probability - Proof: $Xge 0, r>0Rightarrow E(X^r)=rint_0^{infty}x^{r-1}P(X>x)dx$

As the title states, the problem at hand is proving the following:

$X\ge 0, r>0\Rightarrow E(X^r)=r\int_0^{\infty}x^{r-1}P(X>x)dx$

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Attempt/thoughts on a solution

I am guessing this is an application of Fubini's Theorem, but wouldn't that require writing $P(X>x)$ as an expectation? If so, how is this accomplished?

Thoughts and help are appreciated.

Answer

Proof: Consider the expectation of the identity
$$
X^r=r\int_0^{X}x^{r-1}\,\mathrm dx=r\int_0^{+\infty}x^{r-1}\mathbf 1_{X>x}\,\mathrm dx.
$$