Answer
$e-e^{-1}$
Work Step by Step
$Jacobian, J(x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=(1) \cdot (-1) -(1)(1)=-2$
$J(u,v)=|-\dfrac{1}{2}|=\dfrac{1}{2}$
Let us consider that $u=x+y$ and $v=x-y$
Now, $\iint_R e^{x+y} dA= \int_{-1}^{1} \int_{-1}^{1} (\dfrac{1}{2})(e^u) du dv$
or, $\iint_R e^{x+y} dA= \int_{-1}^1 (\dfrac{1}{2}) |e^u|_{-1}^1 dv= \dfrac{1}{2}(e-e^{-`1}) \int_{-1}^1 dv$
Hence, our results is:
$\iint_R e^{x+y} dA=e-e^{-1}$
