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quiz2sol - -xe-x ´ ´ ´ 1 Z 1 e-x dx ²µ = 2 π ± e-e x...

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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 x ln x dx . Solution: Let u = ln x , dv = x dx . Then du = 1 x dx , v = 2 3 x 3 2 . So x ln x dx = 2 3 x 3 2 ln x - 2 3 x dx = 2 3 x 3 2 ln x - 4 9 x 3 2 + C . 2. (3 pts) Use trigonometric substitution to evaluate dx x 2 1 + x 2 . Solution: Let x = tan θ , - π 2 < θ < π 2 . Then dx x 2 1 + x 2 = sec 2 θ dθ tan 2 θ sec θ = sec θ dθ tan 2 θ = dθ/ cos θ sin 2 θ/ cos 2 θ = cos θ dθ sin 2 θ = 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 = 1 0 2 πx ( e x - e - x ) dx
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Unformatted text preview: -xe-x ´ ´ ´ 1 + Z 1 e-x dx ²µ = 2 π ± e-e x ´ ´ ´ 1 + e-1 + e-x ´ ´ ´ 1 ² = 4 πe-1 . Solution 2: V = Z 1 2 πx ( e x-e-x ) dx = 4 π Z 1 x sinh xdx = 4 π ± x cosh x ´ ´ ´ 1-Z 1 cosh xdx ² = 4 π ± e + e-1 2-sinh x ´ ´ ´ 1 ² = 4 π ± e + e-1 2-e-e-1 2 ² = 4 πe-1 . Solution 3: V = Z e e-1 π (1-ln 2 y ) dy = π ( e-e-1 )-π Z e e-1 ln 2 y dy = π ( e-e-1 )-π ± y ln 2 y ´ ´ ´ e e-1-2 Z e e-1 ln y dy ² = π ( e-e-1 )-π ( e-e-1 ) + 2 π ¶ y ln y-y ·´ ´ ´ e e-1 = 2 π ( e-e + e-1 + e-1 ) = 4 πe-1 ....
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