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Unformatted text preview: Arwikar, Dev Homework 8 Due: Oct 24 2007, 1:00 pm Inst: James Rath 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Homework eight is here Lets hope it is easier Than the last one was (Thx Michael) 001 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 1 e 6 arctan y 1 + y 2 dy . 1. I = 1 6 e 3 / 2 1 6 e 3 / 2 2. I = 1 6 e 3 / 2 + 1 6 e 3 / 2 3. I = 1 7 e 3 / 2 + 1 7 e 3 / 2 4. I = 1 7 e 3 / 2 + 1 7 e 3 / 2 5. I = 1 7 e 3 / 2 1 7 e 3 / 2 6. I = 1 6 e 3 / 2 1 6 e 3 / 2 correct Explanation: Set u = arctan y . Then du = 1 1 + y 2 dy , in which case I = Z / 4 / 4 e 6 u du = h e 6 u 6 i / 4 / 4 . Consequently, I = 1 6 ( e 3 / 2 e 3 / 2 ) . keywords: substitution, inverse trig function, integral 002 (part 1 of 1) 10 points Evaluate the definite integral I = Z 2 1 x 2 + 3 x + 1 dx. Correct answer: 2 . 12186 . Explanation: After division x 2 + 3 x + 1 = ( x 2 1) + 4 x + 1 = x 2 1 x + 1 + 4 x + 1 = x 1 + 4 x + 1 . In this case I = Z 2 1 x 1 + 4 x + 1 dx = h 1 2 x 2 x + 4 ln  x + 1  i 2 1 = 1 1 2 + 4 ln 3 ln 2 . Consequently, I = 1 2 + 4 ln 3 2 = 2 . 12186 . keywords: division, log, integral 003 (part 1 of 1) 10 points Evaluate the integral I = Z / 4 sec 2 x { 3 + 4 sin x } dx. 1. I = 7 2 2 2. I = 1 + 4 2 correct 3. I = 1 2 2 Arwikar, Dev Homework 8 Due: Oct 24 2007, 1:00 pm Inst: James Rath 2 4. I = 7 + 4 2 5. I = 7 + 2 2 6. I = 1 4 2 Explanation: Since sec 2 x { 3 + 4 sin x } = 3 sec 2 x + 4 sec x sin x cos x , we see that I = Z / 4 { 3 sec 2 x + 4 sec tan x } dx. But d dx tan x = sec 2 x, while d dx sec x = sec x tan x. Consequently, I = h 3 tan x + 4 sec x i / 4 = 1 + 4 2 . keywords: definite integral, tan integral, sec integral 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 2 2 x 2 1 x 2 dx. 1. I = 4 1 8 2. I = 8 1 4 3. I = 2 1 4. I = 4 1 2 correct 5. I =  1 2 6. I = 2 1 4 Explanation: Set x = sin u . Then dx = cos udu, p 1 x 2 = cos u, while x = 0 = u = 0 , x = 1 2 = u = 4 . In this case I = Z / 4 2 sin 2 u cos u cos u du = 2 Z / 4 sin 2 udu = Z / 4 1 cos 2 u du. Thus I = h u 1 2 sin 2 u i / 4 . Consequently, I = 1 4  1 2 . keywords: definite integral, trig. substitution, halfangle identity 005 (part 1 of 1) 10 points Stewart Chap. 8, sect. 5, Ex 5 page 545 Evaluate the definite integral I = Z 3 r 2 + x 2 x dx. 1. I = 1 3 + 1 Arwikar, Dev Homework 8 Due: Oct 24 2007, 1:00 pm Inst: James Rath 3 2. I = 2 3  3 3. I = 1 3  3 4. I = 1 3 + 3 5. I = 2 3 + 1 correct 6. I = 2 3  1 Explanation: Rationalizing the numerator we see that r 2 + x 2 x = 2 + x 2 x 2 + x = 2 + x 4 x 2 ....
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This note was uploaded on 02/25/2010 for the course M 54363 taught by Professor Olsen during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Olsen

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