ReviewExam2SolnsSp07

# - Create assignment 57611 Homework 14 Mar 23 at 9:27 pm This print-out should have 23 questions Multiple-choice questions may continue on the next

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Create assignment, 57611, Homework 14, Mar 23 at 9:27 pm 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. version 888 CalC8e58s 54:06, calculus3, multiple choice, > 1 min, wording-variable. 001 Determine the indefnite integral I = Z 2 x ln (2 + x ) dx. 1. I = ( x 2 - 4) ln (2 + x ) + 2 x + 1 2 x 2 + C 2. I = 1 2 ( x 2 - 4) ln (2 + x ) - x - 1 4 x 2 + C 3. I = 1 2 ( x 2 - 4) ln (2 + x ) + x - 1 4 x 2 + C 4. I = 2( x 2 - 4) ln (2 + x ) - 4 x + x 2 + C 5. I = ( x 2 - 4) ln (2 + x ) + 2 x - 1 2 x 2 + C correct 6. I = 2( x 2 - 4) ln (2 + x ) + 4 x - x 2 + C Explanation: Integrate frst by parts. Then I = x 2 ln (2 + x ) - Z x 2 2 + x dx. To evaluate the integral on the right we divide: x 2 2 + x = 2 x + x 2 - 2 x 2 + x = x - 2 x 2 + x = x - 2 + 4 2 + x . In this case, Z x 2 2 + x dx = Z n x - 2 + 4 2 + x o dx = 1 2 x 2 - 2 x + 4 ln (2 + x ) . Consequently, I = ( x 2 - 4) ln (2 + x ) + 2 x - 1 2 x 2 + C with C an arbitrary constant. keywords: indefnite integral, integration by parts, natural log, polynomial long division CalC8a08d 54:02, calculus3, multiple choice, > 1 min, wording-variable. 002 ±ind the indefnite integral Z 3 x (ln x ) 2 dx. 1. 3 2 x 2 (ln x ) 2 + ln x + 1 2 · + C 2. 6 x 2 (ln x ) 2 - ln x + 1 2 · + C 3. 3 2 x 2 (ln x ) 2 - ln x + 1 2 · + C correct 4. 3 x 2 (ln x ) 2 - ln x + 1 2 · + C 5. 3 2 x 2 (ln x ) 2 - ln x - 1 2 · + C Explanation: AFter integration by parts, Z x (ln x ) 2 dx = 1 2 x 2 (ln x ) 2 - Z x 2 1 x ln xdx = 1 2 x 2 (ln x ) 2 - Z x ln xdx. But aFter integration by parts once again, Z x ln xdx = 1 2 x 2 ln x - 1 2 Z x 2 1 x dx = 1 2 x 2 ln x - 1 2 Z xdx = 1 2 x 2 ln x - 1 4 x 2 + C.

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Create assignment, 57611, Homework 14, Mar 23 at 9:27 pm 2 Thus Z x (ln x ) 2 dx = 1 2 x 2 (ln x ) 2 - 1 2 x 2 ln x + 1 4 x 2 + C. Consequently, Z 3 x (ln x ) 2 dx = 3 2 x 2 (ln x ) 2 - ln x + 1 2 · + C . keywords: integration by parts, log function CalC8b22exam 54:03, calculus3, multiple choice, > 1 min, wording-variable. 003 Evaluate the deFnite integral I = Z π/ 4 0 4 tan 4 xdx. 1. I = π - 8 3 correct 2. I = 2 π + 4 3 3. I = 2 π - 8 3 4. I = π + 8 3 5. I = 4 π - 4 3 6. I = 4 π + 4 3 Explanation: Since tan 2 x = sec 2 x - 1 , it follows that tan 4 x = tan 2 x (sec 2 x - 1) = tan 2 x sec 2 - tan 2 x. Thus, using the same trig identity as before, we see that tan 4 x = (tan 2 x - 1) sec 2 x + 1 , in which case I = 4 Z π/ 4 0 n (tan 2 x - 1) sec 2 x + 1 o dx. The whole point of this use of trig identities is that d dx tan x = sec 2 x, so set u = tan x . Then du = sec 2 xdx, while x = 0 = u = 0 , x = π 4 = u = 1 . In this case, I = 4 Z 1 0 ( u 2 - 1) du + 4 h x i π/ 4 0 = 4 h 1 3 u 3 - u i 1 0 + π .
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## This note was uploaded on 06/17/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.

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- Create assignment 57611 Homework 14 Mar 23 at 9:27 pm This print-out should have 23 questions Multiple-choice questions may continue on the next

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