homework 5 solutions

# 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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

### Page1 / 10

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online