x)− F (x) must be constant, say c. ThenG (x) = F (x) + c.

Hence if F (x) is an antiderivative of f (x), then any antiderivatives of f (x) is in the form F (x)+ c where c is an arbitraryconstant. All family of antiderivatives of f (x) is called indefinite integral and represented by∫

f (x) dx = F (x) + c.

Example 5.2.1∫

cosxdx = sinx+ c(ddx (sinx) = cosx

).

Example 5.2.2∫

1√1−x2 dx = arcsinx+ c

(ddx (arcsinx) = 1√

1−x2

).

Example 5.2.3∫s2ds = s3

3 + c(dds

(s3

3

)= s2

).

Example 5.2.4∫etdt = et + c

(ddt (et) = et

).

We can make an integral table just by reversing a table of derivatives.

1.∫

0dx = c

2.∫xndx = xn+1

n+1 + c (n 6= 1)

3.∫1xdx = ln |x|+ c

4.∫exdx = ex + c

5.∫axdx = ax

ln a + c

6.∫

sinxdx = − cosx+ c

7.∫

cosxdx = sinx+ c

8.∫

1sin2 x

dx =∫

csc2 xdx = − cotx+ c (u 6= kπ, k ∈ Z)

9.∫

1cos2 xdx =

∫sec2 xdx = tanx+ c

(u 6= π

2 + kπ, k ∈ Z)

10.∫

tanxdx = ln | secx|+ c

11.∫

tan2 xdx = tanx− x+ c

12.∫

cotxdx = 12 ln (2− 2 cos 2x) + c

13.∫

cot2 xdx = 1cos 2x−1 (x+ sin 2x− x cos 2x) + c

14.∫

secx tanxdx = secx+ c

15.∫

cscx cotxdx = − cscx+ c

85

16.∫

1a2+x2 dx = 1

a arctan xa + c

17.∫

1a2−x2 dx = − 1

2a ln∣∣∣a+xa−x

∣∣∣+ c

18.∫

1√a2−x2 dx = arcsin x

a + c

19.∫

1√x2±a2 dx = ln

∣∣x+√x2 ± a2

∣∣+ c

20.∫

1x√1−x2 dx = arcsec |x|+ c

Since (F (x) +G (x))′

= F ′ (x) +G′ (x) and (cF (x))′

= cF ′ (x), indefinite integral satisfy following properties :

1.∫

(f (x) + g (x)) dx =∫f (x) dx+

∫g (x) dx.

2.∫cf (x) dx = c

∫f (x) dx.

86

5.3 Fundamental Theorems of Calculus

The Fundamental Theorem of Calculus (FTC) connects the two branches of calculus: differential calculus and integralcalculus.Motivation :Think about the function f (x) = 1

2x on the interval [0, 1]. For any x ∈ [0, 1], let A (x) =x∫0

f (t) dt ( area under the

graph of f (t) from 0 to x.)

A(x)

x

f(x)

y = 12x

A (x) = xf (x) =1

2x

(1

2x

)=

1

4x2

Notice thatd

dxA (x) =

d

dx

(1

4x2 + c

)=

1

2x = f (x)

that meansd

dx

x∫0

f (t) dt = f (x) .

Derivative of the area function A (x) under the graph of f (x) from 0 to x is f (x). Following theorem says that it happensall the time.

Theorem 5.3.1 (First Fundamental Theorem of Calculus) Suppose f is continuous on the interval I, a, x ∈ I anda ≤ x. Then:

d

dx

x∫a

f (t) dt = f (x)

i.e.,

F (x) =

x∫a

f (t) dt

is an antiderivative for f . ( F ′ (x) = f (x) ).

This theorem says that indefinite integral of f can be obtained with the help of definite integral of f . Sincex∫a

f (t) dt

is an antiderivative of f , we have ∫f (x) dx =

x∫a

f (t) dt+ c.

Example 5.3.2 ddx

x∫1

cos tdt = cosx.

Example 5.3.3 ddx

x∫5

cos tdt = cosx.

87

Example 5.3.4 Let F (x) =x∫1

ln tdt. Find F (1) , F ′(e2)and F ′′ (2) .

F (1) =

x∫1

ln tdt = F (x) =

1∫1

ln tdt = 0

F ′(e2)

=d

dx

x∫1

ln tdt

∣∣∣∣∣∣x=e2

= (lnx)|e2 = 2

F ′′ (2) =d

dx(lnx)|2 =

1

x

∣∣∣∣x=2

=1

2

Example 5.3.5 ddx

1∫x

sin tt dt = d

dx

(−

x∫1

sin tt dt

)= − d

dx

x∫1

sin tt dt = − sin xx

Example 5.3.6 ddx

x2∫1

ln tdt =? If we write F (x) =x∫1

ln tdt, and g (x) = x2 then the question is

d

dx(F (g (x))) =?

or equivalentlyF ′ (g (x)) g′ (x) =?

.

F ′ (x) =d

dx

x∫1

ln tdt = ln |x|

g′ (x) = 2x

we haveF ′ (g (x)) g′ (x) = ln |x2|.2x.

This example is an illustration of following rule:Chain Rule :

d

dx

u(x)∫v(x)

f (t) dt = f (u (x)) .u′ (x)− f (v (x)) .v′ (x) .

Example 5.3.7 ddx

x3∫1

t2dt =(x3)2 (

x3)′

= x63x2 = 3x8.

Example 5.3.8 ddx

0∫e2x

sin2 tdt = − sin2(e2x).(e2x)′

= − sin2(e2x)

2.ex.

Example 5.3.9 ddx

2x∫cos x

cos(t2)dt = cos

((2x)

2). (2x)

′ − cos(

(cosx)2). (cosx)

′= 2 cos

(4x2)

+ sinx. cos(

(cosx)2).

Suppose f (x) is a continuous on [a, b]. Then 1.FTC tells us thatx∫a

f (t) dt is an antiderivative for f (x) on [a, b]. Any

other antiderivative F (x) for f (x) on [a, b] differs fromx∫a

f (t) dt by some constant :

F (x) =

x∫a

f (t) dt+ c.

88

Now notice that

F (b)− F (a) =

b∫a

f (t) dt+ c

− a∫a

f (t) dt+ c

=

b∫a

f (t) dt+ c− (0 + c)

=

b∫a

f (t) dt.

Hence we proved the following theorem.

Theorem 5.3.10 (Second Fundamental Theorem of Calculus) If f (x) is a continuous function and F (x) is anantiderivative of f (x) on [a, b], then

b∫a

f (x) dx = F (b)− F (a)

= F (x)|ba

This theorem gives us an effi cient way to find the area under the graph of function f (x) from a to b. With the helpof indefinite integral of f , we can calculate the definite integral of f .

Example 5.3.11 In the Example 5.1.1, we calculate the1∫0

x2dx by using the definition of Definite Integral and it was

hard even if the function is not complicated. Here we use 2. FTC to evaluate this integral

1∫0

x2dx =x3

3

∣∣∣∣10

=1

3− 0 =

1

3

Example 5.3.122π∫0

cosxdx = sinx|2π0 = sin (2π)− sin 0 = 0− 0 = 0.

Example 5.3.1312∫− 12

1√1−x2 dx = arcsinx|

12

− 12

= arcsin 12 − arcsin

(− 12)

= π6 −

(−π6)

= π3 .

Example 5.3.14π4∫0

tanxdx = ln | secx||π40 = ln | sec π

4 | − ln | sec 0| = ln∣∣√2

∣∣− ln |1| = 12 ln 2.

Example 5.3.15−1∫−e

1xdx = ln |x||−1−e = ln | − 1| − ln | − e| = 0− 1 = −1. (What does it mean!!!)

Example 5.3.16 Find the are under the graph of f (x) = 11+x2 from −1 to 1.

­5 ­4 ­3 ­2 ­1 0 1 2 3 4 5

0.5

1.0

x

y

y = 11+x2

89

1∫−1

1

1 + x2dx = arctanx| 1

−1 = arctan 1− arctan (−1)

=π

4−(−π

4

)=π

2.

MEAN VALUE THEOREM FOR INTEGRALS :The average value of finitely many numbers y1, y2, . . . , yn is defined as

yaverage =y1 + y2 + . . .+ yn

n.

The average value has the property that if each of the numbers y1, y2, . . . , yn is replaced by yaverage, their sum remainsthe same :

y1 + y2 + . . .+ yn =

n times︷ ︸︸ ︷yaverage + yaverage + · · ·+ yaverage

Recall the Mean Value Theorem for derivative :Let y = f (x) be a continuous on [a, b] and differentiable on (a, b), then there exists a number c in [a, b] such that

f ′ (c) =f (b)− f (a)

b− a .

There is a similar mean value theorem for integrals.

Theorem 5.3.17 (Mean Value Theorem for Integral) If f is continuous on [a, b], then there exists a number c in[a, b] such that

f (c) =

∫ baf (x) dx

b− a ,

i.e., ∫ b

a

f (x) dx = f (c) (b− a) =

∫ b

a

f (c) dx.

x

y

a b

A

c

f(c)

Hence the Mean Value Theorem for integral says that if f(x) in the integral is replaced by the average value of f(x) in theinterval [a, b], then the integral remains the same.

Example 5.3.18 What is the average value f (c) of y = f (x) = sinx on the interval [0, π]?Solution : We have

f (c) =1

b− a

∫ b

a

f (x) dx

=1

π − 0

∫ π

0

sinxdx

= − 1

πcosx|π0

= − 1

π(−1− 1)

=2

π

and

c = arcsin

(2

π

).

90

x

y

c

f(c)

2*Pi

y = sinx

Example 5.3.19 What is the average value f (c) of y = f (x) = x2 on the interval [1, 4]?Solution : We have

f (c) =1

b− a

∫ b

a

f (x) dx

=1

4− 1

∫ 4

1

x2dx

=1

3

x3

3

∣∣∣∣41

=1

9

(43 − 13

)= 7

andc =√

7.

0 1 2 3 40

2

4

6

8

10

12

14

16

x

y

c

f(c)

y = x2

91

5.4 Substitution Rule

The substitution rule is a trick for evaluating integrals. It is based on the following identity between differentials whereu = g (x) :

du = g′ (x) dx = u′dx.

Hence we can write ∫f (u)u′dx =

∫f (u) du

or using a slightly different notation : ∫f (g (x)) g′ (x) dx =

∫f (u) du.

Only problem in using this method of integration is finding the right substitution. We need to look at the integrandas a function of some expression (which we will later identify with u) multiplied by the derivative of that expression.

Example 5.4.1 Find∫

cos(x2)

2xdx.Solution : u = x2 ⇒ du = 2xdx ∫

cos(x2)

2xdx =

∫cosudu

= sinu+ c

= sin(x2)

+ c.

Example 5.4.2 Find∫xe−x

2

dx.Solution : t = −x2 ⇒ dt = −2xdx⇒ xdx = − 12dt∫

xe−x2

dx =

∫et(−1

2

)dt

= −1

2

∫etdt

= −1

2et + c

= −1

2e−x

2

+ c.

Example 5.4.3 Evaluate∫

1x+2dx.

Solution : v = x+ 2⇒ dv = dx ∫1

x+ 2dx =

∫1

vdv = ln v + c

= ln (x+ 2) + c.

Example 5.4.4 Evaluate∫

1(x+2)2

dx.

Solution : u = x+ 2⇒ du = dx ∫1

(x+ 2)2 dx =

∫1

u2du =

∫u−2du

=u−1

−14c

= − (x+ 2)−1

+ c.

Example 5.4.5 Find∫ (

3x2 + 1)10

xdx.Solution : u = 3x2 + 1⇒ du = 6xdx⇒ xdx = 1

6du∫(3x+ 1)

10xdx =

∫u10

1

6du

=1

6

∫u10du

=1

6

u11

11+ c

=

(3x2 + 1

)1166

+ c.

92

Example 5.4.6 Evaluate∫

x3

x2+4dx.

Solution : u = x2 + 4⇒ du = 2xdx⇒ x3dx = x2xdx = (u− 4) 12du∫x3

x2 + 4dx =

∫u− 4

u

1

2du =

1

2

∫ (1− 4

u

)du

=1

2

(∫du− 4

∫1

udu

)=

1

2(u− k ln |u|) + c

=1

2

(x2 + 4− 4 ln |x2 + 4|

)+ c.

Example 5.4.7 Evaluate∫

sin x1−cos xdx.

Solution : u = 1− cosx⇒ du = sinxdx∫sinx

1− cosxdx =

∫1

udu = ln |u|+ c

= ln |1− cosx|+ c.

Sometimes the substitution is hard to see until we make some ingenious transformation in the integrand.

