Solution
= Solution
With $y$ held fixed, the inner <antiderivative> gives
$$
\int_{2y}^2x^2y^2\,dx=\frac{y^2}{3}[x^3]_{2y}^2=\frac83(y^2-y^5).
$$
The remaining <integral> is therefore
$$
\boxed{\frac83\int_0^1(y^2-y^5)\,dy=\frac83\left(\frac13-\frac16\right)=\frac49.}
$$