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Unformatted text preview: y ≤ 1 I= 15
7 6. I = 2 ln 4. I = −3 4 I=
1 1 xy
dydx .
+
yx . chester (crc2876) – HW13 – meth – (91845)
15
2 1. I = 15 ln 3. I = 7 15
2 4. I = 9 15
ln (4)
2 5. I = 8 2. I = 4 ln
3. I = Explanation:
Integrating with respect to x with y ﬁxed,
we see that 4. I = 15 ln (4) correct ln(4) 5. I = 4 ln (15)
6. I = 15
ln (15)
2 1 xy
+
yx e2x−y dx = 0 = Explanation:
Integrating with respect to y keeping x
ﬁxed, we see that
4 4 4 y2
x ln(y ) +
2x dy = = (ln(4)) x + 1
x 15
2 1 . I= (ln(4)) x +
1 x2
2 = 15
2 ln(4) + 1
x 15
ln(x)
2 I= 15
2 =− 0 e−y dy = − I=− 15 1
−1
23 009 10.0 points . Evaluate the double integral 1 Evaluate the iterated integral 1. I = 5 correct =5. 4 e2x−y dx dy . 5 + x2
dxdy
1 + y2 when 10.0 points 0 0 dx A= ln(4) ln(3) 15 − ln(3)
e
−1 .
2 A ln(3) 15 −y
e
2 Consequently, I = 15 ln(4) . 2. I = 6 ln(3) I= I= 42 − 1 − y
e.
2 T...
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 Summer '10
 KUSHNER
 Multivariable Calculus

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