Homework 7 Solutions - Mcbiles, Emily Homework 7 Due: Oct...

Info iconThis preview shows pages 1–4. 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

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: Mcbiles, Emily Homework 7 Due: Oct 17 2006, 3:00 am Inst: David Fonken 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 3 5 x 2 + 1 dx. 1. I = 5ln(2- 3) 2. I = 5(2 + 3 ) 3. I = 3(2- 3 ) 4. I = 3 ln(2 + 3) 5. I = 5(2- 3 ) 6. I = 5ln(2 + 3) correct Explanation: Set x = tan u, then dx = sec 2 udu, x 2 + 1 = sec 2 u, while x = 0 = u = 0 , x = 3 = u = 3 . In this case I = Z 3 5sec 2 u sec u du = Z 3 5sec udu. Now Z sec udu = ln | sec u + tan u | + C 1 , Thus I = 5 h ln | sec u + tan u | i / 3 . Consequently, I = 5ln(2 + 3 ) . keywords: 002 (part 1 of 1) 10 points Evaluate the integral I = Z 1 t 2- t 4 dt. 1. I = 1 2 2. I = 1 8 correct 3. I = 1 4 4. I = 1 8 5. I = 1 2 6. I = 1 4 Explanation: Set t 2 = 2 sin u . Then 2 tdt = 2cos udu, in which case tdt = 1 2 dt. On the other hand, t = 0 = u = 0 , t = 1 = u = 4 . Mcbiles, Emily Homework 7 Due: Oct 17 2006, 3:00 am Inst: David Fonken 2 Thus I = 1 2 Z / 4 cos u 2cos u du = 1 2 h u i / 4 . Consequently, I = 1 8 . keywords: definite integral, substitution, in- verse sin integral, 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 4 2 6 x 2- 6 x + 10 dx. 1. I = 7 2 2. I = 6 3. I = 3 correct 4. I = 5 4 5. I = 5 2 Explanation: By completing the square we see that x 2- 6 x + 10 = ( x 2- 6 x + 9) + 10- 9 = ( x- 3) 2 + 1 . Thus I = Z 4 2 6 ( x- 3) 2 + 1 . Since d dx tan- 1 x = 1 1 + x 2 , this suggests the substitution x- 3 = u . For then dx = du , while x = 2 = u =- 1 , x = 4 = u = 1 . In this case, I = 6 Z 1- 1 1 1 + u 2 du = 6 h tan- 1 u i 1- 1 = 6 tan- 1 1- tan- 1 (- 1) / . But tan- 1 1 = 4 while tan- 1 (- 1) =- 4 . Consequently, I = 3 . keywords: trigonometric substitutions, com- plete square 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 / 2 6 sin- 1 x + 3 1- x 2 dx. 1. I = 2 + 3 2 + 3 2. I = 2 - 3 2 + 3 3. I = + 3 2- 3 4. I = 2 + 3 2- 3 5. I = - 3 2- 3 correct Explanation: Let x = sin ; then dx = cos d and 1- x 2 = 1- sin 2 = cos 2 , while x = 0 = = 0 , x = 1 2 = = 6 . Mcbiles, Emily Homework 7 Due: Oct 17 2006, 3:00 am Inst: David Fonken 3 In this case I = Z / 6 6 + 3 cos cos d = Z / 6 (6 cos + 3) d = 6 Z / 6 cos d + 1 2 ....
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 7 Solutions - Mcbiles, Emily Homework 7 Due: Oct...

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

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