Answer
$$\frac{{{e^2}}}{2} - \frac{7}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^2 {\left( {y{e^x} - x - y} \right)} dxdy} \cr
& \int_0^1 {\left[ {\int_0^2 {\left( {y{e^x} - x - y} \right)} dx} \right]dy} \cr
& {\text{Evaluate inner integral }}\int_0^2 {\left( {y{e^x} - x - y} \right)} dx \cr
& {\text{Integrate with respect to }}x \cr
& \int_0^2 {\left( {y{e^x} - x - y} \right)} dx = \left[ {y{e^x} - \frac{{{x^2}}}{2} - xy} \right]_{x = 0}^{x = 2} \cr
& = \left[ {{e^2}y - \frac{{{{\left( 2 \right)}^2}}}{2} - \left( 2 \right)y} \right] - \left[ {{e^0}y - \frac{{{{\left( 0 \right)}^2}}}{2} - \left( 0 \right)y} \right] \cr
& = {e^2}y - 2y - 2 - y \cr
& = {e^2}y - 3y - 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& \int_0^1 {\int_0^2 {\left( {y{e^x} - x - y} \right)} dxdy} = \int_0^1 {\left( {{e^2}y - 3y - 2} \right)} dy \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {\frac{{{e^2}{y^2}}}{2} - \frac{3}{2}{y^2} - 2y} \right]_0^1 \cr
& = \left[ {\frac{{{e^2}{{\left( 1 \right)}^2}}}{2} - \frac{3}{2}{{\left( 1 \right)}^2} - 2\left( 1 \right)} \right] - \left[ {\frac{{{e^2}{{\left( 0 \right)}^2}}}{2} - \frac{3}{2}{{\left( 0 \right)}^2} - 2\left( 0 \right)} \right] \cr
& {\text{Simplifying}} \cr
& = \frac{{{e^2}}}{2} - \frac{3}{2} - 2 \cr
& = \frac{{{e^2}}}{2} - \frac{7}{2} \cr} $$
