A particle is moving in $\mathbb{R}^3$ so that its acceleration function is $a(t)=\langle 2t,1,0 \rangle$. Find the position function, $r(t)$ of the particle if it starts at the point $(-5,0,2)$ with initial velocity $v(t)=\langle 3,1,-1 \rangle$.
We have $v(t)= \langle 3,1,-1 \rangle$ so $r(t)=\int v(t) dt$= $\int \langle 3,1,-1 \rangle dt$ which is $\langle 3t,t,-t \rangle$
From here I don't know what to do.
Let $$\vec{a}(t)=\langle 2t,1,0 \rangle$$According to the definition $$\frac{dv}{dt} = a$$Integrating with respect to time $$v(t) - v(0) = \int_{0}^ta(\lambda)d\lambda$$ So we have $$\vec{v}(t) = \langle 3 , 1 , -1 \rangle + \langle t^2 , t ,0 \rangle = \langle t^2 + 3 , t+1 , -1 \rangle$$ Similarly $$r(t) - r(0) = \int_{0}^tv(\lambda)d\lambda$$ And then $$\vec{r}(t) = \langle -5 , 0 , 2 \rangle + \langle t^3/3 + 3t , t^2/2+t , -t \rangle = \langle t^3/3 + 3t-5 , t^2/2+t , -t+2 \rangle$$