Suppose you have an integral of a matrix-valued function of the form:$$\int_a^b B A(t) C dt$$
In this case, the notation $A(t)$ is to denote a matrix that depends on the integrated variable $t$ (for example $A(t) = e^{tA}$), where the matrices $B$ and $C$ are independent of $t$. Is the following necessarily true as it would be for the analogous scalar problem?
$$\int_a^b BA(t)Cdt = B\int_a^bA(t)Cdt = \int_a^b BA(t)d t C = B\int_a^b A(t)dt C$$
Yes it is.. Let $A(t) = (a_{ij})(t)_{1 \leq i, j \leq n}$, $B = (b_{ij})_{1 \leq i, j \leq n}$ and $BA(t) = (\beta_{ij}(t))_{1 \leq i, j \leq n}$. We only want to prove it for $BA(t)$. The other case is an analogue. Then, for each $i, j$ we have: $$ \left(\int_a^b BA(t)~\mathrm{d}t\right)_{ij} =\int_a^b \beta_{ij}(t)~\mathrm{d}t = \int_a ^b \sum_{j = 1}^n b_{ij}a_{ij}(t) ~\mathrm{d}t = \sum_{j = 1}^n \int_a ^b b_{ij} a_{ij}(t)~\mathrm{d}t = \sum_{j = 1}^n b_{ij} \int_a^b a_{ij}(t)~\mathrm{d}t = \left( B\int^b_a A(t)~\mathrm{d}t \right)_{ij} $$