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CALCL HWK8

# CALCL HWK8 - Arwikar Dev Homework 8 Due 1:00 pm Inst James...

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Arwikar, Dev – Homework 8 – Due: Oct 24 2007, 1:00 pm – Inst: James Rath 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. Homework eight is here Let’s 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 π/ 2 - 1 6 e - 3 π/ 2 2. I = 1 6 e 3 π/ 2 + 1 6 e - 3 π/ 2 3. I = - 1 7 e 3 π/ 2 + 1 7 e - 3 π/ 2 4. I = 1 7 e 3 π/ 2 + 1 7 e - 3 π/ 2 5. I = 1 7 e 3 π/ 2 - 1 7 e - 3 π/ 2 6. I = 1 6 e 3 π/ 2 - 1 6 e - 3 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 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 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 π/ 4 0 sec 2 x { 3 + 4 sin x } dx . 1. I = 7 - 2 2 2. I = - 1 + 4 2 correct 3. I = - 1 - 2 2

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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 π/ 4 0 { 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 π/ 4 0 = - 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 0 2 x 2 1 - x 2 dx . 1. I = π 4 - 1 8 2. I = π 8 - 1 4 3. I = π 2 - 1 4. I = π 4 - 1 2 correct 5. I = π - 1 2 6. I = π 2 - 1 4 Explanation: Set x = sin u . Then dx = cos u du, 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 u du = 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: 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 3 0 r 2 + x 2 - x dx . 1. I = 1 3 π + 1
Arwikar, Dev – Homework 8 – Due: Oct 24 2007, 1:00 pm – Inst: James Rath 3 2. I = 2 3 π - 3 3. I = 1 3 π - 3 4. I = 1 3 π + 3 5. I = 2 3 π + 1 correct 6. I = 2 3 π - 1 Explanation: Rationalizing the numerator we see that r 2 + x 2 - x = 2 + x 2 - x 2 + x = 2 + x 4 - x 2 . Thus I = Z 3 0 2 + x 4 - x 2 dx . Now set x = 2 sin u . Then dx = 2 cos u du , while x = 0 = u = 0 , x = 3 = u = π 3 . Furthermore, p 4 - x 2 = q 4 (1 - sin 2 u ) = 2 cos u . In this case I = Z π/ 3 0 2(1 + sin u )2 cos u 2 cos u du = 2 Z π/ 3 0 (1 + sin u ) du .

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