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Unformatted text preview: Cheung, Anthony – Homework 6 – Due: Oct 10 2006, 3:00 am – Inst: David Benzvi 1 This printout should have 17 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 definite integral I = Z π/ 4 (7 x 2) sec 2 xdx. 1. I = 7 2 π + 2 + 7 ln 2 2. I = 7 4 π 2 7 2 ln 2 correct 3. I = 7 4 π + 2 7 2 ln 2 4. I = 7 2 π 2 7 ln 2 5. I = 7 2 π + 2 + 7 2 ln 2 Explanation: Since d dx tan x = sec 2 x, integration by parts is suggested. For then, I = h (7 x 2) tan x i π/ 4 Z π/ 4 tan x d dx (7 x 2) dx. = 7 4 π 2 7 Z π/ 4 tan xdx. But Z π/ 4 tan xdx = h ln  sec x  i π/ 4 = ln √ 2 , so I = 7 4 π 2 7 2 ln 2 . keywords: integration by parts, trig function 002 (part 1 of 1) 10 points Determine the indefinite integral I = Z e x cos 2 xdx. 1. I = 1 5 e x ‡ 2 sin 2 x cos 2 x · + C 2. I = 1 4 e x ‡ sin 2 x 2 cos 2 x · + C 3. I = 1 5 e x ‡ 2 sin 2 x +cos 2 x · + C correct 4. I = 1 4 e x ‡ 2 sin 2 x cos 2 x · + C 5. I = 1 4 e x ‡ sin 2 x + 2 cos 2 x · + C 6. I = 1 5 e x ‡ sin 2 x + 2 cos 2 x · + C Explanation: After integration by parts, I = e x cos 2 x Z e x d dx cos 2 xdx = e x cos 2 x + 2 Z e x sin 2 xdx. To reduce this last integral to one having the same form as I , we integrate by parts again for then Z e x sin 2 xdx = e x sin 2 x Z e x d dx sin 2 xdx = e x sin 2 x 2 Z e x cos 2 xdx = e x sin 2 x 2 I . Thus I = e x cos 2 x + 2 n e x sin 2 x 2 I o . Cheung, Anthony – Homework 6 – Due: Oct 10 2006, 3:00 am – Inst: David Benzvi 2 Solving for I we see that ‡ 1 + 4 · I = e x cos 2 x + 2 e x sin 2 x. Consequently I = 1 5 e x ‡ 2 sin 2 x + cos 2 x · + C with C an arbitrary constant. keywords: indefinite integral, integration by parts, exponential function, cosine function 003 (part 1 of 1) 10 points Determine the integral I = Z 6 ln x x 5 dx. 1. I = 3 2 x 4 ‡ ln x + 1 4 · + C correct 2. I = 6 5 x 4 ‡ ln x + 1 4 · + C 3. I = 3 2 x 4 ‡ ln x 1 4 · + C 4. I = 6 5 x 4 ‡ ln x + 1 4 · + C 5. I = 3 2 x 4 ‡ ln x + 1 4 · + C 6. I = 6 5 x 4 ‡ ln x 1 4 · + C Explanation: After integration by parts Z ln x x 5 dx = 1 4 ‡ ln x x 4 + Z 1 x 5 dx · = 1 4 x 4 ‡ ln x + 1 4 · + C . Consequently, I = 3 2 x 4 ‡ ln x + 1 4 · + C with C an arbitrary constant....
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This note was uploaded on 01/27/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.
 Spring '08
 RAdin
 Calculus

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