Show that $\int _0^x\left[\int _0^tf\left(u\right)du\:\right]dt\:=\:\int _0^x\:\:f\left(u\right)\left(x-u\right)du$ Notation confusion
Let $f$ be continuous. Then $\int_{0}^{x}\left[\int_{0}^{t}f\left(u\right)du\right]dt=\int_{0}^{x}f\left(u\right)\left(x-u\right)du$