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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.
 Summer '10
 KUSHNER
 Multivariable Calculus

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