probability theory - Let $X$ be a positive random variable with distribution function $F$. Show that $E(X)=int_0^infty(1-F(x))dx$

Let $X$ be a positive random variable with distribution function $F$. Show that

$$E(X)=\int_0^\infty(1-F(x))dx$$

Attempt

$\int_0^\infty(1-F(x))dx= \int_0^\infty(1-F(x)).1dx = x (1-F(x))|_0^\infty + \int_0^\infty(dF(x)).x$ (integration by parts)

$=0 + E(X)$ where boundary term at $\infty$ is zero since $F(x)\rightarrow 1$ as $x\rightarrow \infty$

Is my proof correct?