How can one prove this property of integrals?$\int_0^bf(x)(b-x)\,dx=\int_0^b\left(\int _0^xf(t)\,dt\right)\,dx$ $$\int_0^bf(x)(b-x)\,dx=\int_0^b\left(\int _0^xf(t)\,dt\right)\,dx$$
I can't understand how this property can be proven but it has held true for everything I have tried. How do you even approach this? I have tried substitution but that gets you no where.

\begin{align}
& \int_0^b\left(\int _0^xf(t)\,dt\right)\,dx \\[10pt]
= {} & \iint\limits_{(t,x)\,:\,0\,<\,t\,<\,x\,<\,b} f(t)\, d(t,x) \\[10pt]
= {} & \int_0^b \left( \int_t^b f(t) \, dx \right) dt \\[10pt]
= {} & \int_0^b f(t)(b-t) \, dt.
\end{align}
