Let $A$ an operator on a Hilbert space $\mathcal{H}$ generate the $C^0$ one-parameter semigroup $T(t)$. In this case, the following identity is well-known, for all $x \in \mathcal{H}$: $$A \int_0^t T(s) x ds = T(t)x - x.$$ The proof is quite straightforward. Using $Ax = \lim_{h \to 0} \frac{1}{h}(T(h)x - x)$, one applies this to the integral, and after a small change of variables and re-arranging, we get $$A \int_0^t T(s) x ds = \lim_{h\to0} \frac{1}{h} \int_t^{t+h} T(s)xds - \frac{1}{h}\int_0^h T(s)x dx,$$ and we get the right-hand side of the identity.
Now, what I am looking for is a similar result, but in the case in which $x= x(s)$, a family of time-dependent vectors, continuous in time (but it is unclear in my application if the time-derivative of $x(s)$ is bounded for a.e. $s$). So, in other words, $$ A \int_0^t T(s) x(s) ds = \:?$$
I have tried to show this replicating the proof above, but that runs into problems. For example, I got something like $$ A \int_0^t T(s) x(s) ds = T(t)x(t) - x(0) + \lim_{h\to0}\frac{1}{h} \int_0^t T(s)[x(s)-x(s-h)]ds,$$ but it is unclear how the extra term on the RHS behaves. On the suggestion of a friend, I also tried a Riemann-sum type application, where approximating $x(s)$ by simple functions, in which one can use the identity for constant $x$ on each sub-rectangle, and I get something like $$A \int_0^t T(s) \sum_{j=1}^n c_j x_j ds = \sum_{j=1}^n (T(t_j) - T(t_{j-1}))c_j x_j.$$
I would appreciate any suggestions or help!