1

Chapter 6: Integration

In this chapter, we will study:

6.1 Indefinite integration and the reverse chain rule

6.2 Finding the area under a curve ( ) 0f x by using integrals

6.3 Application of integration: area below the x axis and the area between two curves

6.4 Application of integration: the area between a curve and the y axis

6.5 Application of integration: Volumes of solids of revolution

6.6 Introduction to improper integrals

6.1 Indefinite integration and the reverse chain rule

A. Indefinite integration:

In chapter 5 we have studied methods to calculate the derivative ' ( )f x of a given function ( )f x .

However, in many scientific problems (such as most problems in mechanics), we are interested in the reverse problem:

the derivative of a function is given (such as when the acceleration ( )a t of a position function ( )x t is given), and we

want to find the original function (in this case the speed and afterwards the position). For a more specific example from

physics: see the free fall – free throw type problems.

Our main goal of this section is to develop a general method to find a function whose derivative is given.

Example:

Find ( )y F x with (1) 2dy

xdx

Note: An equation of type (1) , where we solve for the unknown function ( )y x , is called a differential equation, and

the process of finding ( )y x from such an equation is called integration.

We remember that 2 ' 2x x , therefore we can see that one function 2( ) which solves (1) is ( )y x y x x .

However, since ' 0c , we see that other solutions of (1) are 2 2 21, 2, 10 x x x and so on.

Moreover, note that the general solution of the equation (1) is (2) 2( )y x x c , where c is a generic constant.

3 of such solutions of the form (2) are shown in Figure 6.1 below. All 3 of them are solutions of the differential equation

(1).

2

To find a particular solution of (1), an extra condition needs to be given, such as: (3) (2) 1y

If (3) is given, then using (3) in (2), we find: 24 1 3 ( ) 3c c y x x .

Therefore, the particular solution of the set of equations (1) and (3) is: 2( ) 3y x x (this is shown as the red curve

above).

In general, to solve an equation ( )

( )dF x

f xdx

, we will find a function ( )F x whose derivative is ( )f x . This function

( )F x is called the indefinite integral of ( )f x (or the anti-derivative of ( )f x ) and it is denoted by ( ) ( )F x f x dx .

Therefore: ( )

' ( ) ( )d f x dx

F x f xdx

.

In particular, an important rule, which we deduce from: 1'n nx nx is:

1

(4) , for all n N with 11

nn x

x dx c nn

.

Particular cases of (4) are:

2

32

1

+2

3

dx x c

xx dx c

xx dx c

Note that rule (4) works for negative (except 1n ) and even fractional powers, therefore:

12

31 322 2

1

1

2

3 3

2

xx dx c c

x

xx dx c x c

3

Other rules for ( )f x dx are:

(5) ( ) ( ) ( ) ( )

( ) ( )

f x g x dx f x dx g x dx

k f x dx k f x dx

Note however that in general:

( ) ( ) ( ) ( ) (unless one of ( ) or ( ) is a constant)

( )( )

( ) ( )

f x g x dx f x dx g x dx f x g x

f x dxf xdx

g x g x dx

Example 1 (6.1 page 175 textbook):

Given that 23 4 3dy

x xdx

a) Find the general solution ( )y x of this differential equation

b) Find the equation of the curve with this gradient function which passes through 1,10

Solution:

a) Using (5) and (4), we find that: 3 2

3 2( ) 3 4 3 2 33 2

x xy x x c x x x c

b) Using 3 2(1) 10 6 10 4 ( ) 2 3 4y c c y x x x x

Example 2 (6.2 page 176 textbook):

A curve is such that (6) 2

83

dyx

dx x . Given that the point 4,20 lies on the curve, find the equation of the curve.

Solution:

The solution of the equation (6) is

33122

8 8( ) 3 8 2 2

3 1

2

x xy x c x c x x c

x x

Using that 8

(4) 20 16 2 20 6 ( ) 2 6y c c y x x xx

4

Example 3 (problem 8 page 178 textbook):

Solution:

i) Integrating the given differential equation, we find: 3 2( )y x x x x c . Using the condition

3 2(2) 3 8 4 2 3 2 3 1 ( ) 1y c c c y x x x x

ii) The stationary points are given by: 2

1 2

1'( ) 0 3 2 1 0 3 1 1 0 and 1

3y x x x x x x x

are the two stationary points of the graph . Calculate 1 32

and (1) 03 27

y y

, therefore the two stationary

points are 1 32

, and 1,03 27

.

iii) A monotonicity table of ( )y x indicates that

1 32

, is a local maximum point and that 1,0 is a local minimum point3 27

. Using also the values

and y y ,we sketch the following graph:

