How to show $\int_0^x\Big(\int_0^v[\int_0^u f(t) dt ]du \Big)dv=\dfrac 12 \int_0 ^x (x-t)^2f(t) dt$?

Question

How to show that $\int_0^x\Big(\int_0^v[\int_0^u f(t) dt ]du \Big)dv=\dfrac 12 \int_0 ^x (x-t)^2f(t) dt$ ? I am completely at a loss ; the only vague idea I have is that I have to do a variable i.e. co-ordinate transformation but I don't know what . Please help . Thanks in advance

Answer 1

This question is a particular case of the "Cauchy's repeated integral" (Cauchy's formula).

Hint : The proof is easy by induction.

No need to copy out what is already written here : https://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration

Answer 2

Hint Change the order of integration so that the integral with respect to $x$ is the last one you calculate.