Example 5.4.8 Evaluate∫ √

1− x2dx.Solution : x = sin t⇒ dx = cos tdt ∫ √

1− x2dx =

∫ √1− sin2 t cos tdt

=

∫cos t cos tdt

=

∫cos2 tdt.

Since we do not know yet how to integrate cos2 t, we leave it like this and will be back to it later (after we study integralsof trigonometric functions).

When computing a definite integral using the substitution rule there are two possibilities:

1. Compute the indefinite integral first, then use the 2.FTC :∫f (u)u′dx = F (x)

b∫a

f (u)u′dx = F (b)− F (a) .

2. Use the substitution rule for definite integrals:

b∫a

f (u)u′dx =

u(b)∫u(a)

f (u) du.

The advantage of the second method is that we do not need to undo the substitution.

Example 5.4.9 Find4∫0

√2x+ 1dx.

Solution : Using the first method, we first compute the indefinite integral :u = 2x+ 1⇒ du = 2dx⇒ dx = 1

2du ∫ √2x+ 1dx =

∫ √u

1

2du =

1

2

∫u12 du

=1

2

u32

32

+ c =1

3u32 + c

=1

3(2x+ 1)

32 + c

93

Then we use it for computing the definite integral :

4∫0

√2x+ 1dx =

[1

3(2x+ 1)

32

]∣∣∣∣40

=1

3932 − 1

3132

=26

3

In the second method we compute the definite integral directly adjusting the limits of integration after the substitution :u = 2x+ 1⇒ du = 2dx⇒ dx = 1

2du, u (0) = 1 and u (4) = 9∫ 4

0

√2x+ 1dx =

∫ 9

1

√u

1

2du =

1

2

∫ 9

1

u12 du

=

(1

2

u32

32

)∣∣∣∣∣9

1

=

(1

3u32

)∣∣∣∣91

=1

3932 − 1

3132

=26

3

Example 5.4.10 Evaluate∫ 53

x(30−x2)2 dx.

Solution : u = 30− x2 ⇒ du = −2xdx⇒ xdx = − 12du,∫x

(30− x2)2dx =

∫1

u2

(−1

2

)du

= −1

2

∫u−2du = −1

2

(u−1

−1

)+ c =

1

2u+ c

=1

2 (30− x2) + c

∫ 5

3

x

(30− x2)2dx =

1

2 (30− x2)

∣∣∣∣53

=1

10− 1

42

=8

105

Example 5.4.11 Find∫ 10

√exdx. =

Solution : ∫ 1

0

√exdx =

∫ 1

0

ex2 dx

u = x2 ⇒ du = 1

2dx⇒ dx = 2du, u (0) = 0 and u (1) = 12∫ 1

0

ex2 dx =

∫ 12

0

eu2du = 2

∫ 12

0

eudu

= 2 (eu)|120 = 2

(e12 − e0

)= 2

(√e− 1

).

Example 5.4.12 Evaluate∫ 90

√1 +√xdx.

Solution :u =√

1 +√x⇒ u2 = 1 +

√x⇒ x =

(u2 − 1

)2= u4 − 2u2 + 1

94

⇒ dx =(4u3 − 4u

)du, ∫ √

1 +√xdx =

∫u(4u3 − 4u

)du

= 4

∫ (u4 − u2

)du

= 4

(u5

5− u3

3

)+ c

= 4

(√

1 +√x)5

5−

(√1 +√x)3

3

+ c

∫ 9

0

√1 +√xdx = 4

(√

1 +√x)5

5−

(√1 +√x)3

3

∣∣∣∣∣∣∣9

0

=232

15

95

5.5 Integration by Parts

The method of integration by parts is based on the product rule for differentiation :

[f (x) g (x)]′

= f ′ (x) g (x) + f (x) g′ (x)

which we can write like this:f (x) g′ (x) = [f (x) g (x)]

′ − f ′ (x) g (x) .

By integrating we get: ∫f (x) g′ (x) dx =

∫[f (x) g (x)]

′dx−

∫f ′ (x) g (x) dx

i.e.: ∫f (x) g′ (x) dx = f (x) g (x)−

∫f ′ (x) g (x) dx

Writing u = f (x) , v = g (x) , we have du = f ′ (x) dx, dv = g′ (x) dx, hence∫udv = uv −

∫vdu.

Remark 5.5.1 Remember "L(og)A(rc...)P(oly)T(rig)E(xp)" rule for the choice of u.

Example 5.5.2 Evaluate∫xe6xdx.

Solution :u = x dv = e6xdxdu = dx v =

∫e6xdx = 1

6e6x

∫udv = uv −

∫vdu∫

xe6xdx = x

(1

6e6x)−∫

1

6e6xdx

=1

6xe6x − 1

6

1

6e6x + c.

Example 5.5.3 Compute∫ e1

lnxdx.Solution : First we evaluate

∫lnxdx.

u = lnx dv = dxdu = 1

xdx v =∫dx = x

∫udv = uv −

∫vdu∫

lnxdx = ln (x) .x−∫x

1

xdx

= x lnx− x+ c.

Then ∫ e

1

lnxdx = (x lnx− x)|e1= (e ln e− e)− (ln 1− 1)

= 1

Sometimes we need to combine the methods to evolute integrals.

Example 5.5.4 Evaluate∫

sin (√x) dx

Solution : This one is tricky. First we make substitution.t =√x⇒ dt = 1

2√xdx⇒ dx = 2

√xdt = 2tdt∫

sin√xdx =

∫sin (t) .2tdt = 2

∫t sin tdt.

96

Now, integration by partsu = t dv = sin tdtdu = dt v =

∫sin tdt = − cos t∫

udv = uv −∫vdu

2

∫t sin tdt = 2

(−t cos t−

∫− cos tdt

)= 2 (−t cos t+ sin t) + c.

= −2√x cos

√x+ 2 sin

√x+ c.

Example 5.5.5 Compute∫ex cosxdx.

Solution :u = cosx dv = exdxdu = − sinxdx v =

∫exdx = ex∫

udv = uv −∫vdu∫

ex cosxdx = ex cosx+

∫ex sinxdx (2)

As it seen, it didn’t help much. However, watch what happens when the integration by parts formula is applied to∫ex sinxdx.

u = sinx dv = exdxdu = cosxdx v =

∫exdx = ex∫

udv = uv −∫vdu∫

ex sinxdx = ex sinx−∫ex cosxdx (3)

From (2) and (3), we have ∫ex cosxdx = ex cosx+ ex sinx−

∫ex cosxdx

and ∫ex cosxdx =

1

2ex (sinx+ cosx) + c.

Example 5.5.6 Compute∫x2exdx.

Solution :u = x2 dv = exdxdu = 2xdx v =

∫exdx = ex∫

udv = uv −∫vdu∫

x2exdx = x2ex −∫

2xexdx (4)

We need to apply the integration by parts formula to∫xexdx.

u = x dv = exdxdu = dx v =

∫exdx = ex∫

udv = uv −∫vdu∫

xexdx = xex −∫exdx

= xex − ex + c. (5)

From (4) and (5), we have ∫x2exdx = x2ex − 2xex + 2ex + c.

97

Example 5.5.7 Evaluate∫

arctanxdx.Solution :

u = arctanx dv = dxdu = 1

1+x2 dx v =∫dx = x

∫udv = uv −

∫vdu∫

arctanxdx = x. arctanx−∫

x

1 + x2dx

To evaluate∫

x1+x2 dx, we use substitution rule.

t = 1 + x2 ⇒ dt = 2xdx⇒ xdx = 12dt ∫

x

1 + x2dx =

1

2

∫1

tdt =

1

2ln |t|.

=1

2ln |1 + x2|

Hence we have ∫arctanxdx = x. arctanx− 1

2ln |1 + x2|+ c.

98

5.6 Trigonometric Integrals and Trigonometric Substitution

5.6.1 Trigonometric Integrals

Following equations will help us to evaluate the integrals involving trigonometric expression

cos2 x+ sin2 x = 1

1 + tan2 x = sec2 x

1 + cot2 x = csc2 x

cos2 x =1

2(1 + cos 2x)

sin2 x =1

2(1− cos 2x) .

Example 5.6.1 Find∫

secxdx.Solution : ∫

secxdx =

∫secx

secx+ tanx

secx+ tanxdx

=

∫sec2 x+ secx tanx

secx+ tanxdx

u = secx+ tanx⇒ du =(secx tanx+ sec2 x

)dx∫1

udu = ln |u|+ c

= ln | secx+ tanx|+ c∫cosn x sinxdx or

∫sinn x cosxdx where n is an integer :

Example 5.6.2 Evaluate∫

sin3 x cosxdx =Solution : u = sinx⇒ du = cosx ∫

sin3 x cosx =

∫u3du =

u4

4+ c

=sin4 x

4+ c∫

cosn x sinm xdx where n or m is an odd number : We can use the identity sin2 x + cos2 x = 1 to transform theintegrand into an expression containing only one sine or one cosine.

Example 5.6.3 Evaluate∫

sin6 x cos3 xdx.Solution : ∫

sin6 x cos3 xdx =

∫sin6 x cos2 x cosxdx

=

∫sin6 x.

(1− sin2 x

)cosxdx

u = sinx⇒ du = cosxdx ∫sin6 x cos3 xdx =

∫u6(1− u2

)du =

∫ (u6 − u8

)du

=u7

7− u9

9+ c

=sin7 x

7− sin9 x

9+ c∫

cosn dx or∫

sinn xdx where n is an odd number : In that case, we separate one factor and convert the rest into theother trigonometric function, using the rule sin2 x+ cos2 x = 1.

99

Example 5.6.4 :Evaluate∫

cos3 xdx.Solution : ∫

cos3 xdx =

∫cos2 x cosxdx

=

∫(1− sinx)

2cosxdx

u = sinx⇒ du = cosxdx ∫cos3 xdx =

∫(1− u)

2du

=

∫ (1− 2u+ u2

)du

= u− u2 +u3

3+ c

= sinx− sin2 x+sin3 x

3+ c∫

cosn dx or∫

sinn xdx where n is an even number : If the exponent is even then we use the half-angle identities.

Example 5.6.5 Evaluate∫

cos4 xdx.Solution : ∫

cos4 xdx =

∫ (cos2 x

)2dx =

∫ (1 + cos 2x

2

)2dx

=1

4

∫ (1 + 2 cos 2x+ cos2 2x

)dx

=1

4

(x+ sin 2x+

∫ (1 + cos 4x

2

)dx

)=

1

4

(x+ sin 2x+

x

2+

1

2

sin 4x

4

)+ c

=3x

8+

sin 2x

4+

sin 4x

32+ c∫

cosn x sinm xdx where n and m are even numbers : If all the exponents are even then we use the half-angle identities.

Example 5.6.6 Evaluate∫

sin2 x cos2 xdx.Solution : ∫

sin2 x cos2 xdx =

∫ (1− cos 2x

2

)(1 + cos 2x

2

)dx

=1

4

∫ (1− cos2 2x

)dx

=1

4

∫sin2 2xdx

=1

4

∫ (1− cos 4x

2

)dx

=1

8

(x− sin 4x

4

)=

x

8− sin 4x

32+ c

Some Other Trigonometric Integrals : To compute integrals of the form∫

cosmx sinnxdx,∫

sinmx sinnxdx and∫cosmx cosnxdx, we will use the following identities :

sin a cos b =1

2[sin (a+ b) + sin (a− b)]

cos a cos b =1

2[cos (a+ b) + cos (a− b)]

sin a sin b =1

2[cos (a− b)− cos (a+ b)] .

100

Example 5.6.7 Evaluate∫

sin 7x cos 3xdx.Solution : By using

sin a cos b =1

2[sin (a− b) + sin (a+ b)] .

, we have

sin 7x cos 3x =1

2sin 4x+ sin 10x

and ∫sin 7x cos 3xdx =

∫1

2(sin 4x+ sin 10x) dx

=1

2

(−cos 4x

4− cos 10x

10

)+ c

= −cos 4x

8− cos 10x

20+ c.

5.6.2 Trigonometric Substitution

Here we study substitutions of the form x = some trigonometric function. The following substitutions are useful inintegrals containing the following expressions :

Expression Substitution Identity New Expression

a2 − x2 x = a. sin tdx = a. cos tdt

1− sin2t = cos2 t a2 − x2 = a2 cos2 t

a2 + x2x = a. tan tdx = a. sec2 tdt

1 + tan2t = sec2 t a2 + x2 = a2 sec2 t

x2 − a2 x = sec tdx = a tan t sec tdt

sec2t− 1 = tan2 t x2 − a2 = a2 tan2 t

Example 5.6.8 Evaluate∫

x3√9−x2 dx.

