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September 7th, 2017, 07:22 PM   #1
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Double Integral Limits

What is the solution to the wrong answers in the attachment and how to get it ?
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September 7th, 2017, 09:58 PM   #2
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$C = 3y$

$D=4-y$

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 September 8th, 2017, 04:43 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,244 Thanks: 887 Notice that the integral is $\displaystyle \int\int dxdy$ so that the limits of integration, on the "outer integral", with respect to y, must be numbers while the limits of integration, on the "inner integral", with respect to x, may depend on y. Looking at the graph, y must go from 0 to 1. For each y, x must go from the left side line, which goes from (0, 0) to (3, 1), to the right side line, which goes from (3, 1) to (4, 0). The line from (0, 0) to (3, 1) is, of course, x= 3y. Any (non-horizontal) line can be written "x= ay+ b". Since the second line goes through (3, 1), we must have 3= a+ b. Since it also goes through (4, 0) we must have 4= 0(a)+ b. From the second equation, b= 4 so that 3= a+ 4, a= -1. The equation of the second line is x= -y+ 4 The integral is $\displaystyle \int_0^1\int_{3y}^{-y+ 4} xdxdy$

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