# $\iint e^{x-y}$ over the triangle with vertices at $(0,0),(1,3),(2,2)$

$$\displaystyle\iint e^{x-y}$$ 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

• 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. – StubbornAtom Nov 23 '18 at 17:45

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 $$\iint\limits_R e^{x-y}\;dA=\int_0^1\int_x^{3x}e^{x-y}\;dy\;dx+\int_1^2\int_x^{4-x}e^{x-y}\;dy\;dx$$ $$=\left(\frac{1}{2}+\frac{1}{2e^2}\right)+\left(\frac{1}{2}+\frac{1}{2e^2}\right)=1+\frac{1}{e^2}.$$