# HW 5 - Portillo Tom – Homework 5 – Due Oct 3 2006 3:00...

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Unformatted text preview: Portillo, Tom – 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 – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the integral I = Z 1 3 x 3 + 2 x 2 dx. 1. I = 3 2 ln 3 2 2. I = 3 ln 5 3 3. I = 3 4 ln 3 2 4. I = 3 4 ln 5 3 correct 5. I = 3 ln 3 6. I = 3 2 ln 3 Explanation: Set u = 3 + 2 x 2 ; then du = 4 dx while x = 0 = ⇒ u = 3 x = 1 = ⇒ u = 5 . In this case, I = 3 4 Z 5 3 1 u du = 3 4 h ln | u | i 5 3 . Consequently, I = 3 4 (ln 5- ln 3) = 3 4 ln 5 3 . keywords: definite integral, log integral, poly- nomial substitution 002 (part 1 of 1) 10 points Determine the indefinite integral I = Z 2 x ( x- 3) 2 dx. 1. I = ln( x- 3) 2- 6 x- 3 + C correct 2. I = 6 ( x- 3) 2 + C 3. I = 3 ln( x- 3) 2 + C 4. I =- 2 x- 3 + C 5. I = ln( x- 3) 2 + 6 ( x- 3) 2 + C Explanation: Set u = x- 3 ; then du = dx , so I = 2 Z x ( x- 3)- 2 dx = 2 Z ( u + 3) u- 2 du = 2 Z du u + 6 Z u- 2 du. But 2 Z du u = 2 ln | u | + C = ln u 2 + C, while 6 Z u- 2 du =- 6 u- 1 + C. Consequently, I = ln( x- 3) 2- 6 x- 3 + C . keywords: 003 (part 1 of 1) 10 points Portillo, Tom – Homework 5 – Due: Oct 3 2006, 3:00 am – Inst: David Benzvi 2 Evaluate the definite integral I = Z 4 1 4 √ x ( √ x + 3) dx. 1. I = 8( √ 5- 2) 2. I = 8 ln 3 2 3. I = 4 ln 5 4 4. I = 8( √ 6- 2) 5. I = 4( √ 6- 2) 6. I = 4 ln 3 2 7. I = 8 ln 5 4 correct 8. I = 4( √ 5- 2) Explanation: Set u 2 = x . Then 2 udu = dx , while x = 1 = ⇒ u = 1 x = 4 = ⇒ u = 2 . In this case, I = 8 Z 2 1 1 u + 3 du = 8 h ln | u + 3 | i 2 1 . Thus I = 8 ‡ ln 5- ln 4 · = 8 ln 5 4 . keywords: 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 4 x 2 + x + 4 3 x 2 + 3 dx. 1. I = 8 + ln 2 6 correct 2. I = 4 + ln 3 6 3. I = 8 + ln 2 3 4. I = 4 + ln 2 6 5. I = 8 + ln 3 6 6. I = 8 + ln 3 3 Explanation: After division 4 x 2 + x + 4 3 x 2 + 3 = 4 3 + 1 3 ‡ x x 2 + 1 · . Thus I = Z 1 4 3 dx + 1 3 Z x x 2 + 1 dx = h 4 3 x i 1 + 1 3 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 = 4 3 + 1 6 Z 2 1 1 u du = 4 3 + h 1 6 ln u i 2 1 . Conequently, I = 8 + ln 2 6 . keywords: 005 (part 1 of 1) 10 points Evaluate the definite integral I = Z 3 π/ 4 π/ 4 5 cos x- 2 sin x 5 sin x + 2 cos x dx. Portillo, Tom – Homework 5 – Due: Oct 3 2006, 3:00 am – Inst: David Benzvi 3 1. I =- ln (3) 2. I = ln (3) 3. I = ln µ 7 3 ¶ 4. I =- ln µ 7 3 ¶ correct 5. none of these Explanation: Since the integrand is of the form f ( x ) f ( x ) , f ( x ) = 5 sin x + 2 cos x this suggests the substitution u = 5 sin x + 2 cos x . For then du = (5 cos x- 2 sin x ) dx, while x = π 4 = ⇒...
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## This note was uploaded on 02/18/2010 for the course MATH 408L taught by Professor Benzvi during the Spring '06 term at North Texas.

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HW 5 - Portillo Tom – Homework 5 – Due Oct 3 2006 3:00...

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