Finding the value of the definite integral $\int_0^2{x\int_x^2{\frac{dy}{\sqrt{1+y^3}}}}dx$ If $$f(x) = \int_x^2{\frac{dy}{\sqrt{1+y^3}}}$$
then find the value of $$\int_0^2{xf(x)}dx$$
I have no idea how to solve this question. Please help.

The integration domain can be equivalently written as 
$$
\Omega = \{(x,y): x<y<2 ~~\mbox{and}~~~ 0 < x < 2 \}
$$
or
$$
\Omega = \{(x,y): 0<x<y ~~\mbox{and}~~~ 0 < y < 2 \}
$$
Such that
\begin{eqnarray}
\int_0^2{\rm d}x\int_{x}^2{\rm d}y ~\frac{x}{\sqrt{1 + y^3}} &=& \int_0^2{\rm d}y\int_{0}^y{\rm d}x ~\frac{x}{\sqrt{1 + y^3}} \\
&=& \int_0^2{\rm d}y ~\frac{y^2}{2}\frac{1}{\sqrt{1 + y^3}} \\
&=& \frac{1}{2} \int_0^2{\rm d}y  \frac{y^2}{\sqrt{1 + y^3}} \\
&=&\frac{1}{2}\times\frac{4}{3} = \frac{2}{3}
\end{eqnarray}