Solution :x = 3 sin t⇒ dx = 3 cos tdt∫x3√

9− x2dx =

∫27 sin3 t√

9 cos2 t3 cos tdt

= 27

∫sin3 t

cos tcos tdt = 27

∫sin3 tdt

= 27

∫sin2 t sin tdt = 27

∫ (1− cos2 t

)sin tdt

u = cos t⇒ du = − sin tdt⇒ sin tdt = −du∫x3√

9− x2dx = −27

∫ (1− u2

)du

= −27

(u− u3

3

)+ c

= −27

(cos t− cos3 t

3

)+ c

= −27

√9− x23

−

(√9−x23

)33

+ c.

t

x3 sin t = x3

cos t =√9−x23

101

Example 5.6.9 Evaluate∫

1

(4+x2)32dx.

Solution : x = 2 tan t⇒ dx = 2 sec2 tdt∫1

(4 + x2)32

dx =

∫1

(4 sec2 t)32

2 sec2 tdt

=

∫1

8 sec3 t2 sec2 tdt

=1

4

∫1

sec tdt

=1

4

∫cos tdt

=1

4sin t+ c

=1

4

x√x2 + 4

+ c

t

x

2

tan t = x2

sin t = x√x2+4

Example 5.6.10 Evaluate∫ √

x2−1x dx.

Solution : x = sec t⇒ dx = sec t tan tdt∫ √x2 − 1

xdx =

∫ √tan2 t

sec tsec t tan tdt

=

∫tan2 tdt

=

∫sin2 t

cos2 tdt

=

∫tan2 tdt

= tan t− t+ c

=√x2 − 1− arcsecx+ c

t1

x sec t = x

tan t =√x2 − 1

When we have a quadratic term under the square root, complete the square:

Example 5.6.11 Evaluate∫

1√x2+4x

dx.

Solution :x2 + 4x = x2 + 4x+ 4− 4 = (x+ 2)

2 − 4.

Then, we make a substitution.u = x+ 2 =⇒ du = dx∫

1√x2 + 4x

dx =

∫1√

u2 − 22du

102

Now, we make a trigonometric substitution. u = 2 sec t =⇒ du = 2 sec t tan tdt∫1√

x2 + 4xdx =

∫du√u2 − 22

=

∫1√

4 tan2 t2 sec t tan tdt

=

∫1

2. tan t2 sec t tan tdt =

∫sec tdt

= ln | sec t+ tan t|+ c

= ln |u2

+

√u2 − 4

2|+ c

= ln |x+ 2

2+

√(x+ 2)

2 − 4

2|+ c

103

5.7 Partial Fractions

First we recall following integrals :∫A

x+ adx = A ln |x+ a|+ c∫

A

(x+ a)kdx = A

1

− (k − 1) (x+ a)k−1 + c∫

Ax+B

x2 + a2dx =

A

2ln |x2 + a2|+Ba arctan

x

a∫Ax+B

(x2 + a2)kdx =

A

2

1

− (k − 1) (x2 + a2)k−1 +

B

a2k−1

∫cos2k−2 tdt

where t = arctan(x/a).Recall that a rational function is the ratio of two polynomials :

R (x) =P (x)

Q (x).

Integrating rational functions is relatively simple, because most of these functions can be obtained as sums of even simplerfunctions. If the degree of P (x) is at least as large as the degree of Q(x), then we can divide P (x) by Q(x), getting apolynomial as a quotient, and possibly a remainder. That is, if the degree of P (x) is at least as large as the degree ofQ(x), then there exist polynomials P1(x) and P2(x) such that the degree of P2(x) is less than the degree of Q(x) and

R (x) =P (x)

Q (x)= P1 (x) +

P2 (x)

Q (x).

As P1(x) is a polynomial, it is easy to integrate. Therefore, the diffi culty of integrating R(x) lies in integrating P2(x)Q(x) ,

which is a rational function whose denominator is of higher degree than its numerator.For this reason, in the rest of this section, we focus on integrating rational functions with that property, that is, when

the degree of the denominator is higher than the degree of the numerator.Distinct Linear Factors : The easiest case is when Q(x) factors into the product of polynomials of degree 1, and

each of these terms occurs only once. For each factor of the form x+ a write

A

x+ a.

After determining the numbers A, we can evaluate the integral.

Example 5.7.1 Evaluate∫

1(x+1)(x+2)dx.

Solution :

1

(x+ 1) (x+ 2)=

A

x+ 1+

B

x+ 2

=A (x+ 2) +B (x+ 1)

(x+ 1) (x+ 2)

Hence we haveA (x+ 2) +B (x+ 1) = 1.

and A = 1, B = −1. ∫1

(x+ 1) (x+ 2)dx =

∫1

x+ 1dx−

∫1

x+ 2dx

= ln |x+ 1| − ln |x+ 2|+ c.

Repeated Linear Factors : The next case is when Q(x) factors into linear terms, but some of these terms occurmore than once. In that case for each factor of the form (x+ a)

k write

A1(x+ a)

+A2

(x+ a)2 +

A3

(x+ a)3 + · · ·+ A4

(x+ a)k.

104

Example 5.7.2 Evaluate∫

2x+7(x+1)2(x−1)dx.

Solution :2x+ 7

(x+ 1)2

(x− 1)=

A

x+ 1+

B

(x+ 1)2 +

C

x− 1

=A (x+ 1) (x− 1) +B (x− 1) + C (x+ 1)

2

(x+ 1)2

(x− 1).

We have2x+ 7 = A (x+ 1) (x− 1) +B (x− 1) + C (x+ 1)

2

and A = −2.25, B = −2.5, C = 2.25.∫2x+ 7

(x+ 1)2

(x− 1)dx = −2.25

∫1

x+ 1dx︸ ︷︷ ︸

u = x+ 1du = dx

− 2.5

∫1

(x+ 1)2 dx︸ ︷︷ ︸

u = x+ 1du = dx

+ 2.25

∫1

x− 1dx︸ ︷︷ ︸

u = x− 1du = dx

= −2.25

∫1

udu− 2.5

∫1

u2du+ 2.25

∫1

udu

= −2.25 ln |u| − 2.5u−1

−1+ 2.25 ln |u|+ c

= −2.25 ln |x+ 1|+ 2.5

x+ 1+ 2.25 ln |x− 1|+ c.

Distinct Quadratic Factors : The third case is when the factorization of Q(x) contains some quadratic factors thatare irreducible (i.e., they are not the product of two linear polynomials with real coeffi cients), but none of these irreduciblequadratic factors occurs more than once. For each factor of the form ax2 + bx+ c write

Ax+B

ax2 + bx+ c.

Example 5.7.3 Evaluate∫

4x+2(x2+1)(x+1)dx.

Solution :4x+ 2

(x+ 1) (x2 + 1)=

A

x+ 1+Bx+ C

x2 + 1

=A(x2 + 1

)+ (Bx+ C) (x+ 1)

(x+ 1) (x2 + 1)

Hence we have4x+ 2 = (A+B)x2 + (B + C)x+A+ C

and A = −1, B = 1, C = 3.∫4x+ 2

(x2 + 1) (x+ 1)dx = −

∫1

x+ 1dx+

∫x+ 3

x2 + 1dx

= −∫

1

x+ 1dx︸ ︷︷ ︸

u = x+ 1du = dx

+

∫x

x2 + 1dx︸ ︷︷ ︸

u = x2 + 1du = 2xdx

+ 3

∫1

x2 + 1dx

= −∫

1

udu+

∫ du2

udx+ 3 arctanx

= − ln |u|+ 1

2ln |u|+ +3 arctanx+ c

= − ln |x+ 1|+ 1

2ln |x2 + 1|+ 3 arctanx+ c.

Repeated Quadratic Factors : Finally, it can happen that the factorization Q(x) contains irreducible quadratic

factors, some of which occur more than once. In that case for each factor of the form(ax2 + bx+ c

)kwrite

A1x+B1ax2 + bx+ c

+A2x+B2

(ax2 + bx+ c)2 +

A3x+B3

(ax2 + bx+ c)3 + · · ·+ Akx+Bk

(ax2 + bx+ c)k.

105

Example 5.7.4 Evaluate∫x3+2x2+3x+7

(x2+1)2dx.

Solution :

x3 + 2x2 + 3x+ 7

(x2 + 1)2 =

Ax+B

x2 + 1+

Cx+D

(x2 + 1)2

=(Ax+B)

(x2 + 1

)+ Cx+D

(x2 + 1)2

Then we havex3 + 2x2 + 3x+ 7 = Ax3 +Bx2 + (A+ C)x+ (B +D)

and A = 1, B = 2, C = 2, D = 5∫x3 + 2x2 + 3x+ 7

(x2 + 1)2 dx =

∫x+ 2

x2 + 1dx+

∫2x+ 5

(x2 + 1)2 dx

=

∫x

x2 + 1dx︸ ︷︷ ︸

u = x2 + 1du = 2xdx

+ 2

∫1

x2 + 1dx+ 2

∫x

(x2 + 1)2 dx︸ ︷︷ ︸

u = x2 + 1du = 2xdx

+ 5

∫1

(x2 + 1)2 dx︸ ︷︷ ︸

x = tan tdx = sec2 tdt

=

∫ du2

u+ 2 arctanx+ 2

∫ du2

u2+ 5

∫1

(sec2 t)2 sec2 tdt

=1

2

∫1

udu+ 2 arctanx+

∫1

u2du+ 5

∫1

sec2 tdt

=1

2ln |u|+ 2 arctanx+

u−1

−1+ 5

∫cos2 tdt

=1

2ln |x2 + 1|+ 2 arctanx− 1

x2 + 1+ 5

∫1

2(1 + cos (2t)) dt

=1

2ln |x2 + 1|+ 2 arctanx− 1

x2 + 1+

5

2

(t+

1

2sin (2t)

)+ c

=1

2ln |x2 + 1|+ 2 arctanx− 1

x2 + 1+

5

2

(arctanx+

1

2

2x

x2 + 1

)+ c

t

x

1

tan t = xsin t = x√

x2+1

cos t = 1√x2+1

sin (2t) = 2 sin t cos t = 2xx2+1

Rationalizing Substitutions : There are situations when a function that is not a rational function can be turnedinto one by an appropriate substitution, and then it can be integrated by the methods presented in this section.

Example 5.7.5 Evaluate∫ √

x√x+1

dx.

Solution : We use the substitution√x = u, then 1

2√xdx = du and dx = 2

√xdu = 2udu. This leads to∫ √

x√x+ 1

dx =

∫u

u+ 12udu =

∫2u2

u+ 1du

=

∫ (2u− 2 +

2

u+ 1

)du

= u2 − 2u+ 2 ln |u+ 1|+ c

= x− 2√x+ 2 ln |

√x+ 1|+ c.

106

5.8 More on Areas

Notice that if f (x) ≤ 0 ( rather than f (x) ≥ 0 ) on [a, b], then the definite integral takes the negative value. In general adefinite integral gives the net area between the graph of y = f (x) and the x-axis.

x

y

+­

+a b c d

net area =

∫ d

a

f (x) dx

total area =

∫ d

a

|f (x)| dx

=

∫ b

a

f (x) dx−∫ c

b

f (x) dx+

∫ d

c

f (x) dx

=

∣∣∣∣∣∫ b

a

f (x) dx

∣∣∣∣∣+

∣∣∣∣∫ c

b

f (x) dx

∣∣∣∣+

∣∣∣∣∣∫ d

c

f (x) dx

∣∣∣∣∣Example 5.8.1 Find the net area and total area under the graph of f (x) = x2 − 1 from −1 to 2.Solution :

­1 1 2­1

1

2

3

x

y

y = x2 − 1

net area =

∫ b

a

f (x) dx

=

∫ 2

−1

(x2 − 1

)dx

=

(x3

3− x)∣∣∣∣2−1

= 0

For the total area first we find the points that the function change sign.:

x2 − 1 = 0

(x− 1) (x+ 1) = 0

107

So if −1 < x < 1, then f(x) ≤ 0 and if 1 < x < 2,then f(x) ≥ 0. Hence

total area =

∫ b

a

|f (x)| dx

=

∫ 2

−1

∣∣x2 − 1∣∣ dx

= −∫ 1

−1

(x2 − 1

)dx+

∫ 2

1

(x2 − 1

)dx

= −(x3

3− x)∣∣∣∣1−1

+

(x3

3− x)∣∣∣∣2

1

= −(−4

3

)+

4

3

=8

3

Example 5.8.2 Find the net area and total area under the graph of x = f (y) = y3 from −3 to 4.Solution :

x

y

x = y3

net area =

∫ b

a

f (y) dy

=

∫ 4

−3y3dy

=y4

4

∣∣∣∣4−3

=175

4.

