Question

given r(0)=<0,1,-1> and v(t)=<2t^2+1,3t^2-1,t+1> find its position at time t

Answer 1

Given \(\vec{v}(t)=\langle 2t^2+1,3t^2-1,t+1\rangle\), you have that the position vector is given by the antiderivative:
\[\vec{r}(t)=\int \vec{v}(t)~dt=\left\langle \int(2t^2+1)~dt,\int(3t^2-1)~dt,\int(t+1)~dt\right\rangle+\vec{c}\]
where \(\vec{c}\) is a constant vector.