Exam 2 Solutions 2011

# Exam 2 Solutions 2011 - Corrected Solutions to MAT 193 Test...

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Corrected Solutions to MAT 193 Test. Oct 24, 2011. Jennings Show your work. This is a closed book, closed note test. Please put phones, computers, and other communications devices away. It’s OK to use calculators, but you probably won’t need them. Please use your own paper. If you have a question about anything, ask! Some useful formulas: cos 2 ( θ ) + sin 2 ( θ ) = 1 1 + tan 2 ( θ ) = sec 2 ( θ ) 1. (10 points) Find i xe 2 x dx . Solution: Use integration by parts. Let u = x dv = e 2 x dx du = dx v = e 2 x 2 so i xe 2 x dx = xe 2 x 2 - i e 2 x 2 dx = xe 2 x 2 - e 2 x 4 + C 2. (10 points) Find i x 5 ln( x ) dx . Solution: (corrected) Integration by parts. Let u = ln( x ) dv = x 5 dx du = dx x v = x 6 6 so i x 5 ln( x ) dx = p x 6 6 P (ln( x )) - i p x 6 6 P dx x = x 6 ln( x ) 6 - 1 6 i x 5 dx = x 6 ln( x ) 6 - x 6 36 + C 3. (10 points) Use integration by parts twice to ±nd i e x sin( x ) dx. Solution: Let u = e x dv = sin( x ) dx du = e x dx v = - cos( x )

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i e x sin( x ) dx = - e x cos( x ) - i - cos( x ) e x dx = - e x cos( x ) + i cos( x ) e x dx (1) Integrating by parts again, let u = e x dv = cos( x ) dx du = e x dx v = sin( x ) so i cos( x ) e x dx = e x sin( x ) - i sin(
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