Identity for triple integrals I found an identity but I haven't been able to prove it.

This is the identity
$\int _{ 0 }^{ x }{ \int _{ 0 }^{ y }{ \int _{ 0 }^{ z }{ f(t)dtdzdy\quad =\quad \frac { 1 }{ 2 } \int _{ 0 }^{ x }{ (x-t)^{ 2 }f(t)dt }  }  }  }  $

A teacher told me that the first step to achieve the result is to change $x$ for $t$ and then draw the solid. I have no idea how to draw the solid and what to do with that. Any ideas?

The region $z\in[0,y]$ and $t\in[0,z]$ is a triangular region in the $t$-$z$ plane bounded by $t=0$, $z=y$, and $z=t$.  
By changing the order of integration we can write
$$\int_0^y\int_0^zf(t)\,dt\,dz=\int_0^y \int_t^yf(t)\,dy\,dt=\int_0^y(y-t)f(t)\,dt$$
Can you proceed by evaluating the integral $\int_0^x \int_0^y(y-t)f(t)\,dt\,dy$ by changing the order of integration in an analogous way?
