HW8 - Valenica Daniel – Homework 8 – Due 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 π/ 2 + 1 7 e- 3 π/ 2 2. I = 1 6 e 3 π/ 2- 1 6 e- 3 π/ 2 correct 3. I = 1 6 e 3 π/ 2 + 1 6 e- 3 π/ 2 4. I =- 1 6 e 3 π/ 2- 1 6 e- 3 π/ 2 5. I = 1 7 e 3 π/ 2- 1 7 e- 3 π/ 2 6. I = 1 7 e 3 π/ 2 + 1 7 e- 3 π/ 2 Explanation: Set u = arctan y . Then du = 1 1 + y 2 dy , in which case I = Z π/ 4- π/ 4 e 6 u du = h e 6 u 6 i π/ 4- π/ 4 . Consequently, I = 1 6 ( e 3 π/ 2- e- 3 π/ 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 π/ 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 π/ 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 π/ 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 = π- 1 2 2. I = π 4- 1 2 correct 3. I = π 2- 1 4 4. I = π 4- 1 8 5. I = π 2- 1 6. I = π 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 = π 4 . In this case I = Z π/ 4 2 sin 2 u cos u cos u du = 2 Z π/ 4 sin 2 udu = Z π/ 4 ‡ 1- cos 2 u · du. Thus I = h u- 1 2 sin 2 u i π/ 4 . Consequently, I = 1 4 π- 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 π- 2 √ 3 2. I = 4 3 π + 2 Valenica, Daniel – Homework 8 – Due: Oct 24 2007, 1:00 pm – Inst: Cheng 3 3. I = 2 3 π + 2 4. I = 4 3 π- 2 correct 5. I = 2 3 π- 2 √ 3 6. I = 2 3 π + 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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This homework help was uploaded on 04/16/2008 for the course CALC 303L taught by Professor Cheng during the Fall '07 term at University of Texas.

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HW8 - Valenica Daniel – Homework 8 – Due 1:00 pm –...

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