5

B. The reverse chain rule:

The generalized power rule learned in section 5.7 is (7) 1

( ) ' ( ) '( )nnu x n u x u x

,

In particular, (7) implies: (8) 1

'n n

ax b a n ax b

Reversing this formula, we obtain: (9)

1

1

n

n ax bax b dx c

a n

Example 1:

Evaluate:

a) 3

4 2x dx

b) 6

28 2 5x dx

c)

4

1

2 1dx

x

d) 8

1 8dx

x

Note: In integration, if you are not certain if you have calculated a certain indefinite integral correctly, you can always

calculate the derivative of your answer, and check to see if it produces the original function. If that is the case, then the

anti-derivative is correct.

Example 2:

Past papers:

1. June 2002

6

2. November 2002

3. June 2006

6.2 Finding the area under a curve ( ) 0f x by using integrals

In the previous section we developed formulas to calculate the anti-derivatives of functions of the form n

ax b or a

combinations of these.

Aside from the applications of these functions in differential equations and mechanics, we will see in this chapter that

indefinite integrals are also used as the main tool to calculate areas of planar regions and volumes of solids of

revolutions.

In this section we develop a simple method to calculate the area under a curve ( ) 0f x , such as the area shown in

Figure 6.2 below:

Figure 6.2: Area below the graph of a function ( ) 0f x

7

The main formula which we will use is:

If ( ) 0f x , then the area between the graph of ( )f x , the x axis, and the vertical lines and x a x b is:

(10) ( ) ( ) ( ), where '( ) ( )

b

a

Area f x dx F b F a F x f x , that is: ( ) ( )F x f x dx , so

( )F x is an anti-derivative of ( )f x .

( )

b

a

f x dx is called the definite integral of ( )f x .

Formula (10) is the most important formula in this chapter, and it is usually regarded as the most important formula in

Calculus. It is called the fundamental theorem of Calculus.

It gives the area of a plane region as shown in Figure 6.2 in terms of the antiderivative ( )F x of function ( )f x .

Examples:

1. Consider ( ) 2f x x .

Calculate the area under the graph of ( )f x , with 0 x b in two ways : first geometrically, and then using

formula (1) above. Do you obtain the same value? Which method do you prefer ?

2. Consider ( ) 3f x x .

Calculate the area under the graph of ( )f x , with a x b in two ways : first geometrically, and then using

formula (1) above. Do you obtain the same value? Which method do you prefer ?

3. Consider ( ) 3f x x .

Calculate the area under the graph of ( )f x , with a x b in two ways : first geometrically, and then using

formula (1) above. Do you obtain the same value? Which method do you prefer ?

4.

8

5.

6.

7.

8. Calculate the following definite integrals:

iii)

9

9. Do problems 18 to 21 from Exercise set 6B in the textbook.

6.3 Application of integration: area below the x axis and the area between two curves

A. Area below the x axis.

If ( ) 0f x for a x b , then the area between the graph of ( )f x , the x axis, and the vertical lines

and x a x b is:

(11) ( ) ( ) ( )

b

a

Area f x dx F b F a where ( ) ( )F x f x dx is an anti-derivative of ( )f x .

This is normal , since ( ) 0 ( ) 0f x f x and therefore ( )

b

a

f x dx represents a positive number.

Another way to view the formula (11) is that ( )

b

a

f x dx , when ( ) 0f x on ,a b gives the negative of the area

between the function, the x-axis and the lines and x a x b .

Examples:

1. Consider ( ) 3f x x .

Calculate the area between the graph of ( )f x and the x axis, with 1 3x in two ways : first geometrically,

and then using formula (11) above. Do you obtain the same value? Which method do you prefer ?

2. Consider ( ) 3f x x .

Calculate the area under the graph of ( )f x , with 1 3x in two ways : first geometrically, and then using

formula (1) above. Do you obtain the same value? Which method do you prefer ?

3.

When sketching the graph of ( )f x with 1 2x we obtain the following graph:

10

Therefore the required area is: 0 2 0 2

2 2

1 0 1 0

( ) ( ) 3 3Area A B f x dx f x dx x x dx x x dx

and so on .

This procedure remains the same for any function which changes the sign in the required interval: when we want to

calculate the area between the graph of the function and the x axis between a function which changes the sign in the

interval, split the interval at the point where the function changes its sign.

B. Area between two curves.

More generally, consider that we want to calculate the area between the graph of ( )f x and the graph of ( )g x ,

with a x b , and (12) ( ) ( ), for f x g x a x b , as shown below:

Figure 6.3: Area between two curves

In this case, the area of the yellow region is: (13) ( ) ( )

b

a

Area f x g x dx .

