Answer
$$\frac{1}{{40}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_{{x^2}}^x {x{y^2}} dydx} \cr
& = \int_0^1 {\left[ {\int_{{x^2}}^x {x{y^2}} dy} \right]dx} \cr
& {\text{solve the inner integral and treat }}x{\text{ as a constant}} \cr
& \int_{{x^2}}^x {x{y^2}} dy = x\int_{{x^2}}^x {{y^2}} dy \cr
& {\text{using the power rule }} \cr
& = x\left( {\frac{{{y^3}}}{3}} \right)_{{x^2}}^x \cr
& {\text{evaluating the limits in the variable }}y \cr
& = x\left( {\frac{{{{\left( x \right)}^3}}}{3} - \frac{{{{\left( {{x^2}} \right)}^3}}}{3}} \right) \cr
& {\text{simplifying}} \cr
& = x\left( {\frac{{{x^3}}}{3} - \frac{{{x^6}}}{3}} \right) \cr
& = \frac{{{x^4}}}{3} - \frac{{{x^7}}}{3} \cr
& = \frac{{{x^4} - {x^7}}}{3} \cr
& {\text{then}} \cr
& \int_0^1 {\left[ {\int_{{x^2}}^x {x{y^2}} dy} \right]dx} = \int_0^1 {\left( {\frac{{{x^4} - {x^7}}}{3}} \right)dx} \cr
& {\text{integrating}} \cr
& = \frac{1}{3}\left[ {\frac{{{x^5}}}{5} - \frac{{{x^8}}}{8}} \right]_0^1 \cr
& = \frac{1}{3}\left[ {\frac{{{{\left( 1 \right)}^5}}}{5} - \frac{{{{\left( 1 \right)}^8}}}{8}} \right] - \frac{1}{3}\left[ {\frac{{{{\left( 0 \right)}^5}}}{5} - \frac{{{{\left( 0 \right)}^8}}}{8}} \right] \cr
& = \frac{1}{3}\left( {\frac{3}{{40}}} \right) \cr
& = \frac{1}{{40}} \cr} $$
