Prove that if $f$ is continuous, then
\int_0^x f(u)(x-u)\,du = \int_0^x \left(\int_0^u
f(t)\,dt \right)\,du
Solution. Note
that
\frac{d}{dx}\left(\int_0^x \left(\int_0^u f(t)\,dt
\right)\,du \right) =\int_0^{x} f(t)\,dt,
by the Fundamental Theorem of Calculus, while
$\begin{eqnarray}
\frac{d}{dx}\left[\int_0^x f(u)(x-u)\,du \right] & = &
\frac{d}{dx} \left[x\int_0^x f(u)\,du \right] - \frac{d}{dx}
\left[\int_0^x f(u)\,du \right] \\
&=& \int_0^x f(u)\,du + x f(x) - f(x) x \\
&=& \int_0^x f(u)\,du,
\end{eqnarray}$
where we used the product rule on the first term. Therefore,
$\int_0^x f(u)(x-u)\,du = \int_0^x \left(\int_0^u f(t)\,dt \right)\,du
+ C$. Setting $x=0$ gives $C=0$.