Continuous r.v. expectation and double integrals

The second way I was shown is kind of confusing (given I have not really done double integral stuff yet):

$E[X]=\int_{0}^{\infty}P(X>x)dx=\int_{0}^{\infty}\int_{x}^{\infty}f_{X}(u)dudx=\int_{0}^{\infty}\int_{0}^{u}f_{X}(u)dxdu\\=\int_{0}^{\infty}f_{X}(u)\int_{0}^{u}dxdu=\int_{0}^{\infty}f_{X}(u)udu= \int_{0}^{\infty}uf_{X}(u)du=E[X]$

The third inequality is where I am unsure. I can "see" why:
$\int_{0}^{\infty}\int_{x}^{ \infty}dudx=\int_{0}^{\infty} \int_{0}^{u}dxdu$ by taking the same area above the line u=x by varying u, x integration variables in different order.

But why does this imply the second equality?

Thanks.

Is this the one you are having trouble with?

$\displaystyle\int_{0}^{\infty} \int_{0}^{u}f_{X}(u)dxdu=\int_{0}^{\infty}f_{X}(u)\int_{0}^{u}dxdu$

If so, $f_X(u)$ is just a constant like any other as far as x is concerned (i.e. it's independent of x), and hence can be pulled out of the inner integral.


Had to edit the quote a bit, as the LaTex wasn't working properly.

that part is fine, it's the part before which I don't quite follow:

$\displaystyle \int_{0}^{\infty}\int_{x}^{ \infty}f_{X}(u)dudx= \displaystyle \int_{0}^{\infty} \int_{0}^{u}f_{X}(u)dxdu$

That's just changing the order of the integration. Do a sketch to make sense of the limits.

I can sketch and make sense for:

$\displaystyle \int_{0}^{\infty}\int_{x}^{ \infty}dudx= \displaystyle \int_{0}^{\infty} \int_{0}^{u}dxdu$

but can't visually see how to sketch the required things for:
$\displaystyle \int_{0}^{\infty}\int_{x}^{ \infty}f_{X}(u)dudx= \displaystyle \int_{0}^{\infty} \int_{0}^{u}f_{X}(u)dxdu$

Probably missing some obvious stuff as always.

$f_X(u)$ is just a function that represents the height of each volume element, with cross-sectional area, dudx. Its not the region you're integrating within as thats defined by the limits, so afaik, you generally don't need to draw it. If you wanted to, you could draw it, but you'd be drawing a surface/plane, depending on the function and not a line, and it can get horrendously difficult to draw by hand.

If it helps, in $\displaystyle \int_{}^{} \int_{R}^{} dudx, f_X(u)=1$, where R is the region you're integrating over.

thanks. then it does make sense (i think!).