2 thus consequently 008 0 1 2 ln4y e ey 2 thus 4 ln4 1

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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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