Units 7 2xy dx dy 0 1 1 i 003 100 points 2

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Unformatted text preview: value of the iterated integral 3 2 I= I = 72 cu. units . (7 + 2xy ) dx dy . 0 1 1. I = 003 10.0 points 2. I = 57 2 3. I = 65 2 4. I = keywords: double integral, linear function, volume under graph, volume, rectangular region, prism, triangle 53 2 61 2 5. I = 69 correct 2 Find the value of the integral 2 I= f (x, y ) dx 0 when f (x, y ) = x − 4x2 y . 1. I = 2 + 32 2 y 3 Explanation: Integrating with respect to x and holding y ﬁxed, we see that 2 (7 + 2xy ) dx = 1 2. I = 2 − 4y 3. I = y − 4y 7 x + x2 y x=2 x=1 . Thus 2 4. I = y − 16y 2 5. I = 2 + 32 y 3 6. I = 2 − 32 y correct 3 3 I= (7 + 3y ) dy = 0 3 7y + y 2 2 Consequently, I = 21 + 27 69 . = 2 2 3 0 . chester (crc2876) – HW13 – meth – (91845) 3 keywords: 006 005 10.0 points 10.0 points Evaluate the iterated integral Evaluate the double integral 3 4 I= I= A 1 (2x − 3y ) dxdy 0 1. I = ln when A= (x, y )...
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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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