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