Notice that if −3 < x < 0, then f(y) ≤ 0 and if 0 < x < 4,then f(y) ≥ 0. Then

total area =

∫ b

a

|f (x)| dy

=

∫ 4

−3

∣∣y3∣∣ dy= −

∫ 0

−3y3dy +

∫ 4

0

y3dy

= − y4

4

∣∣∣∣0−3

+y4

4

∣∣∣∣40

=337

4.

Areas Between Curves :

108

The area between the curves y = f(x) and y = g(x) and the lines x = a and x = b where f(x), g(x) continuous andf(x) ≥ g(x) for x in [a, b], is

A =

∫ b

a

f (x) dx−∫ b

a

g (x) dx

=

∫ b

a

[f (x)− g (x)] dx.

Example 5.8.3 Find the area between y = ex and y = x from 0 to 1.Solution :

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

2.5

x

y

y = ex, y = x

First note that ex ≥ x for 0 ≤ x ≤ 1. So

A =

∫ 1

0

(ex − x) dx =

[ex − x2

2

]∣∣∣∣10

=

(e1 − 1

2

)−(e0 − 0

)= e− 3

2.

The area between two curves y = f(x) and y = g(x) that intersect at two points can be computed in the followingway. First find the intersection points a and b by solving the equation f(x) = g(x). Then find the difference:

A =

∫ b

a

[f (x)− g (x)] dx =

∫ b

a

[f (x)− g (x)] dx.

If the result is negative that means that we have subtracted wrong. Just take the result in absolute value.

Example 5.8.4 Find the area between y = f (x) = x2 and y = g (x) = 2− x.Solution : First, find the intersection points by solving x2 − (2− x) = x2 + x− 2 = 0. We get x = −2 and x = 1. Thenevaluate : ∫ 1

−2

[x2 − (2− x)

]dx =

∫ 1

−2

[x2 + x− 2

]dx

=

[x3

3+x2

2− 2x

]∣∣∣∣1−2

= −9

2

Hence the area is 92 .

109

­2 ­1 0 1

1

2

3

4

x

y

y = 2− x, y = x2

The situation becomes slightly more complicated if f (x) ≥ g (x) does not hold throughout the entire interval [a, b].For instance, it could happen that f(x) ≥ g(x) at the beginning of the interval [a, b], and then, from a given pointon, g(x) ≥ f(x). In that case, we split [a, b] up into smaller intervals so that on each of these smaller intervals, eitherf(x) ≥ g(x) or g(x) ≥ f(x) holds.

Example 5.8.5 Let f(x) = x3 + 3x2 + 2x and let g(x) = x3 + x2. Compute the area whose borders are the graphs of fand g and the vertical lines x = −2 and x = 1.Solution :

­2.0 ­1.5 ­1.0 ­0.5 0.5 1.0

­4

­3

­2

­1

1

2

3

4

5

6

x

y

y = x3 + 3x2 + 2x, y = x3 + x2

First we need to compute the intersection points.

f(x) = g(x)(x3 + 3x2 + 2x

)=

(x3 + x2

)2x2 + 2x = 0

2x (x+ 1) = 0

That is, there are only two points where these two curves intersect, namely, at x = −1 and x = 0. If x ≤ −1 or if x ≥ 0,

110

then f(x)− g(x) = 2x(x+ 1) > 0, so f(x) ≥ g(x). If −1 < x < 0,then f(x)− g(x) < 0, so f(x) ≤ g(x). Hence

Area =

∫ 1

−2|f (x)− g (x)| dx

=

∫ −1−2

(f (x)− g (x)) dx−∫ 0

−1(f (x)− g (x)) dx+

∫ 1

0

(f (x)− g (x)) dx

=

∫ −1−2

(2x2 + 2x

)dx−

∫ 0

−1

(2x2 + 2x

)dx+

∫ 1

0

(2x2 + 2x

)dx

=

(2x3

3+ x2

)∣∣∣∣−1−2−(

2x3

3+ x2

)∣∣∣∣0−1

+

(2x3

3+ x2

)∣∣∣∣10

=5

3+

1

3+

5

3

=11

3

Sometimes it is easier or more convenient to write x as a function of y and integrate respect to y

Example 5.8.6 Find the area between the line y = x− 1 and the parabola y2 = 2x+ 6.Solution :

­4 ­2 2 4

­2

­1

1

2

3

4

x

y

y = x− 1, y2 = 2x+ 6

The intersection points between those curves are (−1,−2) and (5, 4), but in the figure we can see that the region extendsto the left of x = −1. In this case it is easier to write

x = f (y) =y2

2− 3 and x = g (y) = y + 1

and integrate from y = −2 to y = 4:

A =

∫ 4

−2[g (y)− f (y)] dy

=

∫ 4

−2

[(y + 1)−

(y2

2− 3

)]dy

=

∫ 4

−2

(−y

2

2+ y + 4

)dy

=

[−y

3

6+y2

2+ 4y

]∣∣∣∣4−2

= 18.

111

6 Applications of Integral

6.1 Volumes

6.1.1 Volumes by Cross Sections

First we study how to find the volume of some solids by the method of cross sections (or “slices”). The idea is to dividethe solid into slices perpendicular to a given reference line. The volume of the solid is the sum of the volumes of its slices.Let R be a solid lying alongside some interval [a, b] of the x-axis. For each x in [a, b] we denote A(x) the area of the

cross section of the solid by a plane perpendicular to the x-axis at x. We divide the interval into n subintervals [xi−1, xi],of length ∆x = (b − a)/n each. The planes that are perpendicular to the x-axis at the points x0, x1, x2, . . . , xn dividethe solid into n slices. If the cross section of R changes little along a subinterval [xi−1, xi], the slab positioned alongsidethat subinterval can be considered a cylinder of height ∆x and whose base equals the cross section A (x∗i ) at some pointx∗i ∈ [xi−1, xi]. So the volume of the slice is

∆Vi ≈ A (x∗i ) ∆x.

The total volume of the solid is

V =

n∑i=1

∆Vi ≈n∑i=1

A (x∗i ) ∆x.

Once again we recognize a Riemann sum at the right. In the limit as n→∞ we get the so called Cavalieri’s principle :

V =

∫ b

a

A (x) dx.

Of course, the formula can be applied to any axis. For instance if a solid lies alongside some interval [a, b] on the yaxis, the formula becomes

V =

∫ b

a

A (y) dy.

Consider the plane region between the graph of the function y = f(x) and the x-axis along the interval [a, b]. Byrevolving that region around the x-axis we get a solid of revolution. Now each cross section is a circular disk of radius y,so its area is

A (x) = πy2 = π [f (x)]2.

Hence, the volume of the solid is

V =

∫ b

a

π [f (x)]2dx.

If the revolution is performed around the y-axis, the roles of x and y are interchanged, so in that case the formulas are

A (y) = πx2 = π [h (y)]2

V =

∫ b

a

π [h (y)]2dy

where g (y) is a function of y.

Example 6.1.1 The region under the graph of y = f (x) = x2 and above the interval [0, 1] is revolved around the x-axis.Compute the volume of the resulting solid.Solution :

0.0 0.5 1.00.0

0.5

1.0

x

y

1z

1

0

­1

0.00­1 1.0

xy0.5

The cross section of the solid at each point x is a circular disk of radius x2. So its area is

A (x) = π [f (x)]2

= π(x2)2

= πx4

112

Hence the volume of the solid is

V =

∫ b

a

π [f (x)]2dx =

∫ 1

0

πx4dx

= π

(x5

5

)∣∣∣∣10

=π

5.

Now find the volume of the solid by revolving the region between the graph of f (x) and y-axis from 0 to 1 around they-axis.

0.0 0.5 1.00.0

0.5

1.0

x

y

­1.0 1­0.50

x y0.0

0.0­11.00.5

z 0.5

1.0

Similarly the cross section of the solid at each point y is a circular disk of radius√y. So its area is

A (y) = π [h (y)]2

= π (√y)2

= πy

and the volume of the solid is

V =

∫ b

a

π [h (y)]2dy =

∫ 1

0

πydy

= π

(y2

2

)∣∣∣∣10

=π

2.

Example 6.1.2 Find the volume of a sphere of radius r.Solution : Think of the sphere as the solid obtained by revolving the region under the graph of y = f (x) =

√r2 − x2 and

above the interval [−r, r] about x-axis.

­11

1

00

0

­1 y

z

x1­1

The area of the cross section of the solid at each point x is

A (x) = π [f (x)]2

= π(√

r2 − x2)2

= π(r2 − x2

)and the volume is ∫ b

a

π [f (x)]2dx =

∫ r

−rπ(r2 − x2

)dx

= π

[r2x− x3

3

]∣∣∣∣r−r

=4

3πr3.

113

If the region being revolved is the area between two curves y = f(x) and y = g(x), then each cross section is an annularring (or washer) with outer radius f(x) and inner radius g(x) (assuming f(x) ≥ g(x) ≥ 0). The area of the annular ring is

A (x) = π[[f (x)]

2 − [g (x)]2]

, hence the volume of the solid will be :

V =

∫ b

0

π[[f (x)]

2 − [g (x)]2]dx.

If the revolution is performed around the y-axis, then :

A (x) = π[[h (y)]

2 − [r (y)]2]

and

V =

∫ b

0

π[[h (y)]

2 − [r (y)]2]dy

where h (y) and r (y) are functions of y.

Example 6.1.3 Revolve the region between the graphs of y = f (x) = x2 and y = g (x) = x3 around the x-axis andcompute the volume of the resulting solid.Solution :

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

x

y

1

1.0

z

0.5

0.0

­0.5

­1.0

0.0 0

x0.5

­11.0

y

First find the intersection points :

x3 = x2

x3 − x2 = 0

x2 (x− 1) = 0

Hence x = 0 and x = 1 are the intersection points. Now find find the cross-section area at x ∈ [0, 1]. Since f (x) ≥ g (x)in this interval :

A (x) = π[[f (x)]

2 − [g (x)]2]

= π[(x2)2 − (x3)]2

= π(x4 − x5

)and the volume is

V =

∫ b

a

π[[f (x)]

2 − [g (x)]2]dx

= π

∫ 1

0

(x4 − x5

)dx

π

[x5

5− x6

6

]∣∣∣∣10

=2π

35.

114

Now find the volume of the solid obtained by revolving the same region around the y-axis.

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

x

y

­1.0 1­0.50.0

yx0 0.51.0­1 0.0

z

1.0

0.5

Here we have the functions x = h (y) =√y and x = r (y) = 3

√y on [0, 1]. Since r (y) ≥ h (y) in this interval, the cross

sectional area at each y ∈ [0, 1] is

A (y) = π[[r (y)]

2 − [h (y)]2]

= π[[ 3√y]2 − [√y]2]

= π(y23 − y

)and the volume is

V =

∫ b

a

π[[r (y)]

2 − [h (y)]2]dy

=

∫ 1

0

π(y23 − y

)dy

= π

[y53

53

− y2

2

]∣∣∣∣∣1

0

=π

10.

115

6.1.2 Volumes by Cylindrical Shells

A cylindrical shell is the region between two concentric circular cylinders of the same height h. If their radii are r1 and r2respectively, then the volume is :

V = πr22h− πr21h = πh(r22 − r21

)= πh (r2 + r1) (r2 − r1)= 2πr̄ht

where r̄ = (r2 + r1) /2 is the average radius, and t = r2 − r1 is the thickness of the shell.Consider the solid generated by revolving around the y-axis the region under the graph of y = f(x) between x = a

and x = b. We divide the interval [a, b] into n subintervals [xi−1, xi] of length ∆x = (b− a)/n each. The volume V of thesolid is the sum of the volumes ∆Vi of the shells determined by the partition. Each shell, obtained by revolving the regionunder y = f(x) over the subinterval [xi−1, xi], is approximately cylindrical. Its height is f(x̄i), where x̄i is the midpointof [xi−1, xi]. Its thickness is ∆x. Its average radius is x̄i. Hence its volume is

∆Vi ≈ 2πx̄if (x̄i) ∆x,

and the volume of the solid is

V =

n∑i=1

∆Vi ≈n∑i=1

2πx̄if (x̄i) ∆x.

As n→∞ the right Riemann sum converges to following integral :

V =

∫ b

a

2πxf (x) dx.

If the region is revolved around the x-axis then the variables x and y reverse their roles :

V =

∫ b

a

2πyh (y) dy

where h (y) a function of y.

