Displacement of an object

The velocity as a function of time is $v(t)=t^2\implies \dfrac{dS}{dt}=t^2\implies S_6-S_2=\dfrac{6^3-2^3}{3}=\dfrac{208}{3}\approx \boxed {69.333}$ .

$v(t) = v^2$ is an incorrect way of notating v(t) is quadratically dependent on time. It should be $v(t) = t^2$ .

Nevertheless, integrating $v(t) = t^2$ over $t=2;6$ , you get 69.33.