quiz2sol

quiz2sol - -xe-x 1 + Z 1 e-x dx = 2 e-e x 1 + e-1 + e-x 1 =...

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Math 126 Calculus II (Fall 2006) Quiz 2 (10 pts) Name: Section (circle one): 8AM 9AM Instructions: Read each question carefully, clearly mark your answers, and remember to show your work . 1. (3 pts) Use integration by parts to evaluate Z x ln xdx . Solution: Let u = ln x , dv = xdx . Then du = 1 x dx , v = 2 3 x 3 2 . So Z x ln xdx = 2 3 x 3 2 ln x - 2 3 Z xdx = 2 3 x 3 2 ln x - 4 9 x 3 2 + C . 2. (3 pts) Use trigonometric substitution to evaluate Z dx x 2 1 + x 2 . Solution: Let x = tan θ , - π 2 < θ < π 2 . Then Z dx x 2 1 + x 2 = Z sec 2 θ dθ tan 2 θ sec θ = Z sec θ dθ tan 2 θ = Z dθ/ cos θ sin 2 θ/ cos 2 θ = Z cos θ dθ sin 2 θ = Z d (sin θ ) sin 2 θ = - 1 sin θ + C = - 1 + x 2 x + C . 3. (4 pts) The region bounded by the curves y = e x , y = e - x , and x = 1 is rotated about the y -axis. Find the volume of the resulting solid. Solution 1: V = Z 1 0 2 πx ( e x - e - x ) dx = 2 π ±Z 1 0 xe x dx - Z 1 0 xe - x dx ² . Using integra- tion by parts separately, V = 2 π ³± xe x ´ ´ ´ 1 0 - Z 1 0 e x dx ² - ±
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Unformatted text preview: -xe-x 1 + Z 1 e-x dx = 2 e-e x 1 + e-1 + e-x 1 = 4 e-1 . Solution 2: V = Z 1 2 x ( e x-e-x ) dx = 4 Z 1 x sinh xdx = 4 x cosh x 1-Z 1 cosh xdx = 4 e + e-1 2-sinh x 1 = 4 e + e-1 2-e-e-1 2 = 4 e-1 . Solution 3: V = Z e e-1 (1-ln 2 y ) dy = ( e-e-1 )- Z e e-1 ln 2 y dy = ( e-e-1 )- y ln 2 y e e-1-2 Z e e-1 ln y dy = ( e-e-1 )- ( e-e-1 ) + 2 y ln y-y e e-1 = 2 ( e-e + e-1 + e-1 ) = 4 e-1 ....
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This note was uploaded on 05/04/2008 for the course MATH 218 taught by Professor Haskell during the Fall '06 term at USC.

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