In a circumstance I had to show that $$ \int_0^t \int_0^x \int_0^y f(z)\,dz\,dy\,dx=\frac{1}{2}\int_0^t f(z)(t-z)^2 \,dz $$ , where $t$ is a constant.
I succeeded in doing so by integration by parts. I let $$ \int f(x) \,dx = F(x),\quad \int F(x)\,dx = \varphi(x),\quad \int \varphi(x)\, dx = \phi(x) $$ and parse $\int_0^t f(z)(t-z)^2 \,dz$ by $\int_0^t (t-z)^2 \,d(F(z)) = \left.F(z)(t-z)^2\right|_0^t - \int F(z)\, d(t-z)^2$ and so on... Then, I evaluated the integral $\int_0^t \int_0^x \int_0^y f(z)\,dz\,dy\,dz$ and confirmed that their results were equal.
Yet this method was obviously not simple. Since that I encountered this problem in my multivariable calculus studying, I wonder if there is a geometry approach or something more multivariable to this...
I noticed by the integration limits that the region on which $f(z)$ is being integrated seems to be $$ \{(x,y,z): 0 \leq z \leq y \wedge 0 \leq y \leq x \wedge 0 \leq x \leq t\} $$ But I can't conclude anything on this because I know nothing about $f(z)$...