Integration using Fubini to prove $ \int_{0}^x \int_0^y \int_0^t f(s)\ ds\ dt\ dy = \int_0^x \frac{(x-s)^2}{2} f(s) ds$.

Question

Let $f$ be integrable on the interval $[0,1]$. Prove using Fubini that for $x\in[0,1]$ the follow holds $$ \int_{0}^x \int_0^y \int_0^t f(s)\ ds\ dt\ dy = \int_0^x \frac{(x-s)^2}{2} f(s) ds. $$ In my attempt I showed that the left part is equal to $$ \int_{y\in[0,x]}\int_{t\in[0,y]} \int_{s\in[0,t]} f(s) = \int_{(y,t,s)\in[0,x]\times[0,y]\times[0,t]} f(s). $$ Now I want to apply Fubini to change order of integration. However the rectangle is dependent on $x,y$ and $t$. This seems to be a problem, especially for $y$ and $t$. Is there a way to rewrite $[0,x]\times[0,y]\times[0,t]$, such that Fubini is appliable?

Answer 1

You can write the integral as $\int_0^{x} \int_s^{x}\int_t^{x} f(s)dydtds$. Evaluate the two inside integrals to finish the proof.

I have used Fubini's Theorem and the following fact: $\{(s,t,y): 0<s<t<y<x\}=\{(s,t,y): t <y< x, s <t <x,0 <s<x\}$