Introduction
Just as we can differentiate vector-valued functions by differentiating each component independently, we can also integrate them the same way:
\int \mathbf{r}(t)\,dt = \left\langle \int f(t)\,dt,\, \int g(t)\,dt,\, \int h(t)\,dt \right\rangle + \mathbf{C}
Where
\mathbf{C}=\langle C_1, C_2, C_3 \rangle
.
This process allows us to find position from velocity, or velocity from acceleration.
Examples
Example 1: Basic Integration
Find
\int \langle 2t,\, 3,\, \cos(t) \rangle\,dt
\begin{align*}
\int \langle 2t,\, 3,\, \cos(t) \rangle\,dt &= \left\langle \int 2t\,dt,\, \int 3\,dt,\, \int \cos(t)\,dt \right\rangle + \mathbf{C} \\
&= \langle t^2,\, 3t,\, \sin(t) \rangle + \langle C_1, C_2, C_3 \rangle
\end{align*}
Example 2: Finding Position from Velocity
A particle has velocity
\mathbf{v}(t) = \langle 3t^2,\, 2t,\, -4 \rangle
and initial position
\mathbf{r}(0) = \langle 1,\, 0,\, 2 \rangle
. Find its position function.
\begin{align*}
\mathbf{r}(t) &= \int \mathbf{v}(t)\,dt + \mathbf{C} \\
&= \left\langle \int 3t^2\,dt,\, \int 2t\,dt,\, \int -4\,dt \right\rangle + \mathbf{C} \\
&= \langle t^3,\, t^2,\, -4t \rangle + \langle C_1, C_2, C_3 \rangle
\end{align*}
Using the initial position:
\begin{align*}
\mathbf{r}(0) &= \langle 0,\, 0,\, 0 \rangle + \langle C_1, C_2, C_3 \rangle = \langle 1,\, 0,\, 2 \rangle \\
\mathbf{C} &= \langle 1,\, 0,\, 2 \rangle \\
\mathbf{r}(t) &= \langle t^3 + 1,\, t^2,\, -4t + 2 \rangle
\end{align*}
Example 3: Work and Position
Calculate the change in position of a particle moving with velocity
\mathbf{v}(t) = \langle \cos(t),\, \sin(t),\, 1 \rangle
from t = 0 to t = 2π.
\begin{align*}
\text{Position} &= \int_0^{2\pi} \mathbf{v}(t)\,dt \\
&= \left\langle \int_0^{2\pi} \cos(t)\,dt,\, \int_0^{2\pi} \sin(t)\,dt,\, \int_0^{2\pi} 1\,dt \right\rangle \\
&= \langle [\sin(t)]\Big|_0^{2\pi},\, [-\cos(t)]\Big|_0^{2\pi},\, [t]\Big|_0^{2\pi} \rangle \\
&= \langle 0,\, 0,\, 2\pi \rangle
\end{align*}
The particle completed one revolution, rising by 2π units.