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Unformatted text preview: 25. Integration by Substitution P. K. Lamm 11/22/11 (12:14) p. 1 / 12 Lecture Notes: 25. Integration by Substitution These classnotes are intended to be supplementary to the textbook and are necessarily limited by the time allotted for classes. For full and precise statements of definitions and theorems, as well as material covering other topics and examples, please consult the textbook. 1. The Chain Rule ... and Reversing the Chain Rule Example 1.1: Recall how we used a substitution when first learning the Chain Rule. For example, if we wanted to find F ( x ) when F ( x ) = ( x 5 + 2) 8 we first wrote the problem as that of finding d dx F ( u ( x )) , where F ( u ) = u 8 , and u ( x ) = x 5 + 2 . The Chain Rule then gives d dx F ( u ( x )) = F ( u )  u = u ( x ) · u ( x ) = ( u 8 ) u = x 5 +2 ( x 5 + 2) = 8 u 7 u = x 5 +2 (5 x 4 ) = 8( x 5 + 2) 7 (5 x 4 ) . Now let’s explore the meaning of the above differentiation when solving an integration problem. From the above, we have immediately that Z 8( x 5 + 2) 7 (5 x 4 ) dx = ( x 5 + 2) 8 + C. In looking at this indefinite integral, it’s not surprising that ( x 5 + 2) 8 appears in our answer; however, at first glance it’s a little disturbing that the factor (5 x 4 ) in the integrand seems to have “disappeared”. However, when we remember how this term suddenly appears when differentiating ( x 5 + 2) 8 using the Chain Rule, its disappearance when reversing the Chain Rule makes some sense. Since a substitution was of help in using the Chain Rule to differentiate a function, it stands to reason that a substitution would also be useful in taking those antiderivatives that are essentially “running the Chain Rule backward”. That is, returning to the above integral Z 8( x 5 + 2) 7 (5 x 4 ) dx, 1 25. Integration by Substitution P. K. Lamm 11/22/11 (12:14) p. 2 / 12 we note that what makes this integral difficult is the fact that a complicated function is being raised to the 7th power in the integrand; to simplify the integrand, we could consider making a substitution for that complicated function. That is, we let u ( x ) = x 5 + 2 so that the expression 8( x 5 + 2) 7 is more simply written as 8 u 7 . Note also that du , the differential in u , is obtained via du = u ( x ) dx = 5 x 4 dx. That is, we can rewrite the complicated integral in x as a simpler integral in the new variable u : Z 8( x 5 + 2) 7 (5 x 4 ) dx = Z 8 u 7 du = u 8 + C, and then returning to the original x variable via u ( x ) = x 5 + 2, Z 8( x 5 + 2) 7 (5 x 4 ) dx = u 8 u = x 5 +2 + C = ( x 5 + 2) 8 + C. The Method of Substitution: Suppose we wish to evaluate the integral Z f ( g ( x )) g ( x ) dx, where we know Z f ( u ) du = F ( u ) + C....
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
 Fall '10
 KIHYUNHYUN
 Calculus, Integration By Substitution

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