Integrating Vector-Valued Functions Integrating Vector-Valued Functions

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.

Definite Integration

The definite integral of a vector-valued velocity function gives the change in position over an interval:

\int_a^b \mathbf{r}'(t)\,dt = \mathbf{r}(b) - \mathbf{r}(a)

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.