Note that, if the inequality (12) reverses at some point inside the interval, as in Figure 6.4 below, then the area between

the two curves for 0 2x is: 1 2

3 33 3

0 1

Area x x dx x x dx , that is, we split the integral at the point

where the inequality reverses in order to apply formula (13) correctly.

Figure 6.4: Area between 3 3( ) and ( ) for 0 2f x x g x x x

11

6.4 Application of integration: the area between a curve and the y axis

In some problems we need to calculate the area of the region between a curve ( )y f x and the axisy , as shown in

Figure 6.5 below.

Figure 6.5 The area between the graph of ( )y f x , , y c y d and the axisy

If the curve is given by an equation (14) ( )y f x , then the first step in calculating the area above is to solve for x in

equation (14), that is to find (15) 1( ) ( )x f y g y from (14).

When (15) is found, then the area of the region above is:

(16) ( )

d

c

Area g y dy

Examples:

1. Calculate the area of the region shown in Figure 6.6 below in three ways:

- geometrically, using a geometrical formula;

- using the method shown in this section;

- using the method shown in section 6.3

Do you obtain the same result? Which method do you prefer?

Figure 6.6 . The area between the graph of

1y x and the axisy

12

2. Find the area between the curve y x and the axisy with 0 3y .

3. As extra practice, do problems from Exercise set 6E in the textbook.

6.5 Application of integration: Volumes of solids of revolution

Definite integrals can be used to calculate volumes, especially volumes of solids of revolution.

A solid of revolution is the solid obtained when a planar region, such as the region shown in Figure 6.7 below, is

rotated by 360o about either the x or the y axis.

A. Volumes of solids of revolution about the x axis.

Such a solid is shown in Figure 6.7 below:

Figure 6.7 A generic solid of revolution about the x axis

The volume of such a solid of revolution is:

( )

b

a

V A x dx , where ( )A x represents the area of a circle of revolution in the x direction (such as the purple

slice above, but of negligible “height”).

Since 2( ) ( )A x f x in our case, then the area of a solid of revolution about the x axis is:

(15) 2 ( )

b

a

V f x dx , where ( )y f x is the curve being revolved about the x axis to produce this solid.

13

Example:

1. Calculate the solid of revolution shown below in Figure 6.8 in two ways:

Geometrically, using the formula for the volume of a cone;

Using the method developed in this section (formula 15 above).

Which method do you prefer?

Figure 6.8 A simple solid of revolution

2. Do Example 6.18 and problem 7 from Exercise set 6H.

B. Volumes of solids of revolution about the y axis.

If the solid of revolution is obtained by rotating a curve ( )y f x 1( ) ( )x f y g y about the y axis, as

in Figure 6.9 below, then the volume of this solid of revolution is:

Figure 6.7 A generic solid of revolution about the y axis

(16) 2 ( )

q

p

V g y dy

(note that the independent variable needs to be y here, that is, if the curve is given by ( )y f x , we need to find

1( ) ( )x f y g y first (that is, we need to solve this equation for x ).

14

Examples:

1. Do Example 6.19 from the textbook.

2. Do problem 6 from Exercise set 6H.

6.6 Introduction to improper integrals (pages 206 – 208 in the textbook)

Integrals which involve are called improper integrals.

In this section, we consider only a brief introduction to this type of integrals, and we will essentially use the result:

(17) 1

0 0c

There are two types of improper integrals:

Type I: When one of the limits of integration is infinite:

Examples:

1. Calculate: 2

1

1dx

x

. We calculate such an integral in the usual manner, but use (17) as needed in the

evaluation of the result.

Therefore: 2

11

1 11 0 1 1

1dx

x x

.

The somewhat surprising result is that, even if one of the limits of integration is infinite, our area under the

curve is still finite. This can be (somewhat) explained by looking at the graph of the function (and the area to be

calculated) in Figure 6.8 below:

Figure 6.8 Graph of 1

for 1y xx

15

Part of the explanation for this result is that, even if x when calculating this area, ( ) 0 as f x x .

2. In a similar way, Calculate

2

3

2dx

x

Type II: When the function to be integrated (the integrand) is infinite (usually at one endpoints of the interval):

Once again, we evaluate the integral as usual.

Examples:

1. Calculate

9

0

1dx

x

Note that 1

( )f xx

is undefined at 0, however, evaluating this integral as usual, produces:

99

00

12 6 0 6dx x

x

2. Do exercises 1,2 and 6 from Exercise set 6G from the textbook.

As a conclusion of this chapter, remember the key points below:

16