Answer
$$\frac{{e - {e^{ - 1}}}}{2}$$
Work Step by Step
$$\eqalign{
& {\text{From the digram below we have}} \cr
& y = x + 1 \to x = y - 1 \cr
& y = - x + 1 \to x = - y + 1 \cr
& {\text{we can define the limits of integration}} \cr
& y - 1 \leqslant x \leqslant - y + 1,{\text{ 0}} \leqslant y \leqslant 1 \cr
& {\text{Therefore}} \cr
& \int_a^b {\int_{g\left( y \right)}^{h\left( y \right)} {f\left( {x,y} \right)} dx} dy \cr
& = \int_0^1 {\int_{y - 1}^{ - y + 1} {{e^{x + y}}} dx} dy \cr
& = \int_0^1 {\left[ {\int_{y - 1}^{ - y + 1} {{e^{x + y}}} dx} \right]} dy \cr
& {\text{Evaluate inner integral}} \cr
& \int_{y - 1}^{1 - y} {{e^{x + y}}} dx = \left[ {{e^{x + y}}} \right]_{x = y - 1}^{x = - y + 1} \cr
& = {e^{ - y + 1 + y}} - {e^{y - 1 + y}} \cr
& = e - {e^{2y - 1}} \cr
& {\text{Therefore,}} \cr
& \int_0^1 {\left[ {\int_{y - 1}^{ - y + 1} {{e^{x + y}}} dx} \right]} dy \cr
& = \int_0^1 {\left( {e - {e^{2y - 1}}} \right)} dy \cr
& {\text{Integrating}} \cr
& = \left[ {ey - \frac{1}{2}{e^{2y - 1}}} \right]_0^1 \cr
& = \left[ {e\left( 1 \right) - \frac{1}{2}{e^{2\left( 1 \right) - 1}}} \right] - \left[ {e\left( 0 \right) - \frac{1}{2}{e^{2\left( 0 \right) - 1}}} \right] \cr
& = \left( {e - \frac{1}{2}e} \right) - \left( { - \frac{1}{2}{e^{ - 1}}} \right) \cr
& = \frac{1}{2}e + \frac{1}{2}{e^{ - 1}} \cr
& = \frac{{e - {e^{ - 1}}}}{2} \cr} $$
