Answer
$$\frac{8}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_{{x^2}}^{2x} {xy} dy} dx \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^2 {\left[ {\frac{{x{y^2}}}{2}} \right]_{{x^2}}^{2x}} dx \cr
& = \int_0^2 {\left[ {\frac{{x{{\left( {2x} \right)}^2}}}{2} - \frac{{x{{\left( {{x^2}} \right)}^2}}}{2}} \right]} dx \cr
& = \int_0^2 {\left[ {\frac{{4{x^3}}}{2} - \frac{{{x^5}}}{2}} \right]} dx \cr
& = \int_0^2 {\left[ {2{x^3} - \frac{{{x^5}}}{2}} \right]} dx \cr
& {\text{Integrate with respect to }}x \cr
& = \left[ {\frac{1}{2}{x^4} - \frac{{{x^6}}}{{12}}} \right]_0^2 \cr
& {\text{Evaluating}} \cr
& = \left[ {\frac{1}{2}{{\left( 2 \right)}^4} - \frac{1}{{12}}{{\left( 2 \right)}^6}} \right] - \left[ {\frac{1}{2}{{\left( 0 \right)}^4} - \frac{1}{{12}}{{\left( 0 \right)}^6}} \right] \cr
& = 8 - \frac{{16}}{3} \cr
& = \frac{8}{3} \cr} $$