Example 6.1.4 Find the volume of the solid obtained by revolving around the y-axis the region between the graph ofy = f (x) = 3x3 − x4 and the x-axis.Solution : 3x3 − x4 = 0⇒ x = 0, x = 3

0 1 2 30

2

4

6

8

10

x

y

V =

∫ 3

0

2πxf (x) dx = 2π

∫ 3

0

x(3x3 − x4

)dx

= 2π

∫ 3

0

(3x4 − x5

)dx

= 2π

(3x5

5− x6

6

)∣∣∣∣30

= 2π

(729

5− 729

6

).

116

Here we find the volume of the solid obtained by revolving around the y-axis the area between two curves y = f(x)and y = g(x) over an interval [a, b]. The computation is similar, but if f(x) ≥ g(x) the shells will have height f (x)−g (x),so the volume will be given by the integral :

V =

∫ b

a

2πx [f (x)− g (x)] dx.

Example 6.1.5 Find the volume of the solid obtained by revolving the region in the first quadrant bounded below by thegraph of x2 + y2 = 4 and above by the graph of y = x around the y-axis.Solution : This circle can be represented as y =

√2− x2 in the first quadrant and the intersection points of these two

curves : √2− x2 = x⇒ x =

√2.

2^(1/2)1 20

1

2

x

y

y1

10

­1x

00­1

2

1z

V =

∫ b

a

2πx (f (x)− g (x)) dx

= 2π

∫ √20

x(√

4− x2 − x)dx

= 2π

(∫ √20

x√

4− x2dx−∫ √20

x2dx

)

u = 4− x2 ⇒ du = −2xdx, xdx = −1

2du

V = 2π

∫u12

(−1

2

)du− 2π

(x3

3

)∣∣∣∣√2

0

= −πu32

32

− 4√

2π

3

= −2π

3

(4− x2

) 32

∣∣∣√20− 4√

2π

3

= −2π

3

(2√

2− 8)− 4√

2π

3

=8π

3

(2−√

2).

Now we revolve the same region around the x-axis and find the volume of resulting solid. Notice that in this case it is

117

more convenient to use cross-sectional areas.

V =

∫ b

a

A (x) dx

=

∫ b

a

π(f (x)

2 − g (x)2)dx

= π

∫ √20

((√4− x2

)2− x2

)dx

= π

∫ √20

(4− 2x2

)dx

= π

(4x− 2x3

3

)∣∣∣∣√2

0

= π

(4√

2− 4√

2

3

)

=8√

2π

3.

Example 6.1.6 Find the volume of the solid obtained by revolving the region limited by the curves y = x and y = x2

around the y-axes.Solution :

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

x

y

z

1.0

­1

0.5

0.01

0.0

y

0.51.0

­1.0­0.50

x

Cross Section : Firstly, we need to rewrite the curve as a function of x : y = x⇒ x = y and y = x2 ⇒ x =√y

V =

∫ b

a

π[[f (y)]

2 − [g (y)]2]dy

=

∫ 1

0

π[[y]

2 − [√y]2]dy

=

∫ 1

0

π[y2 − y

]dy

= −π6

=π

6

Cylindrical Shell:

V =

∫ b

a

2πx [f (x)− g (x)] dx

=

∫ 1

0

2πx[x− x2

]dx

=π

6.

Example 6.1.7 Revolve the region under the graph of y = f (x) = x2 and above the interval [0, 1] around the y-axis andcompute the volume of the resulting solid.

118

Solution :

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

x

y

­0.50

x y0.0

0.51.0

­1.01

­1 0.0

z

1.0

0.5

Cross Section : x = f (y) =√y and x = g (y) = 1

V =

∫ 1

0

π(

[g (y)]2 − [f (y)]

2)dy

=

∫ 1

0

π (1− y) dy

=π

2

Cylindrical Shell: y = f (x) = x2

V =

∫ 1

0

2πxf (x) dx

=

∫ 1

0

2πxx2dx

=

∫ 1

0

2πx3dx

=π

2

119

6.2 Improper Integrals

6.2.1 First Type Improper Integrals

Motivation :Consider the function y = f (x) = 1

x2 .If we integrate f (x) over [1, r] ∫ r

1

1

x2dx = − 1

x

∣∣∣∣r1

= −1

r−(−1

1

)= 1− 1

r.

Now notice that this approaches 1 as r →∞ :

limr→∞

∫ r

1

1

x2dx = lim

r→∞

(1− 1

r

)= 1.

r

A=1

x

y

Graph of y = 1x2

Trying the same thing for y = g (x) = 1x , however, gives

limr→∞

∫ r

1

1

xdx = lim

r→∞ln |r| =∞.

r1

A=infinite

x

y

Graph of y = 1x

We will write these limits as integrals with an upper limit of integration of "∞” and call them "improper integrals".

Definition 6.2.1 The improper integral of f (x) over [a,∞] is defined to be∫ ∞a

f (x) dx = limr→∞

∫ r

a

f (x) dx.

If the limit exists,∫∞af (x) dx is said to converge (or to be convergent). If the limit does not exist,

∫∞af (x) dx is said

diverge (or to be divergent).

120

Hence ∫ ∞1

1

x2dx converges and its value is 1,∫ ∞

1

1

xdx diverges.

Example 6.2.2 For what values of p, is the integral∫∞1

1xp dx convergent.

Solution : We have already seen that this diverges when p = 1. So we now assume that p 6= 1.∫ ∞1

1

xpdx = lim

r→∞

∫ r

1

1

xpdx = lim

r→∞

∫ r

1

x−pdx

= limr→∞

x−p+1

−p+ 1

∣∣∣∣r1

=1

−p+ 1limr→∞

(r−p+1 − 1

)This limit exists only when p > 1 and in this case r−p+1 → 0 as r →∞ so∫ ∞

1

1

xpdx =

1

−p+ 1(0− 1) =

1

1− p when p > 1

thus, ∫ ∞1

1

xpdx =

{converges to 1

1−p if p > 1

diverges if p ≤ 1.

Example 6.2.3 Compute∫∞0

(1− x) e−xdxSolution :

r x

y

∫ ∞0

(1− x) e−xdx = limr→∞

∫ r

0

(1− x) e−xdx

Lets evaluate first∫ r0

(1− x) e−xdx.

u = 1− x dv = e−x

du = −dx v =∫e−xdx = −e−x∫

udv = uv −∫vdu∫

(1− x) e−xdx = − (1− x) e−x −∫e−xdx

= − (1− x) e−x + e−x + c

Hence ∫ r

0

(1− x) e−xdx =(− (1− x) e−x + e−x

)∣∣r0

= − (1− r) e−r + e−r −((− (1− 0) e−0 + e−0

))=

r − 1

er+

1

er+ 1− 1

=r

er.

Then ∫ ∞0

(1− x) e−xdx = limr→∞

r

er=

l′hospital’slimr→∞

1

er= 0

Note that this integral ends up being zero.

121

Example 6.2.4 Compute∫∞0

11+x2 dx.

Solution :

r x

y

∫ ∞0

1

1 + x2dx = lim

r→∞

∫ r

0

1

1 + x2dx

= limr→∞

arctanx|r0= lim

r→∞(arctan r − arctan 0)

=π

2

So the integral converges.

Similarly, one can define ∫ b

−∞f (x) dx = lim

k→−∞

∫ b

k

f (x) dx.

Example 6.2.5 Compute∫ 0−∞

11+x2 dx.

Solution :

x

y

∫ 0

−∞

1

1 + x2dx = lim

k→−∞

∫ 0

k

1

1 + x2dx

= limk→−∞

arctanx|k0= lim

k→−∞(arctan 0− arctan k)

= −(−π

2

)=

π

2

Hence the integral converges.

Finally, we define ∫ ∞−∞

f (x) dx =

∫ b

−∞f (x) dx+

∫ ∞b

f (x) dx

= limk→−∞

∫ b

k

f (x) dx+ limr→∞

∫ ∞b

f (x) dx

where b is any real number ( usually taken to be 0 ).∫∞−∞ f (x) dx said to be convergent if both

∫ b−∞ f (x) dx and

∫∞bf (x) dx are convergent. Otherwise, we say that∫∞

−∞ f (x) dx is divergent.

122

Example 6.2.6 Compute∫∞−∞

11+x2 dx.

Solution :

x

y

∫ ∞−∞

1

1 + x2dx =

∫ 0

−∞

1

1 + x2dx+

∫ ∞0

1

1 + x2dx

=π

2+π

2= π

Hence∫∞−∞

11+x2 dx is convergent.

Example 6.2.7 Compute∫∞−∞ e−xdx.

Solution :

x

y

∫ ∞−∞

e−xdx =

∫ 0

−∞e−xdx+

∫ ∞0

e−xdx

= limk→−∞

∫ 0

k

e−xdx+ limr→∞

∫ r

0

e−xdx

limk→−∞

∫ 0

k

e−xdx = limk→−∞

(−e−x

)∣∣ok

= limk→−∞

(−1 + e−k

)=∞

Since∫ 0−∞ e−xdx divergent,

∫∞−∞ e−xdx is divergent.

Remark 6.2.8∫∞−∞ f (x) dx is not the same thing as

limt→∞

∫ t

−tf (x) dx.

For example, ∫ ∞0

xdx = limr→∞

∫ r

0

xdx = limr→∞

x2

2

∣∣∣∣r0

= limr→∞

r2

2=∞

123

obviously diverges so∫∞−∞ f (x) dx diverges. However,

limt→∞

∫ t

−txdx = lim

t→∞

x2

2

∣∣∣∣t−t

= limt→∞

(t2

2− (−t)2

2

)= 0.

All these type of improper integrals called first type (infinite limits of integration). Now we turn to second typeimproper integrals.

124

6.2.2 Second Type Improper Integrals

Motivation :Consider f (x) = 1√

1−x on [0, 1].

0.0 0.2 0.4 0.6 0.8 1.00

1

2

3

4

5

x

y

f (x) is not continuous on 1 so∫ 10

1√1−xdx has not yet been defined. However, for any r in 0 < r < 1,∫ r

0

1√1− x

dx =

∫ r

0

(1− x)− 12 dx = −2 (1− x)

12

= −2√

1− r + 2

and

limr→1−

∫ r

0

1√1− x

dx = limr→1−

(−2√

1− x+ 2)

= 2.

We will define "∫ 10

1√1−xdx " to be the limit. More generally,

Definition 6.2.9 If f (x) is continuous on [a, b), but f (x) → ±∞ as x → b−, then the improper integral of f (x) over[a, b] is ∫ b

a

f (x) dx = limr→b−

∫ r

a

f (x) dx.

If the limit exists, then∫ baf (x) dx is said to converge, otherwise it is said to diverge.

i.e.,∫ 1

0

1√1− x

dx converges ( to 2 ).

Similarly, f (x) is continuous on (a, b], but f (x)→ ±∞ as x→ a+,∫ b

a

f (x) dx = limr→a+

∫ b

r

f (x) dx.

If f (x) is continuous on [a, b] except x = c in (a, b), then∫ b

a

f (x) dx =

∫ c

a

f (x) dx+

∫ b

c

f (x) dx

= limr→c−

∫ r

a

f (x) dx+ lims→c+

∫ b

s

f (x) dx

Note∫ baf (x) dx is convergent if and only if

∫ caf (x) dx and

∫ bcf (x) dx are convergent.

Example 6.2.10 Compute∫ 21

11−xdx.

Solution :

125

­2 ­1 1 2 3

­4

­2

2

4

x

y

∫ 2

1

1

1− xdx = limr→1+

∫ 2

r

1

1− xdx

= limr→1+

− (ln |1− x|)|2r= lim

r→1+ln |1− r|

This limit does not exist. Hence the integral diverges.

Example 6.2.11 Compute∫ 1−1

1x2 dx.

Solution :

­2 ­1 0 1 2

2

4

6

8

10

x

y

∫ 1

−1

1

x2dx =

∫ 0

−1

1

x2dx+

∫ 1

0

1

x2dx

limr→0−

∫ r

−1

1

x2dx = lim

r→0−

∫ r

−1x−2dx = lim

r→0−

(x−1

−1

)∣∣∣∣r−1

= limr→0−

[−1

r− 1

]=∞

Since∫ 0−1

1x2 dx diverges,

∫ 1−1

1x2 dx diverges.

