# lec26 - 26 Substitution in Denite Integrals Lecture Notes P...

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26. Substitution in Definite Integrals P. K. Lamm 11/27/11 (11:52) p. 1 / 6 Lecture Notes: 26. Substitution in Definite Integrals 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. Substitution in Definite Integrals Example 1.1(a): Evaluate: Z 3 0 x 2 x 3 + 9 dx Using the Fundamental Theorem of Calculus, Part 1, we need only find an antiderivative F ( x ) of f ( x ) = x 2 x 3 + 9 and then evaluate Z 3 0 x 2 x 3 + 9 dx = F (3) - F (0) . But the determination of an antiderivative of f ( x ) requires a substitution. To this end, we define u = x 3 + 9 = du = 3 x 2 dx = x 2 dx = 1 3 du, so that Z x 2 x 3 + 9 dx = Z u 1 / 2 · 1 3 du = 1 3 Z u 1 / 2 du = 1 3 u 3 / 2 3 / 2 + C = 2 9 u 3 / 2 + C = 2 9 ( x 3 + 9) 3 / 2 + C . Thus we may let F ( x ) = 2 9 ( x 3 + 9) 3 / 2 , and evaluate Z 3 0 x 2 x 3 + 9 dx = 2 9 ( x 3 + 9) 3 / 2 3 0 = 2 9 h (3 3 + 9) 3 / 2 - (0 3 + 9) 3 / 2 i = 2 9 36 3 / 2 - 9 3 / 2 1

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26. Substitution in Definite Integrals P. K. Lamm 11/27/11 (11:52) p. 2 / 6 = 2 9 6 3 - 3 3 = 2 9 (216 - 27) = 2 · 189 9 = 42 . The process in this last example has been the following: 1. First find an antiderivative of f ( x ): (1) Make a change of integration variable from x to u ; (2) Find the antiderivative with respect to the u variable; (3) Convert the antiderivative back to its original variable x ; 2. Then evaluate the definite integral using the original variable x .
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