Answer
$${e^2} - 3$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\left( {y + 1} \right)}{e^{x\left( {y + 1} \right)}}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 1,\,\,\,\, - 1 \leqslant y \leqslant 1} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_{ - 1}^1 {\int_0^1 {\left( {y + 1} \right){e^{x\left( {y + 1} \right)}}} } dxdy \cr
& = \int_{ - 1}^1 {\left[ {\int_0^1 {\left( {y + 1} \right){e^{x\left( {y + 1} \right)}}} dx} \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = \int_0^1 {\left( {y + 1} \right){e^{x\left( {y + 1} \right)}}} dx \cr
& {\text{integrating}} \cr
& = \left( {{e^{x\left( {y + 1} \right)}}} \right)_0^1 \cr
& {\text{evaluating the limits for the variable }}x \cr
& = {e^{1\left( {y + 1} \right)}} - {e^{0\left( {y + 1} \right)}} \cr
& = {e^{y + 1}} - 1 \cr
& \int_{ - 1}^1 {\left[ {\int_0^1 {\left( {y + 1} \right){e^{x\left( {y + 1} \right)}}} dx} \right]} dy = \int_{ - 1}^1 {\left( {{e^{y + 1}} - 1} \right)} dy \cr
& {\text{integrating}} \cr
& = \left( {{e^{y + 1}} - y} \right)_{ - 1}^1 \cr
& {\text{evaluate}} \cr
& = \left( {{e^{1 + 1}} - 1} \right) - \left( {{e^{ - 1 + 1}} - \left( { - 1} \right)} \right) \cr
& = {e^2} - 1 - {e^0} - 1 \cr
& = {e^2} - 3 \cr} $$
