Double integral over triangle, what bounds should be chosen?

Question

Question

In an exercise, I am supposed to evaluate the integral $$ \iint_D(x -y)e^{x+y} $$ where $D$ is the triangle with the three corner points $(0,0)$, $(0,2)$ and $(2,2)$.

I tried computing $$ \int_0^2\int_0^x f(x,y)~dydx $$ and I got the given correct answer, but with the opposite sign eg. $(-1)\cdot$the correct answer. This can happen when the bounds are flipped for an integral as can be seen in this S.E question for example. Have I flipped on any of the bounds? It seems to me, that these are the obvious bounds to choose.

You can see my solution below, if this is not the problem.

My solution

$$ \begin{align} f &= (x-y)e^{x+y}\\ \iint_D f~dydx &= \int_0^2\int_0^x f~dydx\\ \int_0^x f~dy &= \left[(x-y)e^{x+y} + e^{x+y} \right]_0^x\\ &= e^{2x} - xe^x - e^x\\ \Rightarrow \int_0^2\int_0^x f(x,y)~dydx &= \left[\frac{e^{2x}}{2} - xe^x \color{grey}{+e^x - e^x} \right]_0^x\\ = e^4/2 - 2e^2 - 1/2 &=\frac{1}{2}(e^4 - 4e^2 - 1) \end{align} $$ It's supposed to be $-1/2(e^4 - 4e^2 - 1)$. I have double checked with wolfram alpha and it seems that my calculation of the integral is correct.

Answer 1

In the region $D$, $x\leqslant y\leqslant 2$, so the limits for $y$ should be $x$ and $2$ rather than $0$ and $x$. With your limits, you are finding the integral over the triangle with vertices $(0,0), (2,0)$ and $(2,2)$.

Answer 2

The triangle you are integrating over has vertices $(0,0),(0,2)$ and $(2,2)$. Now what this geometrically means, is that you integrate over the set $\{(x,y)\in\mathbb{R}^2|0\leq x,y\leq2\;and\;y\geq x\}$. Make a sketch if you do not see this directly, but it is not very hard to see this. This has the implication, however that we should integrate with respect to $y$ not from $0$ to $x$ but from $x$ to $2$ (Since $x\geq y\geq 2$). With these new boundaries, your integral should give the right result.