Recovering position from velocity via antidifferentiation

Apply the reverse power rule term by term to $v(t) = 3t^2 - 12t + 5$: the antiderivative of $3t^2$ is $t^3$, of $-12t$ is $-6t^2$, and of $5$ is $5t$. This gives the general antiderivative $s(t) = t^3 - 6t^2 + 5t + C$, where $C$ is an arbitrary constant.

Use the initial condition $s(0) = 4$: substituting $t=0$ gives $s(0) = 0 - 0 + 0 + C = C$, so $C = 4$. Therefore the position function is $s(t) = t^3 - 6t^2 + 5t + 4$.