Answer
$\iint_R f(x+y) dA=\int_0^1u f(u)$
Work Step by Step
Plug $u=x+y$ and $v=x-y$
$J(x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} 1&1\\1& -1\end{vmatrix}=-2$
Then, we have $J(u,v)=|-\dfrac{1}{2}|=\dfrac{1}{2}$
Now,we have $\iint_R f(x+y) dA=(\dfrac{1}{2}) \int_{0}^{1} \int_{-u}^{u} f(u) dv du$
This gives: $\iint_R f(x+y) dA=(\dfrac{1}{2}) \int_{0}^1 [u-(-u)] f(u) du$
This implies that
$\iint_R f{x+y} dA=(\dfrac{1}{2}) \int_0^1(2u) f(u)=\int_0^1u f(u)$
Hence, $\iint_R f(x+y) dA=\int_0^1u f(u)$ (Proved)
