Answer
$$\frac{{{{\left( {e - 1} \right)}^2}}}{2}$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {y{e^{{y^2} + x}}}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \leqslant x \leqslant 1,\,\,\,\,\,\,\,0 \leqslant y \leqslant 1 \cr
& {\text{Replacing the limits for the region }}R \cr
& = \int_0^1 {\int_0^1 {y{e^{{y^2} + x}}dx} dy} \cr
& = \int_0^1 {\left[ {\int_0^1 {y{e^{{y^2} + x}}dx} } \right]dy} \cr
& {\text{solve the inner integral treat }}y{\text{ as a constant}} \cr
& = \int_0^1 {y{e^{{y^2} + x}}dx} \cr
& = y\int_0^1 {{e^{{y^2} + x}}dx} \cr
& {\text{integrate }} \cr
& = y\left( {{e^{{y^2} + x}}} \right)_0^1 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = y\left( {{e^{{y^2} + 1}} - {e^{{y^2} + 0}}} \right) \cr
& {\text{simplifying}} \cr
& = y{e^{{y^2} + 1}} - y{e^{{y^2}}} \cr
& \cr
& \int_0^1 {\left[ {\int_0^1 {y{e^{{y^2} + x}}dx} } \right]dy} = \int_0^1 {\left( {y{e^{{y^2} + 1}} - y{e^{{y^2}}}} \right)dx} \cr
& = \frac{1}{2}\int_0^1 {\left( {2y{e^{{y^2} + 1}} - 2y{e^{{y^2}}}} \right)dx} \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left( {{e^{{y^2} + 1}} - {e^{{y^2}}}} \right)_0^1 \cr
& {\text{evaluate}} \cr
& = \frac{1}{2}\left( {{e^{{1^2} + 1}} - {e^{{1^2}}}} \right) - \frac{1}{2}\left( {{e^{{1^2} + 0}} - {e^0}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\left( {{e^2} - e} \right) - \frac{1}{2}\left( {e - 1} \right) \cr
& = \frac{1}{2}\left( {{e^2} - e - e + 1} \right) \cr
& = \frac{1}{2}\left( {{e^2} - 2e + 1} \right) \cr
& = \frac{{{{\left( {e - 1} \right)}^2}}}{2} \cr} $$
