Suppose that the random variables X and Y have joint p.d.f. f(x,y)={kx(x−y),0<x<2,−x<y<x0, ...

Given :
$f(x,y)=\{\begin{array}{l}kx(x-y),0<x<2,-x<y<x\\ 0,elsewhere\end{array}$
To find marginal P.D.F of x
${\displaystyle f_x\left(x\right)=\frac{1}{8}{\int }_{-x}^{x}x(x-y)dy}$
${\displaystyle \frac{1}{8}{\int }_{-x}^x(x^2-xy)dy}$
${\displaystyle =\frac{1}{8}{[x^{2}y-\frac{x{y}^2}{2}]}_{-x}^x}$
${\displaystyle =\frac{1}{8}[x^3-\frac{x^3}{2}+x^3+\frac{x^3}{2}]}$
${\displaystyle =\frac{1}{8}\left(2x^3\right)}$
${\displaystyle =\frac{x^3}{4}}$
To find marginal P.D.F of y
${\displaystyle f_y\left(y\right)=\frac{1}{8}{\int }_0^{2}x(x-y)dx}$
${\displaystyle =\frac{1}{8}{\int }_0^2(x^2-xy)dx}$
${\displaystyle =\frac{1}{8}{[\frac{x^3}{3}-\frac{x^{2}y}{2}]}_0^2}$
${\displaystyle =\frac{1}{8}[\frac{8}{3}-\frac{4y}{2}-0-0]}$
${\displaystyle =\frac{1}{8}(\frac{8}{3}-y)}$
${\displaystyle =\frac{1}{3}-\frac{y}{8}}$
${\displaystyle =\frac{8-3y}{24}}$