Why is $\frac1h\int_0^t\left(T(s+h)-T(s)\right)x\;{\rm d}s=\frac1h\int_t^{t+h}T(s)x\;{\rm d}s-\frac1h\int_0^hT(s)x\;{\rm d}s$ for a semigroup $T$?

Question

Let $E$ be a $\mathbb R$-Banach space and $T:[0,\infty)\to\mathfrak L(E)$ be a $C^0$-semigroup, i.e.

1. $T(0)=\text{id}_E$
2. $T(s+t)=T(s)T(t)$ for all $s,t\ge 0$
3. $[0,\infty)\ni t\mapsto T(t)x$ is continuous for all $x\in E$

I don't understand why $$\frac1h\int_0^t\left(T(s+h)-T(s)\right)x\;{\rm d}s=\frac1h\int_t^{t+h}T(s)x\;{\rm d}s-\frac1h\int_0^hT(s)x\;{\rm d}s\tag 1$$ for all $t,h>0$ and $x\in E$. I guess it's simply an application of the substitution rule, but I don't see why the second integral on the right-hand side of $(1)$ has $h$ (and not $t$) as the upper limit.

Answer 1

$$\begin{align*} \int_0^t (T(s+h)-T(s))x \, ds &= \int_h^{t+h} T(s) x \, ds - \int_0^t T(s) x \, ds \\ &= \int_t^{t+h} T(s) \, x \, ds - \int_0^h T(s) x \, ds.\end{align*}$$