# I 2 ln 4 i 3 4 i 1 1 xy dydx yx chester crc2876

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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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