Answer
$$\frac{7}{6}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^{2 - y} x dxdy} \cr
& \int_0^1 {\left[ {\int_0^{2 - y} x dx} \right]dy} \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^{2 - y} x dx = \left[ {\frac{1}{2}{x^2}} \right]_{x = 0}^{x = 2 - y} \cr
& = \frac{1}{2}\left[ {{{\left( {2 - y} \right)}^2} - {{\left( 0 \right)}^2}} \right] \cr
& = \frac{1}{2}{\left( {2 - y} \right)^2} \cr
& {\text{Therefore,}} \cr
& \int_0^1 {\left[ {\int_0^{2 - y} x dx} \right]dy} = \int_0^1 {\frac{1}{2}{{\left( {2 - y} \right)}^2}dy} \cr
& = - \frac{1}{2}\int_0^1 {{{\left( {2 - y} \right)}^2}\left( { - 1} \right)dy} \cr
& {\text{Integrating}} \cr
& = - \frac{1}{2}\left[ {\frac{{{{\left( {2 - y} \right)}^3}}}{3}} \right]_0^1 \cr
& = - \frac{1}{6}\left[ {{{\left( {2 - y} \right)}^3}} \right]_0^1 \cr
& = - \frac{1}{6}\left[ {{{\left( {2 - 1} \right)}^3} - {{\left( {2 - 0} \right)}^3}} \right] \cr
& = - \frac{1}{6}\left[ {1 - 8} \right] \cr
& = \frac{7}{6} \cr} $$
