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Exam1 solution

# Exam1 solution - Math 2153 Exam I Sept 23 2011 Name Score...

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Math 2153, Exam I, Sept. 23, 2011 Name: Score: Read the problems carefully before you begin. Show all your work neatly and concisely, and indicate your final answer clearly. 1. (15 points) Evaluate the integral Z 1 0 y e 2 y dy Solution Notice that Z 1 0 y e 2 y dy = Z 1 0 ye - 2 y dy. We will use integration by parts. Set f ( y ) = y, f 0 ( y ) = 1 g ( y ) = e - 2 y - 2 , g 0 ( y ) = e - 2 y Then Z 1 0 y e 2 y dy = Z 1 0 f ( y ) g 0 ( y ) dy = f ( y ) g ( y ) | 1 0 - Z 1 0 f 0 ( y ) g ( y ) dy = y e - 2 y - 2 | 1 0 - Z 1 0 e - 2 y - 2 dy = e - 2 - 2 - 0 - e - 2 y ( - 2) 2 | 1 0 = - 1 2 e - 2 - 1 4 ( e - 2 - 1) = - 3 4 e 2 + 1 4 1

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2. (15 points) Evaluate the integral Z π - π sin 3 θ cos 3 θ dθ Solution 1 Define u = sin θ . Then du = cos θ dθ and Z sin 3 θ cos 3 θ dθ = Z sin 3 θ cos 2 θ cos θ dθ = Z sin 3 θ (1 - sin 2 θ ) cos θ dθ = Z u 3 (1 - u 2 ) du = Z ( u 3 - u 5 ) du = u 4 4 - u 6 6 + C = sin 4 θ 4 - sin 6 θ 6 + C Therefore Z π - π sin 3 θ cos 3 θ dθ = sin 4 θ 4 - sin 6 θ 6 π π = sin 4 π 4 - sin 6 π 6 - sin 4 ( - π ) 4 - sin 6 ( - π ) 6 = 1 4 - 1 6 - 1 4 - 1 6 = 0 Solution 2 Of course you can set u = cos θ . The rest is similar to Solution 1 .
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Exam1 solution - Math 2153 Exam I Sept 23 2011 Name Score...

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