Homework 5 solutions - Mcbiles Emily – Homework 5 – Due Oct 3 2006 3:00 am – Inst David Fonken 1 This print-out should have 20 questions

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Unformatted text preview: Mcbiles, Emily – Homework 5 – Due: Oct 3 2006, 3:00 am – Inst: David Fonken 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 x 5- 4 x 2 dx. 1. I = ln 5 2. I = 1 8 ln 5 correct 3. I = 1 8 ln 5 4 4. I = ln 9 5 5. I = 1 4 ln 9 5 6. I = 1 4 ln 5 4 Explanation: Set u = 5- 4 x 2 ; then du =- 8 dx while x = 0 = ⇒ u = 5 x = 1 = ⇒ u = 1 . In this case, I =- 1 8 Z 1 5 1 u du = 1 8 Z 5 1 1 u du = 1 8 h ln | u | i 5 1 . Consequently, I = 1 8 (ln 5- ln 1) = 1 8 ln 5 . 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 = 6 ( x- 3) 2 + C 2. I =- 2 x- 3 + C 3. I = ln( x- 3) 2- 6 x- 3 + C correct 4. I = 3 ln( x- 3) 2 + 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 Mcbiles, Emily – Homework 5 – Due: Oct 3 2006, 3:00 am – Inst: David Fonken 2 Evaluate the definite integral I = Z 4 1 2 √ x ( √ x + 4) dx. 1. I = 4 ln 7 5 2. I = 4( √ 7- √ 5) 3. I = 2( √ 6- √ 5) 4. I = 2 ln 7 5 5. I = 4( √ 6- √ 5) 6. I = 4 ln 6 5 correct 7. I = 2 ln 6 5 8. I = 2( √ 7- √ 5) Explanation: Set u 2 = x . Then 2 udu = dx , while x = 1 = ⇒ u = 1 x = 4 = ⇒ u = 2 . In this case, I = 4 Z 2 1 1 u + 4 du = 4 h ln | u + 4 | i 2 1 . Thus I = 4 ‡ ln 6- ln 5 · = 4 ln 6 5 . keywords: 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 4 x 2 + 2 x + 4 4 x 2 + 4 dx. 1. I = 2 + ln 2 4 2. I = 4 + ln 2 2 3. I = 4 + ln 3 4 4. I = 4 + ln 2 4 correct 5. I = 4 + ln 3 2 6. I = 2 + ln 3 4 Explanation: After division 4 x 2 + 2 x + 4 4 x 2 + 4 = 1 + 1 2 ‡ x x 2 + 1 · . Thus I = Z 1 1 dx + 1 2 Z x x 2 + 1 dx = h 1 x i 1 + 1 2 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 = 1 + 1 4 Z 2 1 1 u du = 1 + h 1 4 ln u i 2 1 . Conequently, I = 4 + ln 2 4 . keywords: 005 (part 1 of 1) 10 points Evaluate the definite integral I = Z 3 π/ 4 π/ 4 4 cos x- sin x 4 sin x + cos x dx. Mcbiles, Emily – Homework 5 – Due: Oct 3 2006, 3:00 am – Inst: David Fonken 3 1. I = ln µ 7 3 ¶ 2. I =- ln µ 5 3 ¶ correct 3. none of these 4. I =- ln µ 7 3 ¶ 5. I = ln µ 5 3 ¶ Explanation: Since the integrand is of the form f ( x ) f ( x ) , f ( x ) = 4 sin x + cos x this suggests the substitution u = 4 sin x + cos x . For then du = (4 cos x- sin...
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This note was uploaded on 03/19/2008 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

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Homework 5 solutions - Mcbiles Emily – Homework 5 – Due Oct 3 2006 3:00 am – Inst David Fonken 1 This print-out should have 20 questions

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