Answer
$$\frac{1}{{48}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_{{x^2}}^x {{x^3}ydy} dx} \cr
& = \int_0^1 {\left[ {\int_{{x^2}}^x {{x^3}ydy} } \right]dx} \cr
& {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr
& = {x^3}\int_{{x^2}}^x {ydy} = {x^3}\left[ {\frac{{{y^2}}}{2}} \right]_{{x^2}}^x \cr
& {\text{evaluating the limits in the variable }}y \cr
& = {x^3}\left[ {\frac{{{{\left( x \right)}^2}}}{2} - \frac{{{{\left( {{x^2}} \right)}^2}}}{2}} \right] \cr
& {\text{simplifying}} \cr
& = {x^3}\left[ {\frac{{{x^2}}}{2} - \frac{{{x^4}}}{2}} \right] \cr
& = \frac{1}{2}\left( {{x^5} - {x^7}} \right) \cr
& \cr
& \int_0^1 {\left[ {\int_{{x^2}}^x {{x^3}ydy} } \right]dx} = \frac{1}{2}\int_0^1 {\left( {{x^5} - {x^7}} \right)} dx \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left[ {\frac{{{x^6}}}{6} - \frac{{{x^8}}}{8}} \right]_0^1 \cr
& = \frac{1}{2}\left( {\frac{{{{\left( 1 \right)}^6}}}{6} - \frac{{{{\left( 1 \right)}^8}}}{8}} \right) - \left( {\frac{{{{\left( 0 \right)}^6}}}{6} - \frac{{{{\left( 0 \right)}^8}}}{8}} \right) \cr
& = \frac{1}{{48}} \cr} $$
