According to the data given,

$a=3t^2+2t+1$

We know that,

$$\begin{gathered} a=\frac{\text{d}v}{\text{d}t} \\ \Rightarrow \text{d}v=a·\text{d}t \end{gathered}$$

Integrating both sides with given limits,

 ${\int }_0^t\mathrm{dv}={\int }_0^t(3t^2+2t+1)\mathrm{dt}$ 

${\left[v\right]}_0^v =3{\left[\frac{t^3}{3}\right]}_0^t + 2{\left[\frac{t^2}{2}\right]}_0^t +{\left[t\right]}_0^t$

Using the property of definite integration,

$v=t^3+t^2+t$

From the definition of instantaneous velocity,

$$\begin{gathered} v=\frac{\text{d}s}{\text{d}t} \\ \Rightarrow \text{d}s=v\text{d}t \end{gathered}$$

Putting the values with proper limits,

$$\begin{gathered} {\int }_0^s\mathrm{ds}={\int }_0^2\left(t^3+t^2+t\right)\mathrm{dt} \\ \Rightarrow {\left[S\right]}_0^s={\left[\frac{t^4}{4}+\frac{t^3}{3}+\frac{t^2}{2} \right]}_0^2 \end{gathered}$$

After $2 \mathrm{s}$ of start the total displacement is,$S=\frac{26}{3}$