HW8 - Valenica, Daniel Homework 8 Due: Oct 24 2007, 1:00 pm...

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Unformatted text preview: Valenica, Daniel Homework 8 Due: Oct 24 2007, 1:00 pm Inst: Cheng 1 This print-out should have 24 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 definite integral I = Z 1- 1 e 6 arctan y 1 + y 2 dy . 1. I =- 1 7 e 3 / 2 + 1 7 e- 3 / 2 2. I = 1 6 e 3 / 2- 1 6 e- 3 / 2 correct 3. I = 1 6 e 3 / 2 + 1 6 e- 3 / 2 4. I =- 1 6 e 3 / 2- 1 6 e- 3 / 2 5. I = 1 7 e 3 / 2- 1 7 e- 3 / 2 6. I = 1 7 e 3 / 2 + 1 7 e- 3 / 2 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 ln3- ln2 . 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 { 2- sin x } dx. 1. I = 3 + 2 2. I = 1- 2 3. I = 3- 2 correct 4. I = 1 + 1 2 2 5. I = 3 + 1 2 2 Valenica, Daniel Homework 8 Due: Oct 24 2007, 1:00 pm Inst: Cheng 2 6. I = 1- 1 2 2 Explanation: Since sec 2 x { 2- sin x } = 2 sec 2 x- sec x sin x cos x , we see that I = Z / 4 { 2 sec 2 x- sec tan x } dx. But d dx tan x = sec 2 x, while d dx sec x = sec x tan x. Consequently, I = h 2 tan x- sec x i / 4 = 3- 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 = - 1 2 2. I = 4- 1 2 correct 3. I = 2- 1 4 4. I = 4- 1 8 5. I = 2- 1 6. I = 8- 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, half-angle identity 005 (part 1 of 1) 10 points Stewart Chap. 8, sect. 5, Ex 5 page 545 Evaluate the definite integral I = Z 2 3 r 4- x 4 + x dx. 1. I = 4 3 - 2 3 2. I = 4 3 + 2 Valenica, Daniel Homework 8 Due: Oct 24 2007, 1:00 pm Inst: Cheng 3 3. I = 2 3 + 2 4. I = 4 3 - 2 correct 5. I = 2 3 - 2 3 6. I = 2 3 + 2 3 Explanation: Rationalizing the numerator we see that r 4- x 4 + x = 4- x 4 + x 4- x = 4- x 16- x 2 ....
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HW8 - Valenica, Daniel Homework 8 Due: Oct 24 2007, 1:00 pm...

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