The expected value can be defined in the following 2 ways
$$ E(x)=\int x.f(x) dx $$ $$ E(x)=\int_0^1 x dF(x) $$ I have an intuitive understanding of both of them. The first I can visualize moving along the x axis and multiplying x by some weight which is very small for most values. Where it is not so small, those values of x have a higher representation in the total sum. The second I can visualize moving along the y axis, pausing at infinitesimal regular intervals and noting the value of x where my current y value intersects. I will encounter similar values of x when the slope of the function is very steep and so those values will have a higher representation in the total sum.
How do I show, analytically, that these 2 function are equivalent?
The procedure is Integration By Substitution other wise known as Change of Variable.
Since $F:\Bbb R\to [0;1]$ we must change the interval to correspond when we change the variable of integration using $y=F(x)$, $x=F^{-1}(y)$ , and $\frac{\operatorname d F(x)}{\operatorname d x}\operatorname d x = \operatorname d y$ .
$$\begin{align}\mathsf E(X) & = \int_\Bbb R x\;f(x)\operatorname d x \\[1ex] & = \int_\Bbb R x\;\frac{\operatorname d F(x)}{\operatorname d x}\operatorname d x \\[1ex] & =\int_0^1 F^{-1}(y)\operatorname d y \end{align}$$