5 INTEGRALS INTEGRALS 5 5 The Substitution Rule

5 INTEGRALS

INTEGRALS 5. 5 The Substitution Rule In this section, we will learn: To substitute a new variable in place of an existing expression in a function, making integration easier.

INTRODUCTION Due to the Fundamental Theorem of Calculus (FTC), it’s important to be able to find antiderivatives.

INTRODUCTION Equation 1 However, our antidifferentiation formulas don’t tell us how to evaluate integrals such as

INTRODUCTION To find this integral, we use the problemsolving strategy of introducing something extra. § The ‘something extra’ is a new variable. § We change from the variable x to a new variable u.

INTRODUCTION Suppose we let u be the quantity under the root sign in Equation 1, u = 1 + x 2. § Then, the differential of u is du = 2 x dx.

INTRODUCTION Notice that, if the dx in the notation for an integral were to be interpreted as a differential, then the differential 2 x dx would occur in Equation 1.

INTRODUCTION Equation 2 So, formally, without justifying our calculation, we could write:

INTRODUCTION However, now we can check that we have the correct answer by using the Chain Rule to differentiate the final function of Equation 2:

INTRODUCTION In general, this method works whenever we have an integral that we can write in the form ∫ f(g(x))g’(x) dx

INTRODUCTION Equation 3 Observe that, if F’ = f, then ∫ F’(g(x))g’(x) dx = F(g(x)) + C because, by the Chain Rule,

INTRODUCTION If we make the ‘change of variable’ or ‘substitution’ u = g(x), from Equation 3, we have:

INTRODUCTION Writing F’ = f, we get: ∫ f(g(x))g’(x) dx = ∫ f(u) du § Thus, we have proved the following rule.

SUBSTITUTION RULE Equation 4 If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫ f(g(x))g’(x) dx = ∫ f(u) du

SUBSTITUTION RULE Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation. Notice also that, if u = g(x), then du = g’(x) dx. § So, a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.

SUBSTITUTION RULE Thus, the Substitution Rule says: It is permissible to operate with dx and du after integral signs as if they were differentials.

SUBSTITUTION RULE Example 1 Find ∫ x 3 cos(x 4 + 2) dx § We make the substitution u = x 4 + 2. § This is because its differential is du = 4 x 3 dx, which, apart from the constant factor 4, occurs in the integral.

SUBSTITUTION RULE Example 1 Thus, using x 3 dx = du/4 and the Substitution Rule, we have: § Notice that, at the final stage, we had to return to the original variable x.

SUBSTITUTION RULE The idea behind the Substitution Rule is to replace a relatively complicated integral by a simpler integral. § This is accomplished by changing from the original variable x to a new variable u that is a function of x. § Thus, in Example 1, we replaced the integral ∫ x 3 cos(x 4 + 2) dx by the simpler integral ¼ ∫ cos u du.

SUBSTITUTION RULE The main challenge in using the rule is to think of an appropriate substitution. § You should try to choose u to be some function in the integrand whose differential also occurs—except for a constant factor. § This was the case in Example 1.

SUBSTITUTION RULE If that is not possible, try choosing u to be some complicated part of the integrand— perhaps the inner function in a composite function.

SUBSTITUTION RULE Finding the right substitution is a bit of an art. § It’s not unusual to guess wrong. § If your first guess doesn’t work, try another substitution.

SUBSTITUTION RULE Evaluate § Let u = 2 x + 1. § Then, du = 2 dx. § So, dx = du/2. E. g. 2—Solution 1

SUBSTITUTION RULE § Thus, the rule gives: E. g. 2—Solution 1

SUBSTITUTION RULE E. g. 2—Solution 2 Another possible substitution is Then, So, § Alternatively, observe that u 2 = 2 x + 1. § So, 2 u du = 2 dx.

