M408L hwk5 solutions

M408L hwk5 solutions - Pham Quoc Homework 5 Due 3:00 am...

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

View Full Document Right Arrow Icon
Pham, Quoc – Homework 5 – Due: Feb 20 2007, 3:00 am – Inst: Eric Katerman 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points Find the most general ±unction f such that f 0 ( x ) = 6 + 5 4 - x 2 . 1. f ( x ) = 3 - 5 2 tan - 1 x 2 + C 2. f ( x ) = 6 x + 5 2 tan - 1 x 2 + C 3. f ( x ) = 6 - 5 2 tan - 1 x + C 4. f ( x ) = 6 x + 5 sin - 1 x 2 + C correct 5. f ( x ) = 6 x - 5 sin - 1 x 2 + C 6. f ( x ) = 3 x + 5 sin - 1 x + C Explanation: Since d dx sin - 1 x 2 · = 1 4 - x 2 , we see that f ( x ) = 6 x + 5 sin - 1 x 2 + C with C an arbitrary constant. keywords: 002 (part 1 o± 1) 10 points Find the value o± the integral I = Z 4 1 4 9 + ( x - 1) 2 dx. 1. I = 2 3 π 2. I = 3 π 3. I = 2 3 4. I = 3 5. I = 1 3 π correct 6. I = 4 3 π Explanation: Set 3 tan u = x - 1. Then 9 + ( x - 1) 2 = 9 + (3 tan u ) 2 = 9(1 + tan 2 u ) = 9 sec 2 u, while 3 sec 2 udu = Also x = 1 = u = 0 , and x = 4 = u = π 4 . In this case I = Z π/ 4 0 12 sec 2 u 9 sec 2 u du = 4 3 Z π/ 4 0 du. Consequently, I = 4 3 h u i π/ 4 0 = 1 3 π . keywords: 003 (part 1 o± 1) 10 points Evaluate the defnite integral I = Z 1 4 0 1 1 - 4 x 2 Correct answer: 0 . 261799 . Explanation:
Background image of page 1

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

View Full Document Right Arrow Icon
Pham, Quoc – Homework 5 – Due: Feb 20 2007, 3:00 am – Inst: Eric Katerman 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 = 2 x . Then du = 2 dx , and x = 0 = u = 0 , while x = 1 4 = u = 1 2 . In this case I = 1 2 Z 1 2 0 1 1 - u 2 du = 1 2 sin - 1 u 1 2 0 . Consequently, I = 1 2 arcsin µ 1 2 = 0 . 261799 . keywords: 004 (part 1 of 1) 10 points Find the value of the de±nite integral I = Z 3 2 0 8 sin - 1 x 1 - x 2 dx. 1. I = 4 25 π 2 2. I = 1 3 π 2 3. I = 1 9 π 2 4. I = 4 9 π 2 correct 5. I = 1 4 π 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 = 3 2 = u = π 3 . Thus I = 8 Z π 3 0 udu = 4 h u 2 i π 3 0 . Consequently, I = 4 9 π 2 . keywords: 005 (part 1 of 1) 10 points Evaluate the de±nite integral I = Z π π/ 4 3 sin 2 x 1 + cos 2 2 x Correct answer: - 1 . 1781 . Explanation: Set u = cos 2 x . Then du = - 2 sin 2 xdx, while x = π 4 = u = 0 , x = π = u = 1 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 8

M408L hwk5 solutions - Pham Quoc Homework 5 Due 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