Exam1 solution - Math 2153, Exam I, Sept. 23, 2011 Name:...

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Unformatted text preview: 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 y e 2 y dy Solution Notice that Z 1 y e 2 y dy = Z 1 ye- 2 y dy. We will use integration by parts. Set f ( y ) = y, f ( y ) = 1 g ( y ) = e- 2 y- 2 , g ( y ) = e- 2 y Then Z 1 y e 2 y dy = Z 1 f ( y ) g ( y ) dy = f ( y ) g ( y ) | 1- Z 1 f ( y ) g ( y ) dy = y e- 2 y- 2 | 1- Z 1 e- 2 y- 2 dy = e- 2- 2-- e- 2 y (- 2) 2 | 1 =- 1 2 e- 2- 1 4 ( e- 2- 1) =- 3 4 e 2 + 1 4 1 2. (15 points) Evaluate the integral Z - sin 3 cos 3 d Solution 1 Define u = sin . Then du = cos d and Z sin 3 cos 3 d = Z sin 3 cos 2 cos d = Z sin 3 (1- sin 2 ) cos d = Z u 3 (1- u 2 ) du = Z ( u 3- u 5 ) du = u 4 4- u 6 6 + C = sin 4 4- sin 6 6 + C Therefore Z - sin 3 cos 3 d = sin 4 4- sin 6 6 = sin 4 4- sin 6 6- sin 4 (...
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This note was uploaded on 09/29/2011 for the course MATH 2153 taught by Professor Staff during the Fall '08 term at Oklahoma State.

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Exam1 solution - Math 2153, Exam I, Sept. 23, 2011 Name:...

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