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Unformatted text preview: Su, Yung – Homework 9 – Due: Oct 30 2007, 3:00 am – Inst: Shinko Harper 1 This printout should have 15 questions. Multiplechoice 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 iterated integral I = Z 2 1 n Z 2 1 ( x + y ) 2 dx o dy . 1. I = ln 2 3 2. I = 2 ln 3 2 3. I = ln 3 2 correct 4. I = 1 2 ln 2 3 5. I = 1 2 ln 3 2 6. I = 2 ln 2 3 Explanation: Integrating the inner integral with respect to x keeping y fixed, we see that Z 2 1 ( x + y ) 2 dx = h 1 x + y i 2 = n 1 y 1 2 + y o . In this case I = Z 2 1 n 1 y 1 2 + y o dy = h ln y ln(2 + y ) i 2 1 . Consequently, I = ln ‡ (2)(1 + 2) (2 + 2) · = ln 3 2 . keywords: iterated integral, rational function, log integral 002 (part 1 of 1) 10 points Evaluate the iterated integral I = Z ln 5 ˆ Z ln 4 e 2 x y dx ! dy . 1. I = 7 2. I = 5 3. I = 6 correct 4. I = 4 5. I = 8 Explanation: Integrating with respect to x with y fixed, we see that Z ln 4 e 2 x y dx = 1 2 h e 2 x y i ln 4 = 1 2 ‡ e 2 ln 4 y e y · = ‡ 4 2 1 2 · e y . Thus I = 15 2 Z ln 5 e y dy = 15 2 h e y i ln 5 = 15 2 ‡ e ln 5 1 · . Consequently, I = 15 2 ‡ 1 5 1 · = 6 . keywords: 003 (part 1 of 1) 10 points Su, Yung – Homework 9 – Due: Oct 30 2007, 3:00 am – Inst: Shinko Harper 2 Determine the value of the double integral I = Z Z A 3 xy 2 9 + x 2 dA over the rectangle A = n ( x, y ) : 0 ≤ x ≤ 4 , 2 ≤ y ≤ 2 o , integrating first with respect to y . 1. I = 8 ln ‡ 9 25 · 2. I = 4 ln ‡ 9 25 · 3. I = 4 ln ‡ 25 9 · 4. I = 8 ln ‡ 25 9 · correct 5. I = 8 ln ‡ 25 18 · 6. I = 4 ln ‡ 25 18 · Explanation: The double integral over the rectangle A can be represented as the iterated integral I = Z 4 µZ 2 2 3 xy 2 9 + x 2 dy ¶ dx , integrating first with respect to y . Now after integration with respect to y with x fixed, we see that Z 2 2 3 xy 2 9 + x 2 dy = h xy 3 9 + x 2 i 2 2 = 16 x 9 + x 2 . But Z 4 16 x 9 + x 2 dx = h 8 ln(9 + x 2 ) i 4 . Consequently, I = 8 ln ‡ 25 9 · . keywords: 004 (part 1 of 1) 10 points Evaluate the integral, I , of the function f ( x, y ) = 2 xe 3 xy over the rectangle A = { ( x, y ) : 0 ≤ x ≤ 3 , ≤ y ≤ 1 } . 1. I = 2 9 ‡ e 9 10 · correct 2. I = 1 9 ‡ e 9 10 · 3. I = 2 9 ‡ e 9 8 · 4. I = 2 9 ‡ e 9 9 · 5. I = 1 9 ‡ e 9 8 · 6. I = 1 9 ‡ e 9 9 · Explanation: The integral is given by I = Z Z A 2 xe 3 xy dxdy. Since the integral with respect to y can be evaluated easily using substitution (or di rectly making the substitution in one’s head), while the integral with respect to x requires integration by parts, this suggests that we should represent the double integral as the repeated integral I =...
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 Spring '09
 Harper
 Calculus, dy, Shinko Harper, 1 1 g

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