Find Distance Function from Acceleration Function

The (non-constant) acceleration as a function of time, $a(t)$, is defined and known over $[t_0, t_2]$. It is also known that $a(t)$ is integrable. Also, $a(t)=\frac{dv(t)}{dt}$ and $v(t)=\frac{dx(t)}{dt}$, where $v(t)$ is the velocity function and $x(t)$ is the distance function. $t_1$ is a known time within $[t_0, t_2]$. Given $v(t_1)$ and $x(t_1)$, is it possible to find $x(t)$ over the entire interval? If so, how can this be done? Rigor would be appreciated.

Solutions Collecting From Web of "Find Distance Function from Acceleration Function"

By the Fundamental Theorem of Calculus,
$$
v(t)-v(t_1)=\int_{t_1}^{t}a(s)\ ds.
$$
Similarly,
$$
\begin{align*}
x(t)-x(t_1)&=\int_{t_1}^t v(s)\ ds\\
&=\int_{t_1}^t\left(v(t_1)+\int_{t_1}^{s}a(u)\ du\right)\ ds\\
&= (t-t_1)v(t_1)+\int_{t_1}^t\int_{t_1}^{s}a(u)\ du\ ds.
\end{align*}
$$
Rearranging yields the solution
$$
x(t)=x(t_1)+(t-t_1)v(t_1)+\int_{t_1}^t\int_{t_1}^{s}a(u)\ du\ ds,\qquad t\in [t_0,t_2]
$$