Integration over a triangle Let $\Delta$ be a triangle with vertices $(0,0)$, $(0,1)$ and $(1,0)$ in ${\bf R}^2$.
I want to compute $$I=\int\limits_\Delta x^2\mathrm{e}^{y^2}\;\mathrm{d}A.$$
This is what I've done so far: Note that the hypotenuse is given by the line $y=1-x$. Keep $x\in[0,1]$ fixed, then $y$ is between $1$ and $1-x$ this gives
$I=\int_0^1\int_1^{1-x}x^2\mathrm{e}^{y^2}\;\mathrm{d}y\;\mathrm{d}x$, which can't be computed. But when $y\in[0,1]$ is fixed, $x$ is between $0$ and $1-y$, which gives $$I=\int\limits_0^1\int\limits_1^{1-y}x^2\mathrm{e}^{y^2}\;\mathrm{d}x\;\mathrm{d}y=\frac13\int\limits_0^1(1-y)^3\mathrm{e}^{y^2}\;\mathrm{d}y,$$ which gives the same trouble.
As you can see, I keep ending up with some sort of Gaussian integral, which is impossible to compute. Does anyone know how to compute this integral?
 A: Let's integrate over $x$ first, then $y$, 
$$\begin{eqnarray}
I &=& \int_0^1 dy \ e^{y^2} \int_0^{1-y} x^2 \\
&=& \frac{1}{3} \int_0^1 dy \ e^{y^2}(1-y)^3 \\
&=& \frac{1}{3} \left(\langle 1\rangle - 3\langle y\rangle + 3\langle y^2\rangle - \langle y^3\rangle\right),
\end{eqnarray}$$
where, for convenience, we define
$$\langle y^n\rangle = \int_0^1 dy \ e^{y^2}y^n.$$
Using integration by parts show that the integrals are related in the following way for $n>1$
$$\langle y^n\rangle = \frac{1}{2}e - \frac{n-1}{2}\langle y^{n-2}\rangle.$$
Thus, we need only calculate the integrals 
$$\begin{eqnarray}
\langle 1\rangle &=& \int_0^1 dy \ e^{y^2}  = \frac{\sqrt{\pi}}{2} \mathrm{erfi}(1) \\
\langle y\rangle &=& \int_0^1 dy \ e^{y^2} y = \frac{1}{2}(e-1)
\end{eqnarray}$$
since $\langle y^2\rangle$ and $\langle y^3\rangle$ can be found in terms of them. 
(Above, $\mathrm{erfi}$ is the imaginary error function.)
The integral $\langle 1\rangle$ is the only nontrivial integral we must calculate. 
It is enough to recognize that it is a standard integral. 
Putting this all together we find 
$$I = \frac{1}{3}-\frac{\sqrt{\pi}}{12} \mathrm{erfi}(1).$$
