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CALCL HWK6

# CALCL HWK6 - Arwikar Dev Homework 6 Due Oct 9 2007 3:00 am...

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Arwikar, Dev – Homework 6 – Due: Oct 9 2007, 3:00 am – Inst: James Rath 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. Six geese a laying. October is underway. Will it ever stop? 001 (part 1 of 1) 10 points Evaluate the integral I = Z 8 e ln x x 2 dx . 1. I = 2 e - 1 8 ln 8 + 1 · correct 2. I = 8 e - 1 2 ln 8 + 1 · 3. I = 2 e + 1 8 ln 8 + 1 · 4. I = 2 e - 1 8 ln 8 - 1 · 5. I = 8 e - 1 2 ln 8 - 1 · 6. I = 8 e + 1 2 ln 8 + 1 · Explanation: After integration by parts, I = - h 1 x ln x i 8 e + Z 8 e 1 x 2 dx . Thus I = - h 1 x ln x + 1 x i 8 e . Consequently, I = 2 e - 1 8 ln 8 + 1 · . keywords: integration by parts, log function 002 (part 1 of 1) 10 points Determine the indefinite integral Z ( x 2 + 2) sin 2 x dx . 1. 1 4 2 x sin 2 x - (2 x 2 + 3) cos 2 x · + C cor- rect 2. 1 4 2 x cos 2 x + (2 x 2 + 3) sin 2 x · + C 3. 1 4 2 x sin 2 x + (2 x 2 + 3) cos 2 x · + C 4. 1 2 x 2 sin 2 x - x cos 2 x - 5 2 sin 2 x + C 5. 1 2 2 x sin 2 x - (2 x 2 + 3) cos 2 x · + C 6. - x 2 cos 2 x + x sin 2 x + 5 2 cos 2 x + C Explanation: After integration by parts, Z ( x 2 + 2) sin 2 x dx = - 1 2 ( x 2 + 2) cos 2 x + 1 2 Z cos 2 x n d dx ( x 2 + 2) o dx = - 1 2 ( x 2 + 2) cos 2 x + Z x cos 2 x dx . To evaluate this last integral we need to inte- grate by parts once again. For then Z x cos 2 x dx = 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.

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Arwikar, Dev – Homework 6 – Due: Oct 9 2007, 3:00 am – Inst: James Rath 2 keywords: integration by parts, indefinite integral, trig function, integration by parts twice, 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 9 1 e t dt . 1. I = 6 e 9 + 2 e 2. I = 6 e 3 3. I = 6 e 9 4. I = 4 e 3 correct 5. I = 4 e 3 + 2 e 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 = 3 e . (i) Find the area of the region. 1. area = e - 1 2. area = 3 e ln 3 3. area = 3 e ln 3 + 1 correct 4. area = e ln 3 + 1 5. area = e ln 3 - 1 6. area = 3 e ln 3 - 1 Explanation: The area of the region is given by the inte- gral A = Z 3 e 1 ln x dx . To evaluate this integral we use integration by parts. For then A = Z 3 e 1 ln x dx = h x ln x i 3 e 1 - Z 3 e 1 1 dx = h x ln x - x i 3 e 1 .
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