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Unformatted text preview: INTEGRATION BY SUBSTITUTION 1. Determine the following antiderivatives: (1) Z e x 1 e 2 x dx Solution. Here u = e x makes a good substitution because u = e x appears as a factor in the integrand and the rest can be expressed in terms of u . Since u = e x , we have d dx u = e x , or du = e x dx . Hence, the integral becomes Z du 1 u 2 = arcsin( u ) + c = arcsin( e x ) + c. (2) Z 1 x 2 (2 x 3 + 3) 5 dx Solution. Substituting u = x 3 , we get d dx u = 3 x 2 , or du = 3 x 2 dx . Also, as x goes from 0 to 1, we see that u also goes from 0 to 1. Hence, the integral becomes Z 1 (2 u + 3) 5 3 du. You may guess what the antiderivative of (2 u +3) 5 is at this point or do another substitution. Setting w = 2 u + 3, we get dw = 2 du . Also, as u goes from 0 to 1, w goes from 3 to 5. So, the answer is Z 5 3 w 5 6 dw = w 6 36 5 3 = 5 6 3 6 36 . Alternatively, you could directly do the substitution u = 2 x 3 + 3. Then, du = 6 x 2 dx , or du 6 = x 2 dx ....
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 Fall '09
 BenedictGross
 Antiderivatives, Derivative, Integration By Substitution

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