AnsHW9 - Cheung, Anthony – Homework 9 – Due: Nov 1...

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Unformatted text preview: Cheung, Anthony – Homework 9 – Due: Nov 1 2006, 3:00 am – Inst: David Benzvi 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 double integral I = Z 3 2 µZ 1 7 x 6 y dx ¶ dy. Correct answer: 2 . 5 . Explanation: The double integral I can be written as the repeated integral I = Z 3 2 µZ 1 7 x 6 y dx ¶ dy, integrating first with respect to x . Now after use of the Fundamental theorem of integral calculus, the inner integral becomes h 7 x 7 y 7 i 1 = y. Thus I = Z 3 2 y dy = h‡ y 2 2 ·i 3 2 . Consequently, I = 5 2 . keywords: 002 (part 1 of 1) 10 points Evaluate the integral I = Z 2- 1 Z 4 1 ( x 2- 2 xy ) dydx. 1. I =- 25 2 2. I =- 21 2 3. I =- 27 2 correct 4. I =- 13 2 Explanation: The integral can be written in iterated form I = Z 2- 1 ‰Z 4 1 ( x 2- 2 xy ) dy ¾ dx. Now Z 4 1 ( x 2- 2 xy ) dy = h x 2 y- xy 2 i 4 1 = (4 x 2- 16 x )- ( x 2- x ) = 3 x 2- 15 x. But then I = Z 2- 1 (3 x 2- 15 x ) dx = h x 3- 15 x 2 2 i 2- 1 . Consequently, I =- 27 2 . keywords: definite integral, iterated integral, polynomial function, 003 (part 1 of 1) 10 points Evaluate the iterated integral I = Z 5 1 n Z 5 1 ‡ x y + y x · dy o dx. 1. I = 5ln12 2. I = 24ln12 3. I = 24ln5 correct 4. I = 5ln24 5. I = 12ln5 Cheung, Anthony – Homework 9 – Due: Nov 1 2006, 3:00 am – Inst: David Benzvi 2 6. I = 12ln24 Explanation: Integrating with respect to y keeping x fixed, we see that Z 5 1 µ x y + y x ¶ dy = • x ln y + y 2 2 x ‚ 5 1 = (ln5) x + 12 µ 1 x ¶ . Thus I = Z 5 1 • (ln5) x + 12 µ 1 x ¶‚ dx = •µ x 2 2 ¶ ln5 + 12ln x ‚ 5 1 . Consequently, I = 24ln5 . keywords: 004 (part 1 of 1) 10 points Evaluate the double integral I = Z 3 2 Z 2 e x- y dxdy . 1. I = e- 3- e- 2- e- 1 + 1 correct 2. I = e- 3- e- 2 + e- 1 + 1 3. I = e- 3- e- 2- e- 1- 1 4. I = e- 3 + e- 2- e- 1 + 1 Explanation: After integration with respect to x , I = Z 3 2 £ e x- y / 2 dy = Z 3 2 ( e 2- y- e- y ) dy . But then, after integrating next with respect to y we see that I = £- e 2- y + e- y / 3 2 =- e- 1 + e- 3- (- 1 + e- 2 ) . Consequently, I = e- 3- e- 2- e- 1 + 1 . keywords: 005 (part 1 of 1) 10 points Determine the value of the double integral I = Z Z A 3 xy 2 16 + x 2 dA over the rectangle A = n ( x,y ) : 0 ≤ x ≤ 3 ,- 3 ≤ y ≤ 3 o , integrating first with respect to y . 1. I = 27 2 ln ‡ 25 16 · 2. I = 27 2 ln ‡ 25 32 · 3. I = 27ln ‡ 25 32 · 4. I = 27ln ‡ 16 25 · 5. I = 27ln ‡ 25 16 · correct 6. I = 27 2 ln ‡ 16 25 · Explanation: The double integral over the rectangle A can be represented as the iterated integral I = Z 3 µZ 3- 3 3 xy 2 16 + x 2 dy ¶ dx, Cheung, Anthony – Homework 9 – Due: Nov 1 2006, 3:00 am – Inst: David Benzvi 3 integrating first with respect to y . Now after integration with respect to y with x fixed, we see that Z 3- 3 3 xy 2 16 + x 2 dy = h xy 3 16 +...
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This note was uploaded on 01/27/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.

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AnsHW9 - Cheung, Anthony – Homework 9 – Due: Nov 1...

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