4.4  INTEGRATION  BY  CHANGE  OF  VARIABLES

   We have seen that the sum, constant, and power rules for differentiation can be turned around to give the sum, constant, and power ruler for integration. In this section we shall show how to make use of the Chain Rule for differentiation in problems of integration. The Chain Rule will lead to the important method of integration by change of variables. The basic idea is to try to simplify the function to be integrated by changing from one independent variable to another.

 

If F is an antiderivative of f and we take u as the independent variable, then ∫f(u)du is a family of functions of u,

                          ∫ f(u) du = F(u) + C.

 

But if we take x as the independent variable and introduce u as a dependent variable u=g(x), then du and ∫ f(u) du mean the following:

  du = g' (x) dx,  ∫ f(u)du =  ∫ f (g(x)) g'(x)dx = H(x) + C.

 

The notation ∫ f(u)du always stands for a family of functions of the independent variable, which in some cases is another variable such as x. The next theorem can be used as follows. To integrate a given function of x, properly choose a new variable u= g(x) and integrate a new function with respect to u.

 

DEFINITION

        Let I and J be intervals. We say that a function g maps J into I if for every

        Point x in J, g(x) is defined and belongs to I (Figure 4.4.1).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.4.1

 

THEOREM 1   ( Indefinite Integration by Change of Variables)

 

Suppose I and J are open intervals, f has domain I, g maps J into I, and g is differentiable on J.Assume that when we take u as the independent variable,

 

 

Then when x is the independent variable and u = g(x),

 

 

PROOF let H(x) = F(g(x)). For any x in J, the derivatives g' (x) and F '(g(x))= f(g(x))

exist. Therefore by the Chain Rule,

 

It follows that

 

 

So when u = g(x), we have

 

 

Theorem 1 gives another proof of the general power rule

 

 

Where u is given as a function of the independent variable x, from the simpler power rule

 

 

Where x is the independent variable.

 

EXAMPLE 1 Find

 

 

EXAMPLE 2  Find

 

  

          Let u = 1 + 1/x, Then du = －1/x² dx and thus

 

So

 

 

In a simple problem such as this example, we can save writing by using the term 1+1/x instead of introducing a new letter u,

 

 

In examples such as the above one, the trick is to find a new variable u such that the expression becomes simpler when change variables. This usually must be done by an “educated” trial and error process.

 

One must be careful to express dx in terms of du before integrating with respect to u.

 

EXAMPLE 3  Find  ∫(1+5x)² dx, Let u = 1+5x. For emphasis we shall do it correctly and incorrectly.

Correct:             du =5dx,   dx = ___ du,

                    

 

Incorrect:   

 

Incorrect:

 

 

EXAMPLE 4  Find

 

 

             We try to express the integral in terms of u.

 

            

             Since u=2 －x², x²=2 －u. Therefore

 

 

We next describe the method of definite integration by change of variables. In a definite integral

 

 

It is always understood that x is the independent variable and we are integrating between the limits x= a and x=b. Thus when change to a new independent variable u, we must also change the limits of integration. The theorem below will show that if u = c when x=a and u=d when x=b, then c and d will be the new limits of integration.

 

THEOREM 2  (Definite Integration by Change of Variables)

Suppose I and J are open intervals, f is continuous and has an antiderivative on 1, g has a continuous derivative on J, and g maps J into I. Then for any two points a and b in J,

 

 

PROOF  Let F be an antiderivative of f. Then by Theorem 1, H(x) = F(g(x)) is an antiderivative of h(x)= f(g(x))g'(x). Since f,g, and g' are continuous, h is continuous on J. Then by the Fundamental Theorem of Calculus,

 

 

EXAMPLE 5  Find the area under the line y= 1+3x from x = 0 to x = 1. This can be done either with or without a change of variables.

Without change of variable: ∫ (1+3x) dx = x +3x²/2 + C, so

 

 

With change of variable: Let u = 1+ 3x. Then du=3dx, dx= ___ du.

