Show $E(X^k) = \int^\infty_0 (1-F_X(x))h(x) dx$ [closed]
$E(X^k)=\int_0^\infty (1-F_{X^k}(x))dx$ using the result that for any positive r.v. $Y$, $E(Y)=\int_0^\infty (1-F_Y(y))dy$.
Now $F_{X^k}(x)=P(X^k\leq x)=P(X\leq x^{1/k})=F_X(x^{1/k})$.
Substitute this into the integrand and change the variable $x\mapsto y^k$ to get $h(x)=kx^{k-1}$.