Exam 2 Solutions 2011

# 10 points find x5 lnxdx solution corrected integration

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Unformatted text preview: xe2x dx = 2. (10 points) Find x5 ln(x)dx. Solution: (corrected) Integration by parts. Let u = ln(x) du = dx x dv = x5 dx v= x6 6 so x6 x6 (ln(x)) − 6 6 6 x ln(x) 1 = − x5 dx 6 6 x6 ln(x) x6 − +C = 6 36 x5 ln(x)dx = 3. (10 points) Use integration by parts twice to ﬁnd ex sin(x) dx. Solution: Let u = ex dv = sin(x)dx x du = e dx v = − cos(x) dx x ex sin(x) dx = −ex cos(x) − − cos(x)ex dx x (1) x = −e cos(x) + cos(x)e dx Integrating by parts again, let u = ex dv = cos(x)dx x du = e dx v = sin(x) so cos(x)ex dx = ex sin(x) − sin(x)ex dx (2) Plug equation (2) into equation (1) to obtain ex sin(x) dx = −ex cos(x) + ex sin(x) − add ex sin(x) dx ex sin(x) dx to both sides 2 ex sin(x) dx = −ex cos(x) + ex sin(x) ex sin(x) dx = 1 (−ex cos(x) + ex sin(x)) + C 2 π /...
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## This document was uploaded on 03/05/2014 for the course MAT 193 at CSU Dominguez Hills.

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