Let $\hat T$ be the triangle spanned by $(0,0)$, $(1,0)$ and $(0,1)$. Let $$I(r,s) = \int_{\hat T} x^r y^s d(x,y)$$ with $r,s\in\mathbb N\cup\{0\}$. Prove that $$I(r,s) = \frac{r!s!}{(2+r+s)!}$$
To show this I rewrote the integral to $$\int_0^1 \int_0^{-x+1} x^r y^s dy dx = \frac{1}{s+1}\int_0^1 x^r (1-x)^{s+1}dx$$
but I don't know how to continue from this point. How can I solve this? Or is there another way to prove the proposition?
I think the trick is using integration by parts:
$$I\left[m,n\right] = \int_0^1 x^m (1-x)^{n}dx= \int_0^1 (x^m dx)(1-x)^n$$
Now using the formula$\int{v\ du}= uv-\int{u\ dv}$:
$$I\left[m,n\right]=\frac{x^{m+1}}{m+1}(1-x)^n|_0^1+ \frac{n}{m+1} \int_0^1 x^{m+1}\left(1-x\right)^{n-1}=0+\frac{n}{m+1}I\left[m+1,n-1\right]$$
Now using this formula repeatedly in our case will give us:
$$\frac{1}{s+1}I\left[r,s+1\right]=\frac{1}{r}I\left[r+1,s\right]=\frac{s}{r(r+1)}I\left[r+2,s-1\right]=\frac{s!\ r!}{(r+s+1)!}I\left[r+s+2,0\right]$$
But now,$I[r+s+2,0]$ is simply $\frac{1}{r+s+2}$, ergo:
$$I=\frac{s!\ r!}{(r+s+2)!}$$