126

6.3 Arc Length

In this section we are going to determine the length of a smooth arc. A smooth arc is a continuous function y = f (x)on the interval [a, b] whose derivative f ′ exists and is continuous on [a, b] (so it does not have corner points). Initiallywe’ll need to estimate the length of the curve. We’ll do this by dividing the interval up into n subintervals each of width∆xi = xi − xi−1.

a = x0 < x1 < · · · < xi−1 < xi < · · · < xn = b

The point on the curve at each point (xi, f (xi)) is denoted by Pi. We can then approximate the curve by a series ofstraight lines connecting the points. Denote the length of each of these line segments by |Pi−1Pi| and the length of thecurve will then be approximately,

L ≈n∑i=1

|Pi−1Pi|

and we can get the exact length by taking n larger and larger. In other words, the exact length will be,

L = limn→∞

n∑i=1

|Pi−1Pi|.

Let’s define ∆yi = yi − yi−1 = f (xi)− f (xi−1). We can then compute directly the length of the line segments as follows.

|Pi−1Pi| =

√(∆xi)

2+ (∆yi)

2

=

√√√√(∆xi)2

(1 +

[∆yi∆xi

]2)

=

√√√√(1 +

[f (xi)− f (xi−1)

xi − xi−1

]2)∆xi

Since f ′ is continuous, by the Mean Value Theorem for Derivative we know that on the interval [xi−1, xi], there is a pointx∗i such that

f ′ (x∗i ) =f (xi)− f (xi−1)

xi − xi−1.

Now we can writte

|Pi−1Pi| =√

1 + [f ′ (x∗i )]2∆xi.

and the approximated arc length is

L ≈n∑i=1

|Pi−1Pi| =n∑i=1

√1 + [f ′ (x∗i )]

2∆xi

The exact length of the curve is then

L = limn→∞

n∑i=1

|Pi−1Pi|

= limn→∞

n∑i=1

√1 + [f ′ (x∗i )]

2∆xi

=

∫ b

a

√1 + [f ′ (x)]

2dx or

=

∫ b

a

√1 +

[dy

dx

]2dx.

Example 6.3.1 Find the length of the arc defined by the curve y = f (x) = x32 between the points (0, 0) and (1, 1).

Solution :

127

0.0 0.5 1.00.0

0.5

1.0

x

y

f ′ (x) =3

2x12 is continuous.

L =

∫ b

a

√1 + [f ′ (x)]

2dx

=

∫ 1

0

√1 +

[3

2x12

]2dx

=

∫ 1

0

√1 +

9

4xdx

u = 1 + 94x⇒ du = 9

4dx ∫ √1 +

9

4xdx =

∫ √u

4

9du

=4

9

u32

32

+ c

=8

27

(1 +

9

4x

) 32

+ c

L =8

27

(1 +

9

4x

) 32

∣∣∣∣∣1

0

=8

27

((13

4

) 32

− 1

).

Example 6.3.2 Find the length of the curve y = f (x) = ln | secx| over[0, π4

].

Solution :

x

y

f ′ (x) =(secx)

′

secx=

secx tanx

secx= tanx is continuous.

128

L =

∫ b

a

√1 + [f ′ (x)]

2dx

=

∫ π4

0

√1 + tan2 xdx

=

∫ π4

0

√sec2 xdx

=

∫ π4

0

secxdx

= ln | secx+ tanx||π40

= ln | secπ

4+ tan

π

4| − ln | sec 0 + tan 0|

= ln |√

2 + 1| − ln |1 + 0|

= ln(√

2 + 1).

Similarly, the length of the graph of a function of x = f (y) from y = c to y = d is

L =

∫ d

c

√1 + [f ′ (y)]

2dy or

=

∫ d

c

√1 +

[dx

dy

]2dy

Example 6.3.3 Compute the length of the curve x = f (y) = 18y4 + 1

4y−2 over the interval 1 ≤ y ≤ 4.

Solution :

5 10 15 20 25 30

­4

­2

2

4

x

y

f ′ (y) =1

2y3 − 1

2y−3 is continuous.

[f ′ (y)]2

=

[1

2y3 − 1

2y−3

]2=

1

4y6 − 2

1

2y3

1

2y−3 +

1

4y−6

=1

4y6 − 1

2+

1

4y−6

1 + [f ′ (y)]2

=1

4y6 +

1

4y−6 +

1

2

=

[1

2y3 +

1

2y−3

]2

129

L =

∫ b

a

√1 + [f ′ (y)]

2dy =

∫ 4

1

√[1

2y3 +

1

2y−3

]2dy

=

∫ 4

1

[1

2y3 +

1

2y−3

]dy =

1

2

[y4

4+y−2

−2

]∣∣∣∣41

=1

2

[(256

4+

4−2

−2

)−(

1

4− 1

−2

)]=

2055

64

Example 6.3.4 Find the circumference of the circle with radius 1.Solution :

√1− x2

­1.0 ­0.5 0.5 1.0

­1.0

­0.5

0.5

1.0

x

y

x2 + y2 = 1⇒ y =√

1− x2 ⇒ dy

dx= − x√

1− x2is continuous

L = 2

∫ 1

−1

√1 + [f ′ (x)]

2dx

= 2

∫ 1

−1

√1 +

(− x√

1− x2

)2dx

= 2

∫ 1

−1

1√1− x2

dx

= 2 arcsinx|1−1 =π

2−(−π

2

)= 2π

130

6.4 Surface Area

The surface area of a frustum of a cone is given by,

SA = 2πrl

where r = r1+r22 , r1 is the radius of the right end, r2 is the radius of the left end and l is the length of the slant of the

frustum.In this section we are going to derive the formula for the surface area of a solid obtained by revolving a continuous

function y = f (x) in the interval [a, b] around the x-axis. We’ll start by dividing the interval into subintervals each ofwidth ∆xi = xi − xi−1

a = x0 < x1 < · · · < xi−1 < xi < · · · < xn = b

Let Pi = (xi, f(xi)) and let li = |Pi − Pi| denote the length of the straight line segment from Pi−1 to Pi. Let us revolvey = f (x) and the segments Pi−1Pi. The surface generated by segments is an approximation the surface generated byy = f (x). Each section of the surface generated by segments around the x-axis is the frustum of a cone. The radii of theright end and left end of the ith frustum are f (xi−1) and f (xi), respectively and the length of the slant of the ith frustumis |Pi−1Pi| and we know from arc length

|Pi−1Pi| =√

1 + [f ′ (x∗i )]2∆xi

where x∗i ∈ [xi−1, xi]. Before writing down the formula for the surface area we are going to assume that ∆xi is small andsince f (x) is continuous we can then assume that f (xi) ≈ f (x∗i ) and f (xi−1) ≈ f (x∗i ). So, the surface area of the i

th

frustum

2π

≈f(x∗i )︷ ︸︸ ︷f (xi−1) +

≈f(x∗i )︷ ︸︸ ︷f (xi)

2

√

1 + [f ′ (x∗i )]2∆xi

is approximately,

≈ 2πf (x∗i )

√1 + [f ′ (x∗i )]

2∆xi.

The surface area of the whole solid is then approximately,

SA ≈n∑i=1

2πf (x∗i )

√1 + [f ′ (x∗i )]

2∆xi.

We can get the exact surface area by taking the limit as n→∞.

SA = limn→∞

n∑i=1

2πf (x∗i )

√1 + [f ′ (x∗i )]

2∆xi

SA =

∫ b

a

2πf (x)

√1 + [f ′ (x)]

2dx or

=

∫ b

a

2πy

√1 +

[dy

dx

]2dx.

We have a similar formula for the surface area of a solid obtained by revolving a continuous function x = f (y) in theinterval [c, d] around the y-axis :

SA =

∫ d

c

2πf (y)

√1 + [f ′ (y)]

2dy or

=

∫ d

c

2πx

√1 +

[dx

dy

]2dy.

Example 6.4.1 Compute the surface area of a sphere of radius r.

131

Solution : The sphere can be obtained by revolving the curve y = f (x) =√r2 − x2 over [−r, r] around the x-axis.

1

­1

­1 00

y

z

x0

1

1­1

f ′ (x) =−2x

2√r2 − x2

=−x√r2 − x2

is continuous.

SA =

∫ b

a

2πf (x)

√1 + [f ′ (x)]

2dx

=

∫ r

−r2π√r2 − x2

√1 +

[−x√r2 − x2

]2dx

=

∫ r

−r2π√r2 − x2

√r2

r2 − x2 dx

=

∫ r

−r2πrdx

= 2πrx|r−r= 4r2π.

Example 6.4.2 Compute the surface area of the solid obtained y = 3√x, 1 ≤ x ≤ 8 around the y-axis.

Solution :

0 2 4 6 80.0

0.5

1.0

1.5

2.0

x

y

­5 50 0

yx

0 5­5

1z

2

y = 3√x, 1 ≤ x ≤ 8⇒ x = y3, 1 ≤ y ≤ 2

f ′ (y) = 3y2 is continuous.

SA =

∫ d

c

2πf (y)

√1 + [f ′ (y)]

2dy.

=

∫ 2

1

2πy3√

1 + 9y4dy

(u = 1 + 9y4 ⇒ du = 36y3dy, u(1) = 10, u(2) = 145)

= 2π

∫ 145

10

1

36

√udu

=π

18

u32

32

∣∣∣∣∣145

10

=π

27

(145

32 − 10

32

).

132

6.5 Parametric Equations

6.5.1 Introduction

A parametric curve can be thought of as the trajectory of a point that moves trough the plane with coordinates (x, y) =(f(t), g(t)), where f(t) and g(t) are functions of the parameter t ∈ [t1, t2]. For each value of t we get a point of the curve.For example, a parametric equation for a circle of radius 1 and center (0, 0) is :

x = cos t, y = sin t.

The equationsx = f(t), y = g(t), t ∈ [t1, t2]

are called parametric equations and the points (f (t1) , g (t1)) and (f (t2) , g (t2)) are called initial and terminal points ofthe curve, respectively.The graph of a function f is a particular example of a parametric curve:

x = t, y = f(t)

For a given parametric curve, sometimes we can eliminate t and obtain an equivalent non-parametric equation for thesame curve. For instance t can be eliminated from x = cos t, y = sin t by using the trigonometric relation

cos2 t+ sin2 t = 1

, which yields the (non-parametric) equation for a circle of radius 1 and center (0, 0) :

­1 1

­1.0

­0.5

0.5

1.0

x

y

x2 + y2 = 1

Example 6.5.1 Find a non-parametric equation for the following parametric curve :

x = t2 − 2t, y = t+ 1.

Solution : We eliminate t by isolating it from the second equation :

t = (y − 1) ,

and plugging it in the first equation :

x = (y − 1)2 − 2 (y − 1)

= y2 − 4y + 3,

which is a parabola with horizontal axis.

­2 2 4

­2

2

4

x

y

x = t2 − 2t, y = t+ 1

133

Example 6.5.2 Sketch the parametric curve for the following set of parametric equations :

x = t2 + t, y = 2t− 1, − 2 ≤ t ≤ 1.

Solution : To sketch a parametric curve, we pick some values of t, then plug them into the parametric equations andthen plot the points.

t x y−2 2 −5−1 0 −3− 12 − 14 −20 0 −11 2 1

0.5 1.0 1.5 2.0

­5

­4

­3

­2

­1

1

x

y

134

6.5.2 Tangents with Parametric Equations

Consider a parametric curve x = f(t), y = g(t), where the functions f(t) and g(t) are continuously differentiable and thederivatives f ′ (t) and g′ (t) do not vanish simultaneously for any t. Such parametric curves are called smooth.A smooth parametric curve x = f(t), y = g(t) has a tangent line at any point (x0, y0),and its equation is

dx

dt(y − y0)−

dy

dt(x− x0) = 0

or equivalently

y =dydtdxdt

(x− x0) + y0

where (x0, y0) = (f (t0) , g (t0)).

Example 6.5.3 Find the tangent line(s) to the parametric curve given by

x = t5 − 4t3, y = t2

at (0, 4).Solution : At first we need to determine the value(s) of t which will give point (0, 4).

t5 − 4t3 = 0, t2 = 4⇒ t = ±2.

Since there are two values of t that give the point we will get two tangent lines.

dx

dt(y − y0)−

dy

dt(x− x0) = 0(

5t4 − 12t2)

(y − y0)− 2t (x− x0) = 0

For t = −2 : (5t4 − 12t2

)(y − y0)− 2t (x− x0) = 0

x+ 8y − 32 = 0

For t = 2 : (5t4 − 12t2

)(y − y0)− 2t (x− x0) = 0

−x+ 8y − 32 = 0

­20 ­10 0 10 20

1

2

3

4

5

6

x

y

135

6.5.3 Area with Parametric Equations

In this section we will find a formula for determining the area under a parametric curve given by the parametric equations,

x = f(t), y = g(t), t1 ≤ t ≤ t2.

