$V=\alpha t+\beta t^{2}$

$\frac{ ds }{ dt }=\alpha t +\beta t ^{2}$

$\int_{s_{1}}^{s_{2}} d s=\int_{1}^{2}\left(\alpha t+\beta t^{2}\right) d t$

$S_{2}-S_{1}=\left[\frac{\alpha t^{2}}{2}+\frac{\beta t^{3}}{3}\right]_{1}^{2}$

As particle is not changing direction So distance $=$ displacement.

Distance $=\left[\frac{\alpha[4-1]}{2}+\frac{\beta[8-1]}{3}\right]$

$=\frac{3 \alpha}{2}+\frac{7 \beta}{3}$