Question: Prove that $\int^x_0\big[\int^u_0 f(t) dt\big] du = \int^x_0f(u)(x-u) du$ where $f$ is a continuous function.
Attempt: My lecturer hinted that it'd be helpful to apply the fundamental theorem of calculus to $F(u) = u \int^u_0f(t) dt$.
I know that I can apply the FTC to $F(u)$ since $f$ is continuous, meaning it is also a Riemann integrable function and thus the conditions of the FTC are met.
To find $F'(u)$, I will let $G(u) = u$ and $H(u)=\int^u_0f(t) dt$ be functions such that $F(u)=G(u)H(u)$. Thus, $G'(u)=1$ and, applying the FTC, $H'(u)=f(u)$.
Using the product rule, $F'(u)=G(u)H'(u)+G'(u)H(u)=uf(u)+\int^u_0f(t) dt$.
However, this is where I get stuck and am not sure how to use this to prove the initial equation.
Any help would be greatly appreciated.