When x = 0, u= 1+3·0 = 1. When x =1, u =1+3·1 =4.

 

 

EXAMPLE 5  shows us that ____ (1+3x) dx = ___ (u/3) du; that is, the areas shown in Figure 4.4.2 are the same.

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.4.2

 

 

 

Figure 4.4.3

EXAMPLE 6  Find the area under the curve y=2x/ (x²-3)² from x=2 to x=3

         (Figure 4.3.3).

         Let u = x² -3. Then du = 2x dx. At x=2, u= 2² -3 =1. At x=3,

          u = 3² -3 =6. Then

 

 

EXAMPLE 7  Find ________ xdx, the function _____ x as given is only defined on the closed interval [-1,1]. In order to use Theorem2, we extend it to the open interval J=( －∞,∞)by

 

 

Let u=1－x². Then du= －2 xdx, dx= －du/2x.  At x=0, u=1. At x=1,u=0. Therefore

 

 

We see in Figure 4.4.4 that as x increases from 0 to 1, u decreases from 1 to 0, so the limits become reversed. The areas shown in Figure 4.4.5 are equal.

 

 

 

 

 

 

 

 

 

 

 

Figure 4.4.4

 

 

 

 

 

 

 

 

 

 

Figure 4.4.5

     We can use integration by change of variables to derive the formula for the area of a circle, A = πr², where r is the radius. It is easier to work with a semicircle because the semicircle of radius r is just the region under the curve

 

 

To start with we need to give a rigorous definition of π. By definition, π is the area of a unit circle. Thus π is twice the area of the unit semicircle, which means:

 

DEFINITION

 

 

The area of a semicircle of radius r is the definite integral

 

 

To evaluate this integral we let x=ru. Then dx=r du. When x= ____ r, u = ___ 1.

Thus

 

 

Therefore the semicircle has area π r²/2 and the circle area π r² (Figure 4.4.6).

 

EXAMPLE 8  Find

 

 

 

Let  u= x -x3. Then du =(1 -3x²) dx. When x=0, u=0 -03 =0.

 When x =1, u =1 -13 =0. Then

 

 

As x goes from 0 to 1, u starts at 0, increases for a time, then drops back to 0

(Figure 4.4.7).

 

 

 

 

 

 

 

Figure 4.4.6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.4.7

We do not know how to find the indefinite integrals in this example. Nevertheless the answer is 0 because on changing variables both limits of integration become the same. Using the Addition Property, we can also see that, for instance,

 

 

 

PROBLEMS  FOR  SECTION  4.4

In Problems 1-90, evaluate the integral.

1 

3

5

7

9

11

13

15

17

19

21

23

25

27

29

31

33

35

37

39

41

43

45

47

49

51

53

55

57

59

61

63

65

67

69

71

73

75

77

79

81

83

85

87

89

 

In Problems 91-108, evaluate the definite integral.

91

93

95

97

99

101

103

105______               106______

107_______              108______

109 Find the area of the region below the curve y=1/(10-3x) from x=1 and x=2.

110 Find the area of the region under one arch of the curve y=sin xcos x.

111 Find the area of the region under one arch of the curve y=cos(3x).

112 Find the area of the region below the curve y=4x______between x=0 and x=2

113 Find the area below the curve y=(1+7x)²/3between x=0 and x=1

114 Find the area below the curve y=x/(x²+1) between x=0 and x=3.

□115 Evaluate: ________dx

□116 Evaluate: ___2x_______dx

□117 let f and g have continuous derivatives and evaluate ∫ f ' (g(x))g'(x)dx.

□118 a real function fis said to be even if f(x) =f(-x)for all x. Show that if fis a continuous even function, then __f(x)dx = __ f(x) dx.

□119 an odd function is a real function g such that g(-x) = -g(x) for all x. Prove that for a continuous odd function g, __ g(x)dx=0.