# AnsHW5 - Cheung Anthony Homework 5 Due Oct 3 2006 3:00 am...

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Cheung, Anthony – Homework 5 – Due: Oct 3 2006, 3:00 am – Inst: David Benzvi 1 This print-out should have 20 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 integral I = Z 1 0 3 x 6 + 5 x 2 dx. 1. I = 3 ln 6 2. I = 3 ln 11 6 3. I = 3 5 ln 6 5 4. I = 3 5 ln 6 5. I = 3 10 ln 6 5 6. I = 3 10 ln 11 6 correct Explanation: Set u = 6 + 5 x 2 ; then du = 10 dx while x = 0 = u = 6 x = 1 = u = 11 . In this case, I = 3 10 Z 11 6 1 u du = 3 10 h ln | u | i 11 6 . Consequently, I = 3 10 (ln 11 - ln 6) = 3 10 ln 11 6 . keywords: defnite integral, log integral, poly- nomial substitution 002 (part 1 oF 1) 10 points Determine the indefnite integral I = Z 4 x ( x - 2) 2 1. I = 2 ln( x - 2) 2 + C 2. I = 8 ( x - 2) 2 + C 3. I = - 4 x - 2 + C 4. I = ln( x - 2) 4 + 8 ( x - 2) 2 + C 5. I = ln( x - 2) 4 - 8 x - 2 + C correct Explanation: Set u = x - 2 ; then du = dx , so I = 4 Z x ( x - 2) - 2 dx = 4 Z ( u + 2) u - 2 du = 4 Z du u + 8 Z u - 2 du. But 4 Z du u = 4 ln | u | + C = ln u 4 + C, while 8 Z u - 2 du = - 8 u - 1 + C. Consequently, I = ln( x - 2) 4 - 8 x - 2 + C . keywords: 003 (part 1 oF 1) 10 points Evaluate the defnite integral I = Z 9 1 2 x ( x + 3)

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Cheung, Anthony – Homework 5 – Due: Oct 3 2006, 3:00 am – Inst: David Benzvi 2 1. I = 4 ln 5 4 2. I = 2( 5 - 2) 3. I = 2 ln 5 4 4. I = 2( 6 - 2) 5. I = 4( 5 - 2) 6. I = 2 ln 3 2 7. I = 4( 6 - 2) 8. I = 4 ln 3 2 correct Explanation: Set u 2 = x . Then 2 udu = dx , while x = 1 = u = 1 x = 9 = u = 3 . In this case, I = 4 Z 3 1 1 u + 3 du = 4 h ln | u + 3 | i 3 1 . Thus I = 4 ln 6 - ln 4 · = 4 ln 3 2 . keywords: 004 (part 1 of 1) 10 points Evaluate the deFnite integral I = Z 1 0 2 x 2 + x + 2 5 x 2 + 5 dx. 1. I = 4 + ln 2 10 correct 2. I = 2 + ln 2 10 3. I = 4 + ln 2 5 4. I = 4 + ln 3 10 5. I = 2 + ln 3 10 6. I = 4 + ln 3 5 Explanation: After division 2 x 2 + x + 2 5 x 2 + 5 = 2 5 + 1 5 x x 2 + 1 · . Thus I = Z 1 0 2 5 dx + 1 5 Z x x 2 + 1 dx = h 2 5 x i 1 0 + 1 5 Z x x 2 + 1 dx. To evaluate the last integral we use substitu- tion. Set u = x 2 + 1. Then du dx = 2 x, so I = 2 5 + 1 10 Z 2 1 1 u du = 2 5 + h 1 10 ln u i 2 1 . Conequently, I = 4 + ln 2 10 . keywords: 005 (part 1 of 1) 10 points Evaluate the deFnite integral I = Z 3 π/ 4 π/ 4 6 cos x - 3 sin x 6 sin x + 3 cos x 1. I = - ln (3) correct
Cheung, Anthony – Homework 5 – Due: Oct 3 2006, 3:00 am – Inst: David Benzvi 3 2. I = - ln µ 11 3 3. I = ln µ 11 3 4. none of these 5. I = ln (3) Explanation: Since the integrand is of the form f 0 ( x ) f ( x ) , f ( x ) = 6 sin x + 3 cos x this suggests the substitution u = 6 sin x + 3 cos x . For then du = (6 cos x - 3 sin x ) dx, while x = π 4 = u = 9 2 , x = 3 π 4 = u = 3 2 . Thus I = Z 3 / 2 9 / 2 1 u du = h ln | u | i 3 / 2 9 / 2 .

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AnsHW5 - Cheung Anthony Homework 5 Due Oct 3 2006 3:00 am...

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