Chapter 5: Integration

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Section 5.1 An Area Problem; A Speed-Distance Problem

a. An Area Problem

b. An Area Problem (continued)

c. Upper Sums and Lower Sums

d. Overview of the Speed Distance Problem

Section 5.2 The Definite Integral of a Continuous Function

a. Partitions

b. Partitions; Upper Sum and Lower Sum

c. Example

d. Definition 5.2.3

e. Use of "dummy" Variables

f. Integral of a Constant Function

g. Integral of the Identity Function

h. The Integral as the Limit of Riemann Sums

Section 5.3 The Function

a. Theorem 5.3.1

b. The Effects of Adding Points to a Partition

c. Theorem 5.3.2

d. Properties of Integration

e. Theorem 5.3.5

f. Integration when f > 0

g. Example

Section 5.4 The Fundamental Theorem of Integral Calculus

a. Antiderivatives and the Fundamental Theorem of Integral Calculus

b. Example

c. Some Common Antiderivatives and Examples

d. Linearity of the Integral

e. Example

Chapter 5: Integration

Section 5.5 Some Area Problems

a. Area of Ω

b. Signed Area

Section 5.6 Indefinite Integrals

a. Indefinite integrals

b. Common Indefinite Integrals

c. Linearity Properties

d. Application to Motion Example

Section 5.7 Working Back from the Chain Rule; The u-Substitution

a. Theorem 5.7.1

b. Example

c. Substitution in Definite Integrals

Section 5.8 Additional Properties of the Definite Integral

a. Properties I and II

b. Properties III and IV

c. Property V

d. Property VI

Section 5.9 Mean-Value Theorems for Integrals; Average Value of a Function

a. First Mean-Value Theorem for Integrals

b. Area of Ω

c. Second Mean-Value Theorem for Integrals

( )

^x

( )

a

F x = ∫ f t dt

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An Area Problem; A Speed-Distance Problem

In Figure 5.1.1 you can see a region Ω bounded above by the graph of a continuous

function f, bounded below by the x-axis, bounded on the left by the line x = a, and bounded on the right by the line x = b. The question before us is this: What number, if any, should be called the area of Ω?

To begin to answer this question, we split up the interval [a, b] into a finite number of subintervals

[x ₀ , x ₁ ], [x ₁ , x ₂ ], . . . , [x _n−1 , x _n ] with a = x ₀ < x ₁ < · · · < x _n = b . This breaks up the region Ω into n subregions:

Ω 1 , Ω 2 , . . . , Ω n . (Figure 5.1.2)

We can estimate the total area of Ω by estimating the area

of each subregion Ω _i and adding up the results.

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An Area Problem; A Speed-Distance Problem

Adding up these inequalities, we get on the one hand

and on the other hand

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An Area Problem; A Speed-Distance Problem

A sum of the form

m ₁ Δx ₁ + m ₂ Δx ₂ +· · ·+m _n Δx _n (Figure 5.1.4)

is called a lower sum for f.

A sum of the form

M ₁ Δx ₁ + M ₂ Δx ₂ +· · ·+ M _n Δx _n (Figure 5.1.5)

is called an upper sum for f.

For a number to be a candidate for the title “area of Ω,” it must be greater than or

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An Area Problem; A Speed-Distance Problem

If an object moves at a constant speed for a given period of time, then the total distance traveled is given by the familiar formula

distance = speed × time.

Suppose now that during the course of the motion the speed ν does not remain constant;

suppose that it varies continuously. How can we calculate the distance traveled in that case?

To answer this question, we suppose that the motion begins at time a, ends at time b, and during the time interval [a, b] the speed varies continuously.

As in the case of the area problem, we begin by breaking up the interval [a, b] into a finite number of subintervals:

[t ₀ , t ₁ ], [t ₁ , t ₂ ], . . . , [t _n ₋₁ , t _n ] with a = t ₀ < t ₁ < · · · < t _n = b.

On each subinterval [t _i ₋₁ , t _i ] the object attains a certain maximum speed M _i and a certain minimum speed m _i .

The total distance traveled during the full time interval [a, b], call it s, must be the sum of the distances traveled during the subintervals [t _i ₋₁ , t _i ]; thus we must have

s = s ₁ + s ₂ + · · · + s _n .

lower sum for the speed function, and it must be less than or equal to every upper sum. It

turns out that there is one and only one such number, and this is the total distance traveled.