SUBSTITUTION RULE Thus, E. g. 2—Solution 2

SUBSTITUTION RULE Find § Let u = 1 – 4 x 2. § Then, du = -8 x dx. § So, x dx = -1/8 du and Example 3

SUBSTITUTION RULE Calculate ∫ cos 5 x dx § If we let u = 5 x, then du = 5 dx. § So, dx = 1/5 du. § Therefore, Example 4

SUBSTITUTION RULE Example 5 Find § An appropriate substitution becomes more obvious if we factor x 5 as x 4. x. § Let u = 1 + x 2. § Then, du = 2 x dx. § So, x dx = du/2.

SUBSTITUTION RULE Example 5 Also, x 2 = u – 1; so, x 4 = (u – 1)2:

DEFINITE INTEGRALS When evaluating a definite integral by substitution, two methods are possible.

DEFINITE INTEGRALS One method is to evaluate the indefinite integral first and then use the FTC. § For instance, using the result of Example 2, we have:

DEFINITE INTEGRALS Another method, which is usually preferable, is to change the limits of integration when the variable is changed. Thus, we have the substitution rule for definite integrals.

SUB. RULE FOR DEF. INTEGRALS Equation 5 If g’ is continuous on [a, b] and f is continuous on the range of u = g(x), then

SUB. RULE FOR DEF. INTEGRALS Proof Let F be an antiderivative of f. § Then, by Equation 3, F(g(x)) is an antiderivative of f(g(x))g’(x). § So, by Part 2 of the FTC (FTC 2), we have:

SUB. RULE FOR DEF. INTEGRALS Proof § However, applying the FTC 2 a second time, we also have:

SUB. RULE FOR DEF. INTEGRALS Example 6 Evaluate § Using the substitution from Solution 1 of Example 2, we have: u = 2 x + 1 and dx = du/2 using Equation 5.

SUB. RULE FOR DEF. INTEGRALS Example 6 To find the new limits of integration, we note that: § When x = 0, u = 2(0) + 1 = 1 § When x = 4, u = 2(4) + 1 = 9

SUB. RULE FOR DEF. INTEGRALS Example 6 Thus,

SUB. RULE FOR DEF. INTEGRALS Example 6 Observe that, when using Equation 6, we do not return to the variable x after integrating. § We simply evaluate the expression in u between the appropriate values of u.

SUB. RULE FOR DEF. INTEGRALS Example 7 Evaluate § Let u = 3 – 5 x. § Then, du = – 5 dx, so dx = –du/5. § When x = 1, u = – 2, and when x = 2, u = – 7.

SUB. RULE FOR DEF. INTEGRALS Example 7 Thus,

SYMMETRY The next theorem uses the Substitution Rule for Definite Integrals to simplify the calculation of integrals of functions that possess symmetry properties.

INTEGS. OF SYMM. FUNCTIONS Theorem 6 Suppose f is continuous on [–a , a]. a. If f is even, [f(–x) = f(x)], then b. If f is odd, [f(-x) = -f(x)], then

INTEGS. OF SYMM. FUNCTIONS Proof—Equation 7 We split the integral in two:

INTEGS. OF SYMM. FUNCTIONS Proof In the first integral in the second part, we make the substitution u = –x. § Then, du = –dx, and when x = –a, u = a.

INTEGS. OF SYMM. FUNCTIONS Therefore, Proof

INTEGS. OF SYMM. FUNCTIONS So, Equation 7 becomes: Proof—Equation 8

INTEGS. OF SYMM. FUNCTIONS If f is even, then f(–u) = f(u). So, Equation 8 gives: Proof a

INTEGS. OF SYMM. FUNCTIONS If f is odd, then f(–u) = –f(u). So, Equation 8 gives: Proof b

INTEGS. OF SYMM. FUNCTIONS Theorem 6 is illustrated here.

INTEGS. OF SYMM. FUNCTIONS For the case where f is positive and even, part (a) says that the area under y = f(x) from –a to a is twice the area from 0 to a because of symmetry.

INTEGS. OF SYMM. FUNCTIONS Recall that an integral can be expressed as the area above the x–axis and below y = f(x) minus the area below the axis and above the curve.