The instantaneous velocity of a particle moving in a straight line...

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

$\dfrac{ds}{dt} = \alphat + \beta{t}^{2}$

${∫}_{{S}_{1}}^{{S}_{2}}ds = {∫}_{1}^{2}\left(\alphat + \beta{t}^{2}\right)dt$

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

As the particle is not changing direction So distance = displacement. Distance $= \left[\dfrac{\alpha [ 4 - 1 ]}{2} + \dfrac{\beta [ 8 - 1 ]}{3}\right]$

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