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The Definite Integral of a Continuous Function

Example The sets

{0, 1}, {0, ½, 1}, {0, ¼, ½, 1}, {0, ¹ / 4 , ¹ / 3 , ¹ / 2 , ⁵ / 8 , 1}

are all partitions of the interval [0, 1].

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The Definite Integral of a Continuous Function

If P = {x ₀ , x ₁ , x ₂ , . . . , x _n−1 , x _n } is a partition of [a, b], then P breaks up [a, b]

into n subintervals

[x ₀ , x ₁ ], [x ₁ , x ₂ ], . . . , [x _n−1 , x _n ] of lengths Δx ₁ , Δx ₂ , . . . , Δx _n .

Suppose now that f is continuous on [a, b]. Then on each subinterval [x _i−1 , x _i ]

the function f takes on a maximum value, M _i , and a minimum value, m _i .

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The Definite Integral of a Continuous Function

Example

The function f (x) = 1 + x ² is continuous on [0, 1]. The partition P = {0, ½, ¾, 1}

breaks up [0, 1] into three subintervals

[x ₀ , x ₁ ] = [0, ½], [x ₁ , x ₂ ] = [ ½ , ¾] , [x ₂ , x ₃ ] = [ ¾ , 1]

of lengths

Δx 1 = ½ _{− 0 =} ½ , Δ ^x 2 = ¾ ₋ ½ = ¼, Δ ^x 3 = 1 − ¾ = ¼ . Since f increases on [0, 1], it takes on its maximum value at the right endpoint of each subinterval:

The minimum values are taken on at the left endpoints:

Thus

( )

1 2 3

1 5 3 25

, , 1 2

2 4 4 16

M = f       = M = f       = M = f =

( )

1 2 3

1 5 3 25

0 1, ,

2 4 4 16

m = f = m = f       = m = f       =

( ) ^{5 1} ^{25 1} ² ¹ ⁹⁷ ^1.52

U P = M ∆ + x M ∆ + x M ∆ = x     +     +     = ≅

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The Definite Integral of a Continuous Function

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The Definite Integral of a Continuous Function

In the expression

the letter x is a “dummy variable”; in other words, it can be replaced by any letter not already in use. Thus, for example,

all denote exactly the same quantity, the definite integral of f from a to b.

From the introduction to this chapter, you know that if f is nonnegative and

continuous on [a, b], then the integral of f from x = a to x = b gives the area below the graph of f from x = a to x = b:

( )

b

a f x dx

∫

( ) ( ) ^,

b

a f x d x

∫ ∫ a ^b f t dt ^{( )} ^, ∫ a ^b f z dz ^{( )}

( ) ^.

A = ∫ b f x dx

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The Definite Integral of a Continuous Function

The integral of a constant function as shown in Figure 5.2.3:

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The Definite Integral of a Continuous Function

The integral of the identity function as shown in Figure 5.2.5:

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The Integral as the Limit of Riemann Sums

Figure 5.2.8 illustrates the idea that the definite integral of a continuous function is the limit of Riemann sums . Here the base interval is broken up into eight

subintervals. The point is chosen from [x ₀ , x ₁ ], from [x ₁ , x ₂ ], and so on.

which in expanded form reads

*

x 1 x ₂ ^*

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( )

The Function ( )

x

a

F x = ∫ f t dt

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( )

The Function ( )

x

a

F x = ∫ f t dt

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( )

The Function ( )

x

a

F x = ∫ f t dt

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Until now we have integrated only from left to right: from a number a to a number b greater than a. We integrate in the other direction by defining

The integral from any number to itself is defined to be zero:

( )

The Function ( )

x

a

F x = ∫ f t dt

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( )

The Function ( )

x

a

F x = ∫ f t dt

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F(x) = area from a to x and F(x + h) = area from a to x + h. Therefore F(x + h) – F(x) = area from x to x + h. For small h this is approximately f (x) h. Thus

( ) ( )

F x h F x h

+ − is approximately ^{f x h} ( ) ^{f x} ( )

h =

( )

The Function ( )

x

a

F x = ∫ f t dt

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Example

Set for all real numbers x.

(a) Find the critical points of F and determine the intervals on which F increases and the intervals on which F decreases.

(b) Determine the concavity of the graph of F and find the points of inflection (if any).

(c) Sketch the graph of F.

Solution

(a) To find the intervals on which F increases and the intervals on which F decreases, we examine the first derivative of F. By Theorem 5.3.5,

for all real x.

