Fubini's theorem for dependent integral bords

Question

Why we can swap these two integrals in the following way: $$\int_0^t \left(\int_0^s d\tau \right)\, ds=\int_0^t \left(\int_\tau^t \, ds \right)\, d\tau $$ I found this in a proof of convolution formula for inhomogeneous differential equations. $$\int_0^t \left(\int_0^s AS(s-\tau) f(\tau) d\tau \right)\, ds=\int_0^t \left(\int_\tau^t AS(s-\tau)f(\tau)\, ds \right)\, d\tau$$ where $t\mapsto S(t)$ denote a semigroup of bounded operators, and $A$ is its infinitisemal generator.

Answer 1

Both of these bounds represent the triangular region defined by $(0,0), (t, 0), (t, t)$, where the $x$ axis is $\tau$ and the $y$ axis is $s$.

The integral from 0 to $s$ over $\tau$ followed by the integral from 0 to $t$ over $s$ is like going along the $x$ axis first, and the integral from 0 to $t$ over $\tau$ is like going along the $y$ axis first.