Suppose r.v X~f where the p.d.f,  f, is given by

f(x) =4x^3, 0<=x<=1
     = 0 ,  otherwise
Use the probability Integral Transformation Theorem to show that r.v X^4~U[0,1]

Please give detailed explanation.. Thanks

The function f gives the probability for an interval of infinitesimal
width dx, so that the definition of f is P(x<X<x+dx) = f(x) dx for
0<x<1.

The integral of f is F = x^4, which is the cumulative probability,
i.e. P(X<x) = F(x), for 0<x<1.

The function F is monotonic over the support [0, 1], so P(X < x)
implies P(F(X) < F(x)). Define u = F(x). This gives P(F(X) < u ) = u.
This is true for 0<x<1 so for 0<u<1. This is the definition of a
uniform (0,1) rv. So, defining a new rv U = F(X), this shows that U is
uniform(0,1).