Since F ´ (x) > 0 for all real x, F increases on (−∞,∞); there are no critical points.

(b) To determine the concavity of the graph and to find the points of inflection, we use the second derivative

The sign of F ´´ and the behavior of the graph of F are as follows:

( ) ₂

0 1 1

F x x dt

= t

∫ +

( ) ¹ ₂

F x 1

′ = x +

( ) ( ¹ ² ^x ² ) ²

F x

x

′′ = − +

( )

The Function ( )

x

a

F x = ∫ f t dt

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The Fundamental Theorem of Integral Calculus

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The Fundamental Theorem of Integral Calculus

Example Evaluate

Solution

As an antiderivative for f (x) = x ² , we can use the function G(x) = ⅓x ³ .

By the fundamental theorem,

NOTE: Any other antiderivative of f (x) = x ² has the form H(x) = ⅓x ³ + C for some constant C. Had we chosen such an H instead of G, then we would have had

4 2

1 x dx

∫

( ) ( ) ( ) ( )

4 2 3 3

1 1 1 64 1

4 1 4 1 21

3 3 3 3

x dx = G − G = − = − =

∫

   

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The Fundamental Theorem of Integral Calculus

Some examples:

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The Linearity of the Integral

I. Constants may be factored through the integral sign:

II. The integral of a sum is the sum of the integrals:

III. The integral of a linear combination is the linear combination of the integrals:

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The Linearity of the Integral

Example Evaluate

[ ]

4 0 ^π sec x 2 tan x − 5sec x dx

∫

Solution

[ ]

[ ] [ ]

[ ]

4 4

2 0 0

4 4

2 0 0

4 4

0 0

sec 2 tan 5sec 2sec tan 5sec

2 sec tan 5 sec

2 sec 5 tan

2 sec sec 0 5 tan tan 0

4 4

2 2 1 5 1 0 2 2 7

x x x dx x x x dx

x x dx x dx

x x

π π

 

− =  − 

= −

   

=   −   −   −  

 

=  − =  − = −

∫ ∫

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Some Area Problems

Look at the region Ω shown in Figure 5.5.4. The upper boundary of Ω is the graph of a nonnegative function f and the lower boundary is the graph of a nonnegative function g.

We can obtain the area of Ω by calculating the area of Ω ₁ and subtracting off the area of Ω ₂ . Since

( ) ( )

1 2

area of ^b and area of ^b

a f x dx a g x dx

Ω = ∫ Ω = ∫

we have

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Some Area Problems

The area between the graph of f and the x-axis from x = a to x = e is the sum area of Ω ₁ + area of Ω ₂ + area of Ω ₃ + area of Ω ₄

This area is

( ) ( ) ( ) ( ) ^.

b c d e

a f x dx − b f x dx + c f x dx − d f x dx

∫ ∫ ∫ ∫

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Indefinite Integrals

Consider a continuous function f . If F is an antiderivative for f on [a, b], then

If C is a constant, then

Thus we can replace (1) by writing

If we have no particular interest in the interval [a, b] but wish instead to emphasize that F is an antiderivative for f , which on open intervals simply means that F ´ = f , then we omit the a and the b and simply write

( ) ( )

(1) ^b ^b

a f x dx =   F x   a

∫

( ) ^b _a ( ) ( ) ( ) ( ) ( ) ^b _a

F x C F b C F a C F b F a F x

 +  =  +  −  +  = − =  

       

( ) ( ) ^.

b b

a f x dx =   F x +  C  a

∫

( ) ( )

f x dx = F x + C

∫

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Indefinite Integrals

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Indefinite Integrals

The linearity properties of definite integrals also hold for indefinite integrals.

Example

Calculate ∫   5 x ^{3 / 2} − 2 csc ² x dx   Solution

( )

3 / 2 2 3 / 2 2

5 / 2

1 2

5 / 2

5 2 csc 5 2 csc

5 2 2 cot

x x dx x dx xdx

x C x C

x x C

 −  = −

 

=     + − − +

 

= + +

∫ ∫ ∫

(writing C for C + C )

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Indefinite Integrals

Application to Motion Example

An object moves along a coordinate line with velocity v(t) = 2 − 3t + t ² units per second.

Its initial position (position at time t = 0) is 2 units to the right of the origin. Find the position of the object 4 seconds later.

