Changing a double integral into a single integral - Volterra-type integral equations I have a question regarding a calculation that i stumbled upon when proving that a Cauchy problem can be converted in a Volterra-type integral equation. Specifically, this equality:
\begin{equation*}
\int_0^t\int_0^sy(t) dt ds = \int_0^t (t-s) y(s) ds \, .
\end{equation*}
There seems to be some geometric intuition behind this that I am missing. The generalization, which is also a problem for me, is the following:
$$
\int_0^t ds\int_0^s ds_1 ... \int_0^{s_{n-1}}ds_n y(s_n)
 = \frac{1}{n!}\int_0^t (t-s)^n y(s)ds \, .
$$
These integrals also come up when trying to say that a Volterra-type integral equation of the second kind has one and only one solution in $C([a,b])$, using the fixed point theorem.
Thanks to everyone for reading.

There is an issue of confusion between the dummy variable of integration $t$ and the upper limit on the outer integral.  Instead, write
$$F(t)=\int_0^t\int_0^s y(x)\,dx\,ds\tag1$$
Then, note that the region $0\le x\le s$, for $0\le s\le t$ is a triangular shaped region with vertices in the $(x,s)$-plane at $(0,0)$, $(0,t)$, and $(t,t)$.
So, this triangular region is also defined by $x\le s\le t$, for $0\le x\le t$.  Thus, we can write $(1)$ as
$$F(t)=\int_0^t\int_x^t y(x)\,ds\,dx\tag2$$
But note that in $(2)$, $y(x)$ is independent of $s$.  So, we can "take $y(x)$ outside the inner integral" to obtain
$$F(t)=\int_0^t y(x)\int_x^t (1)\,ds\,dx=\int_0^t (t-x)y(x)\,dx$$
