Answer
$$\frac{4}{3}$$
Work Step by Step
$$\eqalign{
& {\text{We have the points }}\left( {2,0} \right){\text{ and }}\left( {0,2} \right) \cr
& {\text{The slope is}} \cr
& m = \frac{{2 - 0}}{{0 - 2}} = - 1 \cr
& {\text{The }}y{\text{ - intercept is }}b = 2,{\text{ The equation of the line is}} \cr
& y = mx + b \cr
& y = - x + 2 \cr
& {\text{From the graph we can define the limits of integration}} \cr
& 0 \leqslant y \leqslant - x + 2,{\text{ 0}} \leqslant x \leqslant 2 \cr
& {\text{Therefore}} \cr
& \int_a^b {\int_{g\left( x \right)}^{h\left( x \right)} {f\left( {x,y} \right)} dy} dx \cr
& = \int_0^2 {\int_0^{ - x + 2} x dy} dx \cr
& = \int_0^2 {x\left[ {\int_0^{ - x + 2} {dy} } \right]} dx \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^{ - x + 2} {dy} = \left[ y \right]_{y = 0}^{y = - x + 2} \cr
& = - x + 2 \cr
& {\text{Therefore,}} \cr
& \int_0^2 {x\left[ {\int_0^{ - x + 2} {dy} } \right]} dx = \int_0^2 {x\left( { - x + 2} \right)} dx \cr
& = \int_0^2 {\left( {2x - {x^2}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \left[ {{x^2} - \frac{1}{3}{x^3}} \right]_0^2 \cr
& = \left[ {{{\left( 2 \right)}^2} - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - \left[ {{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr
& = \frac{4}{3} \cr} $$