Solution

Let x(t) be the position (coordinate) of the object at time t. We are given that x(0) = 2. Since x ´ (t) = v(t),

Since x(0) = 2 and x(0) = 2(0) − ³ ∕ ² (0) ² + ⅓(0) ³ + C = C, we have C = 2 and x(t) = 2t − ³ ∕ ² ^t ² + ⅓t ³ + 2.

The position of the object at time t = 4 is the value of this function at t = 4:

x (4) = 2(4) − ³ ∕ ² ⁽⁴⁾ ² ⁺ ⅓(4) ³ + 2 = 7 ⅓

At the end of 4 seconds the object is 7 ⅓ units to the right of the origin.

The motion of the object is represented schematically in Figure 5.6.1.

( ) ( ) ( ^{2 3} ² ) ² ³ ² ¹ ³

2 3

x t = ∫ v t dt = ∫ − + t t dt = t − t + t + C

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The u-Substitution

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The u-Substitution

Example Calculate Solution

Set u = 3 + 5x, du = 5 dx. Then

( ) ²

1 3 5

dx + x

∫

( )

( ) ( )

2 2 2

2 1

2 1 1 1 1

5 5

3 5

1 1 1 1

5 5 5 3 5

3 5

dx du u du

x u

dx u du u C C

x x

−

− −

 

=     = +

= = − + = − +

+ +

∫ ∫

and

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The u-Substitution

The Definite Integral

This formula is called the change-of-variables formula. The formula can be used to evaluate provided that u ´ is continuous on [a, b]

and f is continuous on the set of values taken on by u on [a, b]. Since u is continuous, this set is an interval that contains a and b.

( ( ) ) ^{( )}

b

a f u x u x dx ′

∫

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Additional Properties of the Definite Integral

I. The integral of a nonnegative continuous function is nonnegative:

The integral of a positive continuous function is positive:

II. The integral is order-preserving: for continuous functions f and g,

and

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Additional Properties of the Definite Integral

III. Just as the absolute value of a sum of numbers is less than or equal to the sum of the absolute values of those numbers,

|x ₁ + x ₂ +· · ·+ x _n | ≤ |x ₁ | + |x ₂ |+· · ·+|x _n |,

the absolute value of an integral of a continuous function is less than or equal to the integral of the absolute value of that function:

IV. If f is continuous on [a, b], then

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Additional Properties of the Definite Integral

V. If f is continuous on [a, b] and u is a differentiable function of x with values in [a, b], then for all u(x) ∈ (a, b)

Example Find

Solution

At this stage you probably cannot carry out the integration: it requires the

natural logarithm function. (Not introduced in this text until Chapter 7.) But for our purposes, that doesn’t matter. By (5.8.7),

3

0 1 1 d x

dx t dt

 

 + 

 ∫ 

3 2

2 3 3

0 1 1 3

1 1 3 1

d x x

dt x

dx t x x

  = =

 +  + +

 ∫ 

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Additional Properties

VI. Now a few words about the role of symmetry in integration. Suppose that f is

continuous on an interval of the form [−a, a], a closed interval symmetric about

Chapter 5: Integration

Main Menu

Section 5.1 An Area Problem; A Speed-Distance Problem

a. An Area Problem

b. An Area Problem (continued)

c. Upper Sums and Lower Sums

d. Overview of the Speed Distance Problem

Section 5.2 The Definite Integral of a Continuous Function

a. Partitions

b. Partitions; Upper Sum and Lower Sum

c. Example

d. Definition 5.2.3

e. Use of "dummy" Variables

f. Integral of a Constant Function

g. Integral of the Identity Function

h. The Integral as the Limit of Riemann Sums

Section 5.3 The Function

a. Theorem 5.3.1

b. The Effects of Adding Points to a Partition

c. Theorem 5.3.2

d. Properties of Integration

e. Theorem 5.3.5

f. Integration when f > 0

g. Example

Section 5.4 The Fundamental Theorem of Integral Calculus

a. Antiderivatives and the Fundamental Theorem of Integral Calculus

b. Example

c. Some Common Antiderivatives and Examples

d. Linearity of the Integral

e. Example

Chapter 5: Integration

Section 5.5 Some Area Problems

a. Area of Ω

b. Signed Area

Section 5.6 Indefinite Integrals

a. Indefinite integrals

b. Common Indefinite Integrals

c. Linearity Properties

d. Application to Motion Example

Section 5.7 Working Back from the Chain Rule; The u-Substitution

a. Theorem 5.7.1

b. Example

c. Substitution in Definite Integrals

Section 5.8 Additional Properties of the Definite Integral

a. Properties I and II

b. Properties III and IV

c. Property V

d. Property VI

Section 5.9 Mean-Value Theorems for Integrals; Average Value of a Function

a. First Mean-Value Theorem for Integrals

b. Area of Ω

c. Second Mean-Value Theorem for Integrals

( )

( )

F x = ∫ f t dt

An Area Problem; A Speed-Distance Problem

In Figure 5.1.1 you can see a region Ω bounded above by the graph of a continuous

function f, bounded below by the x-axis, bounded on the left by the line x = a, and bounded on the right by the line x = b. The question before us is this: What number, if any, should be called the area of Ω?

