Artificial intelligent assistant

Simple Integration Identity > Prove, by sketching the region of integration and interchanging the order of integration, that > > $$ \int_a^x \int_a^{\xi} f(s) ds d\xi = \int_a^x (x-s)f(s)ds $$ This is probably quite simple but I'm having brain freeze right now...

Note that the integration region is a triangular region with vertices at $(a,a)$, $(a,x)$, and $(x,x)$ in the $s-\xi$-plane.

Thus, if the inner integral is on $s$, we see that for any fixed $\xi$, $s$ begins at $a$ and ends at $\xi$.

If the inner integral is on $\xi$, we see that for any fixed $s$, $\xi$ begins at $s$ and ends at $x$.

Therefore, we can write

$$\begin{align} \int_a^x \int_a^\xi f(s)\,ds\,d\xi&=\int_a^x \int_s^x f(s)\,d\xi\,ds\\\\\\\ &=\int_a^x f(s)\left(\int_s^x (1)\,d\xi\right)\,ds\\\\\\\ &=\int_a^x f(s)(x-s)\,ds \end{align}$$

as was to be shown!

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