A integrating rst with respect to y after integration

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Unformatted text preview: + y ) ln x dy dx, 1 0 1 xy + y 2 l n x 2 Thus (e2x − 1) dx = 3(x + y ) ln x dxdy. A integrating first with respect to y . After integration the inner integral becomes 0 3 1 30 The integral with respect to y can be carried out immediately, whereas the integral with respect to x would require integration by parts, so this suggests that I should be written as a repeated integral xexy dy dx. Now after integration the inner integral becomes 2 = e2x − 1 . exy I= 19 4 Explanation: Since the area of the rectangle A is 30, the average value of f over A is given by xexy dxdy I= 36 19 ln 6 + 5 4 6 0 = 6x ln x + 18 ln x. Now is the time to integrate by parts, for then 6 6x ln x + 18 ln x dx and so I= 012 16 e −7 2 1 . 10.0 points 3 = 3x2 ln x − x2 + 18x ln x − 18x 2 6 1 . Thus Find the average value, I , of the function over the rectangle A = { (x, y ) : 1 ≤ x ≤ 6, 0 ≤ y ≤ 6 }. 1. I = 3 19...
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