Answer
$e-e^{-1}$
Work Step by Step
Plug $u=x+y$ and $v=x-y$
$J(x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=(1) \cdot (-1) -(1)(1)=-2$
and $J(u,v)=|-\dfrac{1}{2}|=\dfrac{1}{2}$
Now, we have
$\iint_R e^{x+y} dA=(\dfrac{1}{2}) \int_{-1}^{1} \int_{-1}^{1} e^u du dv=(\dfrac{1}{2}) \int_{-1}^1 (e^u)_{-1}^1 dv$
This implies that
$\iint_R e^{x+y} dA= \dfrac{1}{2}(e-e^{-`1}) \int_{-1}^1 dv$
Thus, $\iint_R e^{x+y} dA=e-e^{-1}$
