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Unformatted text preview: hus Consequently, 008 0 1 2 ln(4)−y
e
− e−y =
2 Thus
4 ln(4) 1 2x−y
e
2 1. I = (x, y ) : 0 ≤ x ≤ 2,
10
π
3 2. I = 3π 0 3. I = 17
π
6 4. I = 7
π
2 0≤y≤1 . chester (crc2876) – HW13 – meth – (91845)
5. I = 19
π correct
6 5. I = −2(4 − π ) Explanation:
Since
A= 6. I = −2π (x, y ) : 0 ≤ x ≤ 2, Explanation:
By treating I as an iterated integral, integrating ﬁrst with respect to x, we see that 0≤y≤1 is a rectangle with sides parallel to the coordinate axes, the double integral can be represented as the iterated integral
1 2 I=
0 0 2x sin(x + y ) dx = −2x cos(x + y ) 5 + x2
dxdy .
1 + y2 +2 cos(x + y ) dx = 2 −x cos(x + y ) + sin(x + y ) . Now
2
0 5 1
1
5 + x2
dx =
5 x + x3
2
2
1+y
1+y
3 2
0 Thus . π /2 Thus 0 38
I=
3 1
0 38
1
tan−1 (y )
dy =
2
1+y
3 1
0 2x sin(x + y ) dx = −π cos π
+ y − 2 sin(y ) .
2 + 2 sin . π
+y
2 In this...
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 Summer '10
 KUSHNER
 Multivariable Calculus

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