Can we have $A(t)\exp(\int_{t_0}^t A(s)\,ds )=\left(\exp(\int_{t_0}^t A(s)\,ds )\right)A(t)$, when $\,A(t)\,$ is symmetric? 
$A(t)$ is a symmetric matrix for $t\in [t_0,a]$. Can we say $$A(t)\exp\left(\int_{t_0}^t A(s)ds \right)=\exp\left(\int_{t_0}^t A(s)\,ds \right) A(t)$$

My attempt:
it is easy to see that $B(s)=\int_{t_0}^t A(s)\,ds $ is symetric . symetric matrices are diagnosable and also $\exp(B(t))$ is diagnosable. We know that two diagnosable matrices are commutative iff they have same eigenvectors. 
Can someone help me.

Observe that
$$
\exp\left(\int_{t_0}^t A(s)\,ds\right)=\lim_{n\to\infty}\sum_{k=0}^n \frac{\left(\int_{t_0}^t A(s)\,ds\right)^k}{k!}.
$$
Hence it suffices to show that
$$
A(t)\left(\int_{t_0}^t A(s)\,ds\right)^k=\left(\int_{t_0}^t A(s)\,ds\right)^kA(t),
$$
for all $k$. And clearly it suffices to show that
$$
A(t)\left(\int_{t_0}^t A(s)\,ds\right)=\left(\int_{t_0}^t A(s)\,ds\right)A(t)
$$
Next note that 
$$
\int_{t_0}^t A(s)\,ds=\lim_{n\to\infty}\frac{t-t_0}{n}\sum_{k=1}^n 
A\left(t_0+\frac{k}{n}(t-t_0)\right),
$$
and hence it suffices to show that
$$
A(t)A(s)=A(s)A(t), \tag{1}.
$$
If we further assume that 
$$
A(t)\int_{\tau}^t A(s)\,ds=\left(\int_{\tau}^t A(s)\,ds\right)A(t), \tag{2}
$$
for all $\tau,t\in[a,b]$, then differentiating with respect to $\tau$ we obtain that
$$
A(t)A(\tau)=A(\tau)A(t), \quad\text{for all $t,\tau\in[a,b]$} \tag{3}
$$
Thus $(2)$ and $(3)$ are equivalent.
