Solution
= Solution
For $0\leq x\leq1$, integrate the <joint probability density function> over its allowed $y$-range:
$$
f_X(x)=\int_0^{1-x}24xy\,dy=12x(1-x)^2.
$$
Outside this interval it is zero. Therefore the <marginal density> is
$$
\boxed{f_X(x)=\begin{cases}12x(1-x)^2,&0\leq x\leq1,\\0,&\text{otherwise}.\end{cases}}
$$
Its integral is $12(1/2-2/3+1/4)=1$, also checking the original joint normalization. By symmetry $Y$ has the same marginal density.