We need to assume that the curve is traced out exactly once as t increases from t1 to t2.Recall that the area under the graph of y = F (x) on a ≤ x ≤ b :

A =

∫ b

a

F (x) dx =

∫ b

a

ydx.

When we substitute the parametric equation into the integral, the area formula for parametric equations is obtainedas

A =

∫ t2

t1

g (t) f ′ (t) dt.

Example 6.5.4 Determine the area under the parametric curve given by the following parametric equations :

x = t− sin t, y = 1− cos t, 0 ≤ t ≤ 2π.

Solution :

­3 ­2 ­1 0 1 2 3 4 5 6 7 8 9

1

2

x

y

∫ t2

t1

g (t) f ′ (t) dt =

∫ 2π

0

(1− cos t) (t− sin t)′dt

=

∫ 2π

0

(1− cos t) (1− cos t) dt

=

∫ 2π

0

1− 2 cos t+ cos2 t︸ ︷︷ ︸12 (1+cos 2t)

dt

=

∫ 2π

0

(3

2− 2 cos t+

1

2cos 2t

)dt

=

[3

2t− 2 sin t+

1

2

1

2sin t

]∣∣∣∣2π0

= 3π

136

6.5.4 Arc Length with Parametric Equations

Here we describe how to find the length of a smooth curve

x = f(t), y = g(t), t1 ≤ t ≤ t2.

Assume that the curve is traced out exactly once as t increases from t1 to t2. We also need to assume that the curve istraced out from left to right as t increases. This is equivalent to saying,

dx

dt≥ 0

for t1 ≤ t ≤ t2. Recall that the arc length formulas :

L =

∫ b

a

√1 +

[dy

dx

]2dx if y = F (x) , a ≤ x ≤ b

L =

∫ d

c

√1 +

[dx

dy

]2dy if x = F (y) , c ≤ y ≤ d

Since we can write

dx =dx

dtdt

,the first arc length formula becomes

L =

∫ t2

t1

√√√√1 +

[dydxdt dt

]2dx

dtdt =

∫ t2

t1

√√√√1 +

[dydtdxdt

]2dx

dtdt

=

∫ t2

t1

√√√√√[dxdt ]2 +[dydt

]2[dxdt

]2 dx

dtdt =

∫ t2

t1

√[dx

dt

]2+

[dy

dt

]21∣∣∣∣dxdt∣∣∣∣︸︷︷︸

≥0

dx

dtdt

=

∫ t2

t1

√[dx

dt

]2+

[dy

dt

]2dt

If we had assumed thatdy

dt≥ 0

for t1 ≤ t ≤ t2, by using second arc length formula we have gotten the same formula.

Example 6.5.5 Find the arc length of the curve x = t2, y = t3 between (1, 1) and (4, 8).Solution : The given points correspond to the values t = 1 and t = 2 of the parameter, so :

1 2 3 4 5 6 7 8 9

­20

­10

0

10

20

x

y

137

L =

∫ t2

t1

√[dx

dt

]2+

[dy

dt

]2dt

=

∫ 2

1

√(2t)

2+ (3t2)

2dt

=

∫ 2

1

√4t2 + 9t4dt∫ 2

1

√4 + 9t2tdt

u = 4 + 9t2 ⇒ du = 18tdt⇒ tdt = du18 ∫ √

4 + 9t2tdt =

∫ √u

1

18du

=1

18

u32

32

+ c

=1

27

(4 + 9t2

) 32 + c

Then the length is

L =

∫ 2

1

√4 + 9t2tdt

=1

27

(4 + 9t2

) 32

∣∣∣21

=40

27

√40− 13

27

√13.

Example 6.5.6 Find the arc length of the curve x = r cos t, y = r sin t between 0 ≤ t ≤ 2π.Solution : This describes a circle of radius r.

x

y

√[dx

dt

]2+

[dy

dt

]2=√r2 sin2 t+ r2 cos2 t = r

L =

∫ t2

t1

√[dx

dt

]2+

[dy

dt

]2dt

=

∫ 2π

0

rdt = rt|2π0= 2πr.

138

6.5.5 Surface Area with Parametric Equations

We will determine the area of the surface obtained by revolving the parametric curve given by,

x = f(t), y = g(t), t1 ≤ t ≤ t2.

around the x or y-axis. We assume that the curve is traced out exactly once as t increases from t1 to t2.Recall the surface area formulas:

SA =

∫ b

a

2πy

√1 +

[dy

dx

]2dx revolving around x-axis

SA =

∫ d

c

2πx

√1 +

[dx

dy

]2dy revolving around y-axis

Since we know that √1 +

[dy

dx

]2dx =

√[dx

dt

]2+

[dy

dt

]2dt√

1 +

[dy

dx

]2dx =

√[dx

dt

]2+

[dy

dt

]2dt

, we have the following formulas for the surface area :

SA =

∫ t2

t1

2πy

√[dx

dt

]2+

[dy

dt

]2dt revolving around x-axis

SA =

∫ t2

t1

2πx

√[dx

dt

]2+

[dy

dt

]2dt revolving around y-axis

Example 6.5.7 Find the area of the surface obtained by revolving the arch of the cycloid

x = t− sin t, y = 1− cos t, 0 ≤ t ≤ 2π

around the x-axis.Solution :

0 1 2 3 4 5 60.0

0.5

1.0

1.5

2.0

x

y

1 2 3 4 5 6

­2

­1

0

1

2

x

y

139

SA =

∫ t2

t1

2πy

√[dx

dt

]2+

[dy

dt

]2dt

=

∫ sπ

0

2π (1− cos t)

√[1− cos t]

2+ [sin t]

2dt

=

∫ sπ

0

2π (1− cos t)√

1− 2 cos t+ cos2 t+ sin2 tdt

=

∫ sπ

0

2π (1− cos t)√

2− 2 cos tdt

=

∫ sπ

0

2π1− cos t︸ ︷︷ ︸=2 sin2 t

2

√√√√√√2

1− cos t︸ ︷︷ ︸=2 sin2 t

2

dt= 2π

∫ sπ

0

2 sin2t

22 sin

t

2dt = 8π

∫ sπ

0

(1− cos2

t

2

)sin

t

2dt(

u = cost

2⇒ du = −1

2sin

t

2dt

)= 8π

∫ −11

(1− u2

)(−2) du

= −16π

(u− u3

3

)∣∣∣∣−11

=64π

3.

140

6.6 Polar Coordinates

6.6.1 Introduction.

Coordinate systems are a way to define a point in space

(x,y)x

y

Rectangular Coordinates

(r,theta)xyr

theta

Polar Coordinates

To represent all points of a plane, the radial variable has to range over the interval r ∈ [0,∞), while the polar angle takesits values in the interval [0, 2π) because any ray from the origin does not change after rotation about the origin throughthe angle 2π.Polar to Cartesian Conversion Formulas :

x = r cos θ, y = r sin θ

Cartesian to Polar Conversion Formulas

r =√x2 + y2, θ = tan−1

(yx

)Example 6.6.1 Convert

(4, 2π3

)into Cartesian coordinates.

Solution :r = 4, θ =

2π

3

x = r cos θ = 4 cos2π

3= 4.

(−1

2

)= −2

y = r sin θ = 4 sin2π

3= 4.

(√3

2

)= 2√

3

Hence in Cartesian coordinates this point is(−2, 2

√3).

Example 6.6.2 Convert (−1,−1) into polar coordinates.Solution :

x = −1, y = −1

r =√x2 + y2 =

√(−1) + (−1)

2=√

2

θ = tan−1(yx

)= tan−1

(−1

−1

)=π

4

These conversion formulas can be used to convert equations from one coordinate system to the other.

Example 6.6.3 Convert 2x+ x2 = 1 + xy into polar coordinates.Solution :

2x+ x2 = 1 + xy ⇒ 2r cos θ + (r cos θ)2

= 1 + r cos θr sin θ

Example 6.6.4 Convert r = cos θ into Cartesian coordinates.Solution :

r = cos θ ⇒ r2 = r cos θ

⇒ x2 + y2 = x

141

Example 6.6.5 Sketch the polar curve θ = π4 .

Solution :

­4 ­2 2 4

­4

­2

2

4

x

y

θ = π4

Example 6.6.6 Sketch the polar curve r = 2.Solution :

x = 2 cos θ ⇒ x2 = 22 cos2 θ

y = 2 sin θ ⇒ y2 = 22 sin2 θ

⇒ x2 + y2 = 22(cos2 θ + sin2 θ

)⇒ x2 + y2 = 22

­2 ­1 1 2

­2

­1

1

2

x

y

r = 2

Example 6.6.7 Sketch the polar curve r = 2 cos θSolution :

r = 2 cos θ ⇒ r2 = 2r cos θ = 2x

x2 + y2 = r2 ⇒ x2 + y2 = 2x

x2 + y2 − 2x = 0

(x− 1)2

+ y2 = 1

0.5 1.0 1.5 2.0

­1.0

­0.5

0.0

0.5

1.0

x

y

r = 2 cos θ

142

Example 6.6.8 Sketch the polar curve r = 2 sin θ.Solution :

r = 2 sin θ ⇒ r2 = 2r sin θ = 2y

x2 + y2 = r2 ⇒ x2 + y2 = 2y

x2 + y2 − 2y = 0

x2 + (y − 1)2

= 1

­1.0 ­0.5 0.0 0.5 1.0

0.5

1.0

1.5

2.0

x

y

r = 2 sin θ

Example 6.6.9 Sketch the polar curve r = 1 + sin θ (cardioid).Solution :

r = 1 + sin θ ⇒ r2 = r + r sin θ = r + y

x2 + y2 = r2 ⇒ x2 + y2 = r + y

x2 + y2 − y = r(x2 + y2 − y

)2= r2(

x2 + y2 − y)2

= x2 + y2

θ 0 π6

2π6

π2

4π6

5π6 π 7π

64π3

3π2

10π6

11π6 2π

r = 1 + sin θ 1 32

12

√3 + 1 2 1

2

√3 + 1 3

2 1 12 1− 1

2

√3 0 1− 1

2

√3 1

2 1

­1.0 ­0.5 0.5 1.0

0.5

1.0

1.5

2.0

x

y

r = 1 + sin θ.

143

­1.0 ­0.5 0.5 1.0

0.5

1.0

1.5

2.0

x

y

r = 1 + sin θ

­1.0 ­0.5 0.5 1.0

­2.0

­1.5

­1.0

­0.5

xy

r = 1− sin θ

0.5 1.0 1.5 2.0

­1.0

­0.5

0.5

1.0

x

y

r = 1 + cos θ

­2.0 ­1.5 ­1.0 ­0.5

­1.0

­0.5

0.5

1.0

x

y

r = 1− cos θ

­1.5 ­1.0 ­0.5 0.0 0.5 1.0 1.5

1

2

3

x

y

r = 1 + 2 sin θ

­5 ­4 ­3 ­2 ­1

­2

­1

0

1

2

x

y

r = 1− 4 cos θ

144

­0.6 ­0.4 ­0.2 0.2 0.4 0.6

­0.6

­0.4

­0.2

0.2

0.4

0.6

x

y

r = sin 2θ

­1.0 ­0.5 0.5 1.0

­1.0

­0.5

0.5

1.0

x

y

r = cos 2θ

­0.8­0.6­0.4­0.2 0.2 0.4 0.6 0.8

­1.0

­0.5

0.5

x

y

r = sin 3θ

­0.5 0.5 1.0

­0.8­0.6­0.4­0.2

0.20.40.60.8

x

y

r = cos 3θ

­0.8­0.6­0.4­0.2 0.2 0.4 0.6 0.8

­0.8­0.6­0.4­0.2

0.20.40.60.8

x

y

r = sin 4θ

­1.0 ­0.5 0.5 1.0

­1.0

­0.5

0.5

1.0

x

y

r = cos 4θ

145

6.6.2 Tangents with Polar Coordinates

Here we will find tangent lines to polar curves. In this case we are going to assume that the equation is in the formr = f (θ). By using polar to Cartesian conversion formulas

x = r cos θ, y = r sin θ

we obtain the derivative dydx as

dy

dx=

dydθdxdθ

=drdθ sin θ + r cos θ

drdθ cos θ + r (− sin θ)

=drdθ sin θ + r cos θdrdθ cos θ − r sin θ

and the tangent line at the point (x0, y0) is

y =dydθdxdθ

(x− x0) + y0

or equivalentlydx

dθ(y − y0)−

dy

dθ(x− x0) = 0

where (x0, y0) = (r cos θ, r sin θ).

