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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.
 Summer '10
 KUSHNER
 Multivariable Calculus

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