Answer
$s(t)=\frac{1}{3}t^3+\frac{1}{2}t^2-2t+3$
Work Step by Step
$$v(t)=\int a(t) \ dt \Rightarrow v(t)=\int 2t+1 \ dt$$
$$v(t)=t^2+t+C$$
Using the initial condition, we find that
$$-2=0+C \Rightarrow C=-2. Now, $$$$s(t)=\int v(t) \ dt \Rightarrow s(t)=\int t^2+t-2 \ dt$$
$$s(t)=\frac{1}{3}t^3+\frac{1}{2}t^2-2t+C$$
Again, plugging in the initial condition we find $3=0+C \Rightarrow C=3.$
