HW_5_Answers - Miller, Kierste Homework 5 Due: Feb 20 2007,...

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: Miller, Kierste Homework 5 Due: Feb 20 2007, 3:00 am Inst: Gary Berg 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 Find the most general function f such that f ( x ) = 5- 4 4- x 2 . 1. f ( x ) = 5 x- 2 tan- 1 x 2 + C 2. f ( x ) = 5 2 + 2 tan- 1 x 2 + C 3. f ( x ) = 5 2 x- 4 sin- 1 x + C 4. f ( x ) = 5 x- 4 sin- 1 x 2 + C correct 5. f ( x ) = 5 x + 4 sin- 1 x 2 + C 6. f ( x ) = 5 + 2 tan- 1 x + C Explanation: Since d dx sin- 1 x 2 = 1 4- x 2 , we see that f ( x ) = 5 x- 4 sin- 1 x 2 + C with C an arbitrary constant. keywords: 002 (part 1 of 1) 10 points Find the value of the integral I = Z 5 2 1 9 + ( x- 2) 2 dx. 1. I = 3 2. I = 3 3. I = 1 12 correct 4. I = 1 6 5. I = 1 6 6. I = 1 3 Explanation: Set 3 tan u = x- 2. Then 9 + ( x- 2) 2 = 9 + (3 tan u ) 2 = 9(1 + tan 2 u ) = 9 sec 2 u, while 3 sec 2 udu = dx. Also x = 2 = u = 0 , and x = 5 = u = 4 . In this case I = Z / 4 3 sec 2 u 9 sec 2 u du = 1 3 Z / 4 du. Consequently, I = 1 3 h u i / 4 = 1 12 . keywords: 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 7 1 1- 25 x 2 dx. Correct answer: 0 . 159121 . Explanation: Miller, Kierste Homework 5 Due: Feb 20 2007, 3:00 am Inst: Gary Berg 2 Since Z 1 1- x 2 dx = sin- 1 x + C , a change of variable x is needed to reduce I to this form. Set u = 5 x . Then du = 5 dx , and x = 0 = u = 0 , while x = 1 7 = u = 5 7 . In this case I = 1 5 Z 5 7 1 1- u 2 du = 1 5 sin- 1 u 5 7 . Consequently, I = 1 5 arcsin 5 7 = 0 . 159121 . keywords: 004 (part 1 of 1) 10 points Find the value of the definite integral I = Z 1 2 2 sin- 1 x 1- x 2 dx. 1. I = 1 25 2 2. I = 1 36 2 correct 3. I = 1 16 2 4. I = 1 12 2 5. I = 1 9 2 Explanation: Since Z 1 1- x 2 dx = sin- 1 x + C , this suggests the substitution u = sin- 1 x , for then du = 1 1- x 2 dx, while x = 0 = u = 0 , x = 1 2 = u = 6 ....
View Full Document

This note was uploaded on 05/08/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.

Page1 / 8

HW_5_Answers - Miller, Kierste Homework 5 Due: Feb 20 2007,...

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