Let D be a triangle with corners at (0,0), (1,0) and (1,1). To integrate this without substitution I found: $0 \leq x \leq 1$ and $0 \leq y \leq x$. So I will skip the evaluation part because its pretty easy for me. Now this is the part where I am stuck. My calculus books wants me to do a substitution $x = u + v$ and $y = u - v$. I have some trouble with finding the bounds for the new integral, I already saw an answer on the site but it pas posted with little to no explanation.
this is the other post where I don't understand how they found the bounds of u: Change of variables $x=u+v$, $y=u-v$
$$\begin{align}\iint_{0\le y\le x\le1}(x+y)dxdy&=\iint_{0\le v\le u,\;u+v\le1}(2u)(2dudv)\\&=4\int_0^{1/2}\left(\int_v^{1-v}udu\right)dv\\&=\dots\\&=\frac12.\end{align}$$