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Unformatted text preview: Su, Yung – Homework 6 – Due: Oct 9 2007, 3:00 am – Inst: Shinko Harper 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the integral I = Z 7 e ln x x 2 dx. 1. I = 7 e + 1 2 ‡ ln 7 + 1 · 2. I = 2 e 1 7 ‡ ln 7 + 1 · correct 3. I = 2 e 1 7 ‡ ln 7 1 · 4. I = 2 e + 1 7 ‡ ln 7 + 1 · 5. I = 7 e 1 2 ‡ ln 7 1 · 6. I = 7 e 1 2 ‡ ln 7 + 1 · Explanation: After integration by parts, I = h 1 x ln x i 7 e + Z 7 e 1 x 2 dx. Thus I = h 1 x ln x + 1 x i 7 e . Consequently, I = 2 e 1 7 ‡ ln 7 + 1 · . keywords: integration by parts, log function 002 (part 1 of 1) 10 points Determine the indefinite integral Z ( x 2 4) sin 2 xdx. 1. 1 4 ‡ 2 x sin 2 x (2 x 2 9) cos 2 x · + C cor rect 2. 1 2 x 2 sin 2 x x cos 2 x + 7 2 sin 2 x + C 3. 1 2 ‡ 2 x sin 2 x (2 x 2 9) cos 2 x · + C 4. 1 4 ‡ 2 x sin 2 x + (2 x 2 9) cos 2 x · + C 5. x 2 cos 2 x + x sin 2 x 7 2 cos 2 x + C 6. 1 4 ‡ 2 x cos 2 x + (2 x 2 9) sin 2 x · + C Explanation: After integration by parts, Z ( x 2 4) sin 2 xdx = 1 2 ( x 2 4) cos 2 x + 1 2 Z cos 2 x n d dx ( x 2 4) o dx = 1 2 ( x 2 4) cos 2 x + Z x cos 2 xdx. To evaluate this last integral we need to inte grate by parts once again. For then Z x cos 2 xdx = x sin 2 x 2 Z sin 2 x 2 dx = 1 2 x sin 2 x + 1 4 cos 2 x. Consequently, the indefinite integral is 1 4 ‡ 2 x sin 2 x (2 x 2 9) cos 2 x · + C with C an arbitrary constant. keywords: integration by parts, indefinite integral, trig function, integration by parts twice, 003 (part 1 of 1) 10 points Su, Yung – Homework 6 – Due: Oct 9 2007, 3:00 am – Inst: Shinko Harper 2 Evaluate the definite integral I = Z 9 1 e √ t dt. 1. I = 6 e 9 + 2 e 2. I = 4 e 3 2 e 3. I = 6 e 9 4. I = 4 e 3 correct 5. I = 6 e 3 6. I = 4 e 3 + 2 e Explanation: Let w = √ t , so that t = w 2 , dt = 2 w dw . Then I = Z 3 1 2 w e w dw . To evaluate this last integral we use now use integration by parts: I = h 2 w e w i 3 1 2 Z 3 1 e w dw = 6 e 3 2 e 2( e 3 e ) . Consequently, I = 4 e 3 . keywords: substitution, integration by parts, definite integral 004 (part 1 of 2) 10 points The shaded region in is bounded by the graphs of y = ln x, y = 0 , x = 4 e. (i) Find the area of the region. 1. area = e ln 4 + 1 2. area = 4 e ln 4 3. area = e ln 4 1 4. area = 4 e ln 4 1 5. area = 4 e ln 4 + 1 correct 6. area = e 1 Explanation: The area of the region is given by the inte gral A = Z 4 e 1 ln xdx....
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This note was uploaded on 04/15/2009 for the course M 59685 taught by Professor Harper during the Spring '09 term at University of Texas.
 Spring '09
 Harper
 Calculus

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