Let $D \subseteq \mathbb{R}^2$ be the triangle with corners $(0,0),(0,1),(1,0)$ and let $f:[0,1]\to \mathbb{R}$ be continuous. How can I show that for $m, n \in \mathbb{N}$:
$$\int_Df(x+y)x^my^n d\lambda_2(x,y)^T=\frac{m!n!}{(m+n+1)!}\int_{0}^{1}f(t)t^{m+n+1}dt$$
I'm thankful for any hints/tip/solutions!
Presumably, $\lambda_2$ is the Lebesgue measure on $\mathbb{R}^2$. Then, we are to show $$\iint_D\,f(x+y)\,x^m\,y^n\,\text{d}x\,\text{d}y=\frac{m!\,n!}{(m+n+1)!}\,\int_0^1\,f(t)\,t^{m+n+1}\,\text{d}t\,.$$ Rewrite the required integral as $$\iint_D\,f(x+y)\,x^m\,y^n\,\text{d}x\,\text{d}y=\int_0^1\,\int_0^t\,f(t)\,x^m\,(t-x)^n\,\text{d}x\,\text{d}t\,,$$ where $t:=x+y$, noting that the Jacobian determinant of the coordinate transformation $$(x,y)\mapsto (x,x+y)$$ is $1$ (since the Jacobian matrix is $\begin{bmatrix}1&0\\1&1\end{bmatrix}$). This shows that $$\iint_D\,f(x+y)\,x^m\,y^n\,\text{d}x\,\text{d}y=\int_0^1\,f(t)\,t^{m+n+1}\,\int_0^1\,\,s^m\,(1-s)^n\,\text{d}s\,\text{d}t\,,$$ where $s:=\dfrac{x}{t}$ (for $t>0$). Use this to show $$\int_0^1\,\,s^m\,(1-s)^n\,\text{d}s=\frac{m!\,n!}{(m+n+1)!}\,.$$