HW6 Solution

# HW6 Solution - Kim, Jin Homework 6 Due: Oct 9 2007, 3:00 am...

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Unformatted text preview: Kim, Jin Homework 6 Due: Oct 9 2007, 3:00 am Inst: Diane Radin 1 This print-out should have 16 questions. Multiple-choice 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 3 e ln x x 2 dx. 1. I = 2 e- 1 3 ln 3- 1 2. I = 3 e- 1 2 ln 3- 1 3. I = 2 e- 1 3 ln 3 + 1 correct 4. I = 2 e + 1 3 ln 3 + 1 5. I = 3 e- 1 2 ln 3 + 1 6. I = 3 e + 1 2 ln 3 + 1 Explanation: After integration by parts, I =- h 1 x ln x i 3 e + Z 3 e 1 x 2 dx. Thus I =- h 1 x ln x + 1 x i 3 e . Consequently, I = 2 e- 1 3 ln 3 + 1 . keywords: integration by parts, log function 002 (part 1 of 1) 10 points Determine the indefinite integral Z ( x 2- 1) sin 2 xdx. 1. 1 2 x 2 sin 2 x- x cos 2 x + 1 2 sin 2 x + C 2. 1 4 2 x sin 2 x- (2 x 2- 3) cos 2 x + C cor- rect 3. 1 4 2 x sin 2 x + (2 x 2- 3) cos 2 x + C 4. 1 2 2 x sin 2 x- (2 x 2- 3) cos 2 x + C 5.- x 2 cos 2 x + x sin 2 x- 1 2 cos 2 x + C 6. 1 4 2 x cos 2 x + (2 x 2- 3) sin 2 x + C Explanation: After integration by parts, Z ( x 2- 1) sin 2 xdx =- 1 2 ( x 2- 1) cos 2 x + 1 2 Z cos 2 x n d dx ( x 2- 1) o dx =- 1 2 ( x 2- 1) 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- 3) 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 Kim, Jin Homework 6 Due: Oct 9 2007, 3:00 am Inst: Diane Radin 2 Evaluate the definite integral I = Z 4 1 e t dt. 1. I = 2 e 2 + 2 e 2. I = 4 e 2 3. I = 4 e 4 + 2 e 4. I = 2 e 2 correct 5. I = 4 e 4 6. I = 2 e 2- 2 e Explanation: Let w = t , so that t = w 2 , dt = 2 w dw . Then I = Z 2 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 2 1- 2 Z 2 1 e w dw = 4 e 2- 2 e- 2( e 2- e ) . Consequently, I = 2 e 2 . 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 = 2 ln x, y = 0 , x = 3 e. (i) Find the area of the region. 1. area = 2 e ln 3- 2 2. area = 6 e ln 3 + 2 correct 3. area = 2 e- 2 4. area = 6 e ln 3- 2 5. area = 6 e ln 3 6. area = 2 e ln 3 + 2 Explanation: The area of the region is given by the inte- gral A = Z 3 e 1 2 ln xdx....
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## This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.

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HW6 Solution - Kim, Jin Homework 6 Due: Oct 9 2007, 3:00 am...

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