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$\displaystyle\iint e^{x-y}$ over the triangle with vertices at $(0,0),(1,3),(2,2)$ I tried the following change of variables. let $u=x-y$, $v=3x-y$. In $(u,v)$, we get the triangle with vertices at $(0,0),(-2,0),(0,4)$.The Jacobian i calculated is $1/2$. I tried integrating $(1/2)\displaystyle\iint e^u$ over this new region and got $2/e^2$. The answer provided $1+1/e^2$, I'm not too sure why my change of variables is not working. Any help would be greatly appreciated

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  • $\begingroup$ For such simple form of the integrand, changing variables is not necessary. Drawing a sketch of the region is enough to find the limits of integration. $\endgroup$ – StubbornAtom Nov 23 '18 at 17:45
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First of all, I assume you mean the region $R$ enclosed by the triangle, not the triangle itself. Otherwise the integral is just zero.

Instead of bothering with Jacobians, which I personally find super annoying, it may be best simply to break it up into two pieces. Thus we have $$\iint\limits_R e^{x-y}\;dA=\int_0^1\int_x^{3x}e^{x-y}\;dy\;dx+\int_1^2\int_x^{4-x}e^{x-y}\;dy\;dx$$ $$=\left(\frac{1}{2}+\frac{1}{2e^2}\right)+\left(\frac{1}{2}+\frac{1}{2e^2}\right)=1+\frac{1}{e^2}.$$

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