Example 6.6.10 Determine the equation of the tangent line to r = 3 + 8 sin θ at θ = π6 .

Solution :dr

dθ= 8 cos θ

dy

dx=

drdθ sin θ + r cos θdrdθ cos θ − r sin θ

=8 cos θ sin θ + (3 + 8 sin θ) cos θ

8 cos θ cos θ − (3 + 8 sin θ) sin θ

=16 cos θ sin θ + 3 cos θ

8 cos2 θ − 3 sin θ − 8 sin2 θ

The slope of the tangent line is

m =dy

dx

∣∣∣∣θ=π

6

=4√

3 + 3√3

2

4− 32

=11√

3

5

At θ = π6 we have r = 7. Then the corresponding x-y coordinates

x = 7 cosπ

6=

7√

3

2, y = 7 sin

π

6=

7

2

So the tangent line is

y =11√

3

5

(x− 7

√3

2

)+

7

2.

­5 5 10­2

2

4

6

8

10

12

x

y

146

6.6.3 Area with Polar Coordinates

In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need touse the formula for the area of a sector of a circle

A =1

2r2θ

where r is the radius and θ is the radian measure of the central angle. Let A be the region bounded by the polar curver = f(θ) and by the rays θ = α and θ = β, where f is a positive continuous function and where 0 < β − α ≤ 2π.We divide the interval [α, β] into n subintervals each of width ∆θi = θi − θi−1.

α = θ0, θ1, . . . , θn = β

The rays θ = θi divide A into n smaller regions with central angle ∆θi. If we choose θ∗i in the i

th subinterval [θi−1, θi],then the area of the ith region Ai is approximated by the area of the sector of a circle with central angle ∆θ and radiusf (θ∗i ). Thus we have

Ai ≈1

2[f (θ∗i )]

2∆θi.

So an approximation to the total area is

A ≈n∑i=1

1

2[f (θ∗i )]

2∆θi.

The approximation improves as n→∞ and the total area is exactly

A = limn→∞

n∑i=1

1

2[f (θ∗i )]

2∆θi

=

∫ β

α

1

2[f (θ)]

2dθ.

The region bounded by curves with polar equations r = f (θ) , r = g (θ) , θ = α and θ = β where f (θ) ≥ g (θ) ≥ 0 and0 < β − a ≤ 2π is

A =

∫ β

α

1

2

([f (θ)]

2 − [g (θ)]2)dθ

Example 6.6.11 Determine the area of the inner loop of r = 2 + 4 cos θ.Solution :

1 2 3 4 5 6

­3

­2

­1

0

1

2

3

x

y

r = 2 + 4 cos θ

First we find α and β.

2 + 4 cos θ = 0⇒ cos θ = −1

2⇒ θ =

2π

3,

4π

3

147

The area is

A =

∫ β

α

1

2[f (θ)]

2dθ =

∫ 4π3

2π3

1

2[2 + 4 cos θ]

2dθ

=

∫ 4π3

2π3

1

2

[4 + 16 cos θ + 16 cos2 θ

]2dθ

=

∫ 4π3

2π3

[2 + 8 cos θ + 8

1

2(1 + cos 2θ)

]2dθ

=

[2θ + 8 sin θ + 4

(θ +

1

2sin 2θ

)]∣∣∣∣ 4π32π3

= 4π − 6√

3.

Example 6.6.12 : Find the area enclosed by one loop of the four-leaved rose r = cos 2θ.Solution :

­1.0 ­0.5 0.5 1.0

­1.0

­0.5

0.5

1.0

x

y

r = cos 2θ

Notice that the region enclosed by the right loop is swept out by a ray that rotates from -π4 toπ4 . Hence

∫ π4

−π412 [cos 2θ]

2dθ =

18π = 1

6π −116

√3

A =

∫ β

α

1

2[f (θ)]

2dθ =

∫ π4

−π4

1

2[cos 2θ]

2dθ

=

∫ π4

−π4

1

2[cos 2θ]

2dθ =

1

2

∫ π4

−π4

1

2(1 + cos 4θ) dθ

=1

4

(θ +

1

4sin 4θ

)∣∣∣∣π4−π4

=π

8.

Example 6.6.13 Find the area that lies inside r = 3 + 2 sin θ and outside r = 2.Solution :

­3 ­2 ­1 1 2 3

­2

2

4

x

y

148

First we find α and β.

3 + 2 sin θ = 2⇒ sin θ = −1

2⇒ θ =

7π

6,−π

6

Therefore the area is

A =

∫ β

α

1

2

([f (θ)]

2 − [g (θ)]2)dθ =

∫ 7π6

−π6

1

2

([3 + 2 sin θ]

2 − [2]2)dθ

=

∫ 7π6

−π6

1

2

5 + 12 sin θ + 4 sin2 θ︸ ︷︷ ︸12 (1−cos 2θ)

dθ

=1

2

∫ 7π6

−π6(7 + 12 sin θ − 2 cos 2θ) dθ

=1

2

(7θ − 12 cos θ − 2

1

2sin 2θ

)∣∣∣∣ 7π6−π6

=11√

3

2+

14π

3.

149

6.6.4 Arc Length with Polar Coordinates

In this section we’ll look at the arc length of the curve given by

r = f (θ) , α ≤ θ ≤ β

where we also assume that the curve is traced out exactly once. At first we write the curve in terms of a set of parametricequations,

x = r cos θ = f (θ) cos θ

y = r sin θ = f (θ) sin θ

and we can now use the parametric formula

L =

∫ β

α

√[dx

dθ

]2+

[dy

dθ

]2dθ

for finding the arc length. Now we find the derivatives :

dx

dθ= f ′ (θ) cos θ − f (θ) sin θ

=dr

dθcos θ − r sin θ

dy

dθ= f ′ (θ) sin θ + f (θ) cos θ

=dr

dθsin θ + r cos θ

Then the arc length formula for polar coordinates is

L =

∫ β

α

√[dr

dθcos θ − r sin θ

]2+

[dr

dθsin θ + r cos θ

]2dθ

=

∫ β

α

√[dr

dθ

]2cos2 θ − 2r

dr

dθcos θ sin θ + r2 sin2 θ +

[dr

dθ

]2sin2 θ + 2r

dr

dθcos θ sin θ + r2 cos2 θdθ

=

∫ β

α

√[dr

dθ

]2 (cos2 θ + sin2 θ

)+ r2

(cos2 θ + sin2 θ

)dθ

=

∫ β

α

√[dr

dθ

]2+ r2dθ

Example 6.6.14 Find the length of the cardioid r = 1 + sin θ.Solution :

­1.0 ­0.5 0.5 1.0

0.5

1.0

1.5

2.0

x

y

r = 1 + sin θ

150

L =

∫ β

α

√[dr

dθ

]2+ r2dθ =

∫ 2π

0

√[cos θ]

2+ (1 + sin θ)

2dθ

=

∫ 2π

0

√2 (1 + sin θ)dθ =

√2

∫ 2π

0

√(1 + sin θ) (1− sin θ)

1− sin θdθ

=√

2

∫ 2π

0

|cos θ|√1− sin θ

dθ

=√

2

∫ π2

0

cos θ√1− sin θ

dθ −√

2

∫ 3π2

π2

cos θ√1− sin θ

dθ +√

2

∫ 2π

3π2

|cos θ|√1− sin θ

dθ

(u = 1− sin θ ⇒ du = − cos θdθ

⇒∫

cos θ√1− sin θ

dθ = −∫

1√udu = −u

12

12

+ c = −2√

1− sin θ)

L = −2√

2√

1− sin θ∣∣∣π20

+ 2√

2√

1− sin θ∣∣∣ 3π2π2

− 2√

2√

1− sin θ∣∣∣2π3π2

= 8.

151

6.7 Other Problems Related Definite Integral

Here we show how the concept of definite integral can be applied to more general problems

6.7.1 Distance

Now, we study the problem of finding the distance traveled by an object with variable velocity during a certain period oftime.If the velocity v is constant we just multiply it by the time t and

net distance = v × t.

If the velocity v is variable, we can approximate the total distance traveled by dividing the total time interval into smallintervals so that in each of them the velocity varies very little and can be considered approximately constant. So, assumethat the body starts moving at time t0 and finishes at time t1, and the velocity function is v(t) where v (t) is continuous on[t0, t1]. We divide the time interval into n small intervals [ti−1, ti] of length ∆t = t1−t0

n , choose some instant t∗i ∈ [ti−1, ti]Then the distance traveled during that time interval is approximately v (t∗i ) ∆t and the total distance can be approximatedas the sum

n∑i=1

v (t∗i ) ∆t.

The result will be more accurate with larger n and the exact distance traveled will be limit of the above expression as ngoes to infinity :

net distance= limn→∞

n∑i=1

v (t∗i ) ∆t.

That limit turns out to be the following definite integral :

net distance=∫ t1

t0

v (t) dt.

In the computation of the displacement the distance traveled by the object when it moves to the left (while v(t) ≤ 0) issubtracted from the distance traveled to the right (while v(t) ≥ 0). If we want to find the total distance traveled we needto add all distances with a positive sign, and this is accomplished by integrating the absolute value of the velocity :

total distance =

∫ t2

t1

|v (t)| dt.

Example 6.7.1 Find the displacement and the total distance traveled by an object that moves with velocity v(t) = t2−t−6from t = 1 to t = 4.Solution :

net distance =

∫ 4

1

(t2 − t− 6

)dt =

[t3

3− t2

2− 6t

]∣∣∣∣41

= −9

2

In order to find the total distance traveled, we need to separate the intervals in which the velocity takes values of differentsigns.

1 2 3 4 5

­6

­4

­2

0

2

4

6

x

y

152

t2 − t− 6 ≤ 0 =⇒ t ∈ [1, 3]

t2 − t− 6 ≥ 0 =⇒ t ∈ [3, 4]

Hence :

total distance =

∫ 4

1

|v (t)| dt

= −∫ 3

1

(t2 − t− 6

)dt+

∫ 4

3

(t2 − t− 6

)dt

= −[t3

3− t2

2− 6t

]∣∣∣∣31

+

[t3

3− t2

2− 6t

]∣∣∣∣43

= −22

3+

17

6

=61

6.

153

6.7.2 Work

If a constant force of magnitude F is applied in the direction of motion of an object and if, as a result, the object movesa distance d, then the work W done on the object is

W = F.d

Note that in the metric system, distance is measured in meters (m), time is measured in seconds (s), and thereforeacceleration is measured in m

s2 . Mass is measured in kilograms (kg), so force is measured in kg ×ms2 , which are called

newtons (N), and, finally, work is measured in N ×m, which are also called joules (J). One joule is the work that is donewhen a force equal to 1 newton moves an object a distance of 1 meter.

Example 6.7.2 How much work is needed to lift a child of 20 kg to a height of 0.5 meters? Use the fact that gravitationcauses downward acceleration of g = 9.8 m/s2.Solution : In order to lift the child, one needs to overcome the downward acceleration caused by gravity. This means thatan upward force of

F = m.g = 20kg × 9.8m

s2= 198N

has to be exerted across a distance of d = 0.5 meters. This yields

W = F.d = 198N × 0.5m = 99J.

So the work needed is 99J .

Usually, forces applied to objects are not constant.Let a and b be real numbers. If an object is moved from a to b by a force that is equal to f(x) at point x, where f is

continuous on [a, b], then the total work done by the force on [a, b] is

W =

∫ b

0

f (x) dx.

Example 6.7.3 The force needed to extend a given spring x centimeters over its natural length is given by the functionf(x) = 70x.How much work is needed to extend the spring 10 cm over its natural length?Solution :

W =

∫ 0.1

0

f (x) dx =

∫ 0.1

0

70xdx

=

(x2

35

)∣∣∣∣0.10

= 0.35J.

So 0.35J of work is needed to stretch the spring 10 cm over its natural length.

154

6.7.3 Mass

Consider a thin (1-dimensional) metal wire lying along the x-axis from a to b.

a b

If the metal has constant density (homogeneous), say ρ0 (grcm ), then the mass M of the wire is just

M = ρ0 × (b− a) gr.

Suppose, that the metal is inhomogeneous (density varies from point to point along the wire) and density is equal to ρ (x)at point x where ρ is continuous on [a, b]. Then the mass M of the wire from a to b is

W =

∫ b

a

ρ (x) dx.

Example 6.7.4 A wire lies along the x-axis from 0 to1 and has a density at each point ρ (x) = x2 . Find the the mass of

wire.Solution : ∫ 1

0

ρ (x) dx =

∫ 1

0

x

2dx

=x2

4

∣∣∣∣10

=1

4.

155