Homework 08 - Su, Yung Homework 8 Due: Oct 24 2007, 1:00 pm...

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Su, Yung – Homework 8 – Due: Oct 24 2007, 1:00 pm – Inst: Shinko Harper 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Evaluate the defnite integral I = Z 1 - 1 e 2 arctan y 1 + y 2 dy . 1. I = 1 3 e π/ 2 - 1 3 e - π/ 2 2. I = 1 2 e π/ 2 + 1 2 e - π/ 2 3. I = 1 3 e π/ 2 + 1 3 e - π/ 2 4. I = - 1 2 e π/ 2 - 1 2 e - π/ 2 5. I = - 1 3 e π/ 2 + 1 3 e - π/ 2 6. I = 1 2 e π/ 2 - 1 2 e - π/ 2 correct Explanation: Set u = arctan y . Then du = 1 1 + y 2 dy , in which case I = Z π/ 4 - π/ 4 e 2 u du = h e 2 u 2 i π/ 4 - π/ 4 . Consequently, I = 1 2 ( e π/ 2 - e - π/ 2 ) . keywords: substitution, inverse trig Function, integral 002 (part 1 oF 1) 10 points Evaluate the defnite integral I = Z 2 1 x 2 + 2 x + 1 dx. Correct answer: 1 . 7164 . Explanation: AFter division x 2 + 2 x + 1 = ( x 2 - 1) + 3 x + 1 = x 2 - 1 x + 1 + 3 x + 1 = x - 1 + 3 x + 1 . In this case I = Z 2 1 x - 1 + 3 x + 1 · dx = h 1 2 x 2 - x + 3 ln | x + 1 | i 2 1 = 1 - 1 2 · + 3 ln 3 - ln 2 · . Consequently, I = 1 2 + 3 ln 3 2 = 1 . 7164 . keywords: division, log, integral 003 (part 1 oF 1) 10 points Evaluate the integral I = Z π/ 4 0 sec 2 x { 1 + 2 sin x } dx. 1. I = 3 - 2 2. I = - 1 - 2 2 3. I = - 1 + 2 2 correct 4. I = 3 + 2 2 5. I = - 1 - 2
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Su, Yung – Homework 8 – Due: Oct 24 2007, 1:00 pm – Inst: Shinko Harper 2 6. I = 3 + 2 Explanation: Since sec 2 x { 1 + 2 sin x } = sec 2 x + 2 sec x sin x cos x · , we see that I = Z π/ 4 0 { sec 2 x + 2 sec tan x } dx. But d dx tan x = sec 2 x, while d dx sec x = sec x tan x. Consequently, I = h tan x + 2 sec x i π/ 4 0 = - 1 + 2 2 . keywords: defnite integral, tan integral, sec integral 004 (part 1 oF 1) 10 points Evaluate the defnite integral I = Z 1 2 0 2 x 2 1 - x 2 dx. 1. I = π 2 - 1 2. I = π 8 - 1 4 3. I = π - 1 2 4. I = π 2 - 1 4 5. I = π 4 - 1 8 6. I = π 4 - 1 2 correct Explanation: Set x = sin u . Then dx = cos udu, p 1 - x 2 = cos u, while x = 0 = u = 0 , x = 1 2 = u = π 4 . In this case I = Z π/ 4 0 2 sin 2 u cos u cos u du = 2 Z π/ 4 0 sin 2 udu = Z π/ 4 0 1 - cos 2 u · du. Thus I = h u - 1 2 sin 2 u i π/ 4 0 . Consequently, I = 1 4 π - 1 2 . keywords: defnite integral, trig. substitution, halF-angle identity 005 (part 1 oF 1) 10 points Stewart Chap. 8, sect. 5, Ex 5 page 545 Evaluate the defnite integral I = Z 2 3 0 r 4 + x 4 - x dx. 1. I = 2 3 π + 2 2. I = 2 3 π - 2 3
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Su, Yung – Homework 8 – Due: Oct 24 2007, 1:00 pm – Inst: Shinko Harper 3 3. I = 4 3 π - 2 4. I = 4 3 π - 2 3 5. I = 2 3 π + 2 3 6. I = 4 3 π + 2 correct Explanation: Rationalizing the numerator we see that r 4 + x 4 - x = 4 + x 4 - x 4 + x = 4 + x 16 - x 2 . Thus
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Homework 08 - Su, Yung Homework 8 Due: Oct 24 2007, 1:00 pm...

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