I'd love your help with proving that:
For continuous function $f$, $$\int_{0}^{x} \left[\int_{0}^{t}f(u) \; du \right] \; dt = \int_{0}^{x} f(u)(x-u)du$$
I'm not quite sure what I should do with this.
From Newton-Leibniz theorem I get that the left side of the equation is $(F(t)-F(0))x$, while on the right side I wanted to do an integration by parts, but I'm not sure how or if it is the direction.
Thanks a lot for the help.
Let $$\begin{align}F(x)&=\int_{0}^{x}f(u)(x-u)\,du \\&= x \int_{0}^x f(u)\,du -\int_{0}^x uf(u)\,du.\end{align}$$
Now, we know how to take the derivative of all these terms, $x$ and the two integrals, so we can compute:
$$\begin{align}F'(x) &= \int_0^x f(u) \,du + xf(x) - xf(x) \\&= \int_0^x f(u) \,du.\end{align}$$
But that means that $$\begin{align}F(x) &= C + \int_0^x F'(u) \,du \\&= C + \int_0^x \left(\int_0^u f(t)\,dt\right)\,du\end{align}$$ for some constant $C$. But clearly, using $x=0$, $C=0$, so you are done.
You should integrate by parts the LHside with differential factor $\text{d} t$.
In fact, since $f$ is continuous then $F(t):=\int_0^t f(u)\ \text{d} u$ is of class $C^1$ and its derivative is $f(t)$; hence: $$\begin{split} \int_0^x \left[\int_0^t f(u)\ \text{d} u\right]\ \text{d} t &= t\ \int_0^t f(u)\ \text{d} u\Bigg|_0^x - \int_0^x t\ f(t)\ \text{d} t\\ &= x\int_0^x f(t)\ \text{d} t -\int_0^x t\ f(t)\ \text{d} t\\ &= \int_0^x (x-t)\ f(t)\ \text{d} t\; ,\end{split}$$ which is the claim.
HINT
There are couple of ways.
The first one is to change the order of integration using Fubini's theorem and this immediately gives you the answer.
The second way is a bit roundabout. Consider $$F(x) = \int_0^x \int_0^t f(u) du dt - \int_0^x f(u) (x-u) du. $$ Prove that $F'(x) = 0$ using Leibniz integration rule and hence $F(x) = F(0) = 0$ giving us the desired result.
