Answer
$$\frac{{88}}{{15}}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_{{x^2}}^{2x} {\left( {{x^2} + 2y} \right)} } dydx \cr
& = \int_0^2 {\left[ {\int_{{x^2}}^{2x} {\left( {{x^2} + 2y} \right)} dy} \right]} dx \cr
& {\text{Integrate with respect to }}y \cr
& \int_{{x^2}}^{2x} {\left( {{x^2} + 2y} \right)} dy = \left[ {{x^2}y + {y^2}} \right]_{{x^2}}^{2x} \cr
& = \left[ {{x^2}\left( {2x} \right) + {{\left( {2x} \right)}^2}} \right] - \left[ {{x^2}\left( {{x^2}} \right) + {{\left( {{x^2}} \right)}^2}} \right] \cr
& = 2{x^3} + 4{x^2} - {x^4} - {x^4} \cr
& = 2{x^3} + 4{x^2} - 2{x^4} \cr
& \int_0^2 {\left[ {\int_{{x^2}}^{2x} {\left( {{x^2} + 2y} \right)} dy} \right]} dx = \int_0^2 {\left( {2{x^3} + 4{x^2} - 2{x^4}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \left[ {\frac{1}{2}{x^4} + \frac{4}{3}{x^3} - \frac{2}{5}{x^5}} \right]_0^2 \cr
& {\text{Simplifying}} \cr
& = \frac{1}{2}{\left( 2 \right)^4} + \frac{4}{3}{\left( 2 \right)^3} - \frac{2}{5}{\left( 2 \right)^5} - 0 \cr
& = \frac{{88}}{{15}} \cr} $$
