HW 6 - Portillo, Tom Homework 6 Due: Oct 10 2006, 3:00 am...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Portillo, Tom Homework 6 Due: Oct 10 2006, 3:00 am Inst: David Benzvi 1 This print-out should have 17 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 definite integral I = Z / 4 (6 x + 5) sec 2 xdx. 1. I = 3 2 - 5- 3 ln 2 2. I = 3 - 5 + 3 ln 2 3. I = 3 - 5 + 6 ln 2 4. I = 3 2 + 5- 3 ln 2 correct 5. I = 3 + 5- 6 ln 2 Explanation: Since d dx tan x = sec 2 x, integration by parts is suggested. For then, I = h (6 x + 5) tan x i / 4- Z / 4 tan x d dx (6 x + 5) dx. = 3 2 + 5- 6 Z / 4 tan xdx. But Z / 4 tan xdx = h ln | sec x | i / 4 = ln 2 , so I = 3 2 + 5- 3 ln 2 . keywords: integration by parts, trig function 002 (part 1 of 1) 10 points Determine the indefinite integral I = Z e x cos xdx. 1. I = 1 2 e x sin x + cos x + C correct 2. I = e- x sin x- cos x + C 3. I = 1 2 e x sin x- cos x + C 4. I = e x sin x + cos x + C 5. I = e- x sin x- cos x + C 6. I = 1 2 e- x sin x + cos x + C Explanation: After integration by parts, I = e x cos x- Z e x d dx cos xdx = e x cos x + Z e x sin xdx. To reduce this last integral to one having the same form as I , we integrate by parts again for then Z e x sin xdx = e x sin x- Z e x d dx sin xdx = e x sin x- Z e x cos xdx = e x sin x- I . Thus I = e x cos x + n e x sin x- I o . Solving for I we see that 1 + 1 I = e x cos x + e x sin x. Portillo, Tom Homework 6 Due: Oct 10 2006, 3:00 am Inst: David Benzvi 2 Consequently I = 1 2 e x sin x + cos x + C with C an arbitrary constant. keywords: indefinite integral, integration by parts, exponential function, cosine function 003 (part 1 of 1) 10 points Determine the integral I = Z 5 ln x x 4 dx. 1. I =- 5 3 x 3 ln x + 1 3 + C correct 2. I =- 5 4 x 3 ln x + 1 3 + C 3. I = 5 3 x 3 ln x + 1 3 + C 4. I = 5 3 x 3 ln x- 1 3 + C 5. I = 5 4 x 3 ln x- 1 3 + C 6. I = 5 4 x 3 ln x + 1 3 + C Explanation: After integration by parts Z ln x x 4 dx = 1 3 - ln x x 3 + Z 1 x 4 dx =- 1 3 x 3 ln x + 1 3 + C . Consequently, I =- 5 3 x 3 ln x + 1 3 + C with C an arbitrary constant. keywords: indefinite integral, log integral, in- tegration by parts 004 (part 1 of 1) 10 points Determine the indefinite integral I = Z 3 cos(ln x ) dx....
View Full Document

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.

Page1 / 9

HW 6 - Portillo, Tom Homework 6 Due: Oct 10 2006, 3:00 am...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online