584
Chapter 5 | Integration
5.5 | Substitution
Learning Objectives
5.5.1 Use substitution to evaluate indefinite integrals. 5.5.2 Use substitution to evaluate definite integrals.
The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called integration by substitution , to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative. At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to see? We are looking for an integrand of the form f ⎡ ⎣ g ( x ) ⎤ ⎦ g ′( x ) dx . For example, in the integral ⌠ ⌡ ⎛ ⎝ x 2 −3 ⎞ ⎠ 3 2 xdx , we have
f ( x ) = x 3 , g ( x ) = x 2 −3, and g '( x ) =2 x . Then, f ⎡ ⎣ g ( x ) ⎤
3
⎛ ⎝ x 2 −3
⎞ ⎠
⎦ g ′( x ) =
(2 x ),
and we see that our integrand is in the correct form. The method is called substitution because we substitute part of the integrand with the variable u and part of the integrand with du . It is also referred to as change of variables because we are changing variables to obtain an expression that is easier to work with for applying the integration rules. Theorem 5.7: Substitution with Indefinite Integrals Let u = g ( x ), , where g ′( x ) is continuous over an interval, let f ( x ) be continuous over the corresponding range of g , and let F ( x ) be an antiderivative of f ( x ). Then, (5.19) ∫ f ⎡ ⎣ g ( x ) ⎤ ⎦ g ′( x ) dx = ∫ f ( u ) du = F ( u )+ C = F ⎛ ⎝ g ( x ) ⎞ ⎠ + C .
Proof Let f , g , u , and F be as specified in the theorem. Then d dx
F ( g ( x )) = F ′( g ( x ) ⎞
⎠ g ′( x )
= f ⎡
⎤ ⎦ g ′( x ).
⎣ g ( x )
Integrating both sides with respect to x , we see that ∫ f ⎡ ⎣ g ( x ) ⎤ If we now substitute u = g ( x ), and du = g '( x ) dx , we get ∫ f ⎡ ⎣ g ( x ) ⎤
⎦ g ′( x ) dx = F ⎛
⎞ ⎠ + C .
⎝ g ( x )
⎦ g ′( x ) dx = ∫ f ( u ) du = F ( u )+ C = F ⎛ ⎝ g ( x ) ⎞
⎠ + C .
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