To begin to answer this question, we split up the interval [a, b] into a finite number of subintervals

[x 0 , x 1 ], [x 1 , x 2 ], . . . , [x n−1 , x n ] with a = x 0 < x 1 < · · · < x n = b . This breaks up the region Ω into n subregions:

Ω 1 , Ω 2 , . . . , Ω n . (Figure 5.1.2)

We can estimate the total area of Ω by estimating the area

of each subregion Ω i and adding up the results.

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An Area Problem; A Speed-Distance Problem

Adding up these inequalities, we get on the one hand

and on the other hand

An Area Problem; A Speed-Distance Problem

A sum of the form

m 1 Δx 1 + m 2 Δx 2 +· · ·+m n Δx n (Figure 5.1.4)

is called a lower sum for f.

A sum of the form

M 1 Δx 1 + M 2 Δx 2 +· · ·+ M n Δx n (Figure 5.1.5)

is called an upper sum for f.

For a number to be a candidate for the title “area of Ω,” it must be greater than or

Main Menu

An Area Problem; A Speed-Distance Problem

If an object moves at a constant speed for a given period of time, then the total distance traveled is given by the familiar formula

distance = speed × time.

Suppose now that during the course of the motion the speed ν does not remain constant;

[x ₀ , x ₁ ], [x ₁ , x ₂ ], . . . , [x _n−1 , x _n ] with a = x ₀ < x ₁ < · · · < x _n = b . This breaks up the region Ω into n subregions:

of each subregion Ω _i and adding up the results.

m ₁ Δx ₁ + m ₂ Δx ₂ +· · ·+m _n Δx _n (Figure 5.1.4)

M ₁ Δx ₁ + M ₂ Δx ₂ +· · ·+ M _n Δx _n (Figure 5.1.5)

[t ₀ , t ₁ ], [t ₁ , t ₂ ], . . . , [t _n ₋₁ , t _n ] with a = t ₀ < t ₁ < · · · < t _n = b.

On each subinterval [t _i ₋₁ , t _i ] the object attains a certain maximum speed M _i and a certain minimum speed m _i .

The total distance traveled during the full time interval [a, b], call it s, must be the sum of the distances traveled during the subintervals [t _i ₋₁ , t _i ]; thus we must have

s = s ₁ + s ₂ + · · · + s _n .

{0, 1}, {0, ½, 1}, {0, ¼, ½, 1}, {0, ¹ / 4 , ¹ / 3 , ¹ / 2 , ⁵ / 8 , 1}

If P = {x ₀ , x ₁ , x ₂ , . . . , x _n−1 , x _n } is a partition of [a, b], then P breaks up [a, b]

[x ₀ , x ₁ ], [x ₁ , x ₂ ], . . . , [x _n−1 , x _n ] of lengths Δx ₁ , Δx ₂ , . . . , Δx _n .

Suppose now that f is continuous on [a, b]. Then on each subinterval [x _i−1 , x _i ]

the function f takes on a maximum value, M _i , and a minimum value, m _i .

The function f (x) = 1 + x ² is continuous on [0, 1]. The partition P = {0, ½, ¾, 1}

[x ₀ , x ₁ ] = [0, ½], [x ₁ , x ₂ ] = [ ½ , ¾] , [x ₂ , x ₃ ] = [ ¾ , 1]

Δx 1 = ½ _{− 0 =} ½ , Δ ^x 2 = ¾ ₋ ½ = ¼, Δ ^x 3 = 1 − ¾ = ¼ . Since f increases on [0, 1], it takes on its maximum value at the right endpoint of each subinterval:

( ) ^{5 1} ^{25 1} ² ¹ ⁹⁷ ^1.52

( ) ( ) ^,

∫ ∫ a ^b f t dt ^{( )} ^, ∫ a ^b f z dz ^{( )}

( ) ^.