# 15 7 2 i ln 2 dx dy x y 2 3 2 1 i 1 correct 3

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Unformatted text preview: : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 . 15 7 2. I = ln 2 dx dy . (x + y )2 3 2 1. I = 1 correct 3. I = 1 ln 2 15 7 2. I = −1 4. I = 1 ln 2 3 2 3. I = −2 5. I = 2 ln 5. I = 0 Explanation: Since A= is a rectangle with sides parallel to the coordinate axes, the value of I can be found by interpreting the double integral as the iterated integral 1 0 2 0 0 x2 − 3xy 0 2 dx = (x + y )2 =2 3 But now after integration with respect to x keeping y ﬁxed, we see that (2x − 3y ) dx = 4 2 0 0 4y − 3y 2 (4 − 6y ) dy = Consequently, 0 1 1 − . y 4+y 1 1 − y 4+y I=2 1 . dy 3 1 0 1 (3)(1 + 4) (4 + 3) = 2 ln 15 7 . 007 10.0 points Evaluate the iterated integral 4 I=1 . . Consequently, I = 2 ln I= 4 2 x+y = 2 ln(y ) − ln(4 + y ) Thus 2 − In this case (2x − 3y ) dxdy . 2 3 2 correct Explanation: Integrating the inner integral with respect to x keeping y ﬁxed, we see that (x, y ) : 0 ≤ x ≤ 2 , 0...
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## This note was uploaded on 08/25/2013 for the course MATH 408M taught by Professor Kushner during the Summer '10 term at University of Texas at Austin.

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