Integrals of Vector Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 3.2: “Calculus of Vector-Valued Functions” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions |
| Textbook used in class | Stewart, Calculus, Section 13.2: “Derivatives and Integrals of Vector Functions” (Examples 4, 5) |
Opening Scenario
If $\mathbf{r}'(t)$ is the velocity of a particle, integrating gives the position: $\mathbf{r}(t) = \int \mathbf{r}'(t)\,dt$, plus a constant vector determined by the initial position. This mirrors the scalar case exactly: differentiation and integration are component-wise operations on vector functions.
Quick Reference
Indefinite integral: \[ \int \mathbf{r}(t)\,dt = \left\langle \int f(t)\,dt,\; \int g(t)\,dt,\; \int h(t)\,dt\right\rangle + \mathbf{C}, \] where $\mathbf{C} = \langle C_1, C_2, C_3\rangle$ is a constant vector.
Definite integral: \[ \int_a^b \mathbf{r}(t)\,dt = \left\langle \int_a^b f(t)\,dt,\; \int_a^b g(t)\,dt,\; \int_a^b h(t)\,dt\right\rangle. \]
Key Concepts
1. Computing Indefinite and Definite Integrals
Example 1. Find $\displaystyle\int_0^1 \langle t^2, e^t, \cos(\pi t)\rangle\,dt$. (Stewart 13.2, Example 4.)
\[ = \left\langle \frac{t^3}{3}\Big|_0^1,\; e^t\Big|_0^1,\; \frac{\sin(\pi t)}{\pi}\Big|_0^1\right\rangle = \left\langle \frac{1}{3}, e - 1, 0\right\rangle. \]
2. Finding Position From Velocity and Initial Condition
Example 2. A particle has velocity $\mathbf{r}'(t) = \langle 2t, e^t, \cos t\rangle$ and initial position $\mathbf{r}(0) = \langle 1, 0, 1\rangle$. Find $\mathbf{r}(t)$. (Stewart 13.2, Example 5.)
Integrate: \[ \mathbf{r}(t) = \langle t^2 + C_1,\; e^t + C_2,\; \sin t + C_3\rangle. \]
Apply initial condition $\mathbf{r}(0) = \langle 1, 0, 1\rangle$: \[ \langle 0 + C_1,\; 1 + C_2,\; 0 + C_3\rangle = \langle 1, 0, 1\rangle, \] so $C_1 = 1$, $C_2 = -1$, $C_3 = 1$.
\[ \mathbf{r}(t) = \langle t^2 + 1,\; e^t - 1,\; \sin t + 1\rangle. \]
adding a single scalar constant instead of a constant vector. When integrating $\mathbf{r}'(t)$, each component picks up its own constant of integration. The result is a constant vector $\mathbf{C} = \langle C_1, C_2, C_3\rangle$, not a single constant $C$. Determine all three by applying the initial condition to all three components.
Common Misconceptions
integrating a vector function produces a single scalar constant of integration.
This is the input-output-confusion error. Integrating $\mathbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$ component-wise introduces an independent constant in each component: $\mathbf{r}(t) = \langle \int f'\,dt + C_1, \int g'\,dt + C_2, \int h'\,dt + C_3\rangle$. The constant of integration is a vector $\mathbf{C} = \langle C_1, C_2, C_3\rangle$, not a single scalar. Each component constant is determined separately by applying the initial condition to each component individually.
Leveled Practice
Problem 1. Find $\mathbf{r}(t)$ if $\mathbf{r}'(t) = \langle 3t^2, -\sin t, 2\rangle$ and $\mathbf{r}(0) = \langle 0, 1, -1\rangle$.
Show answer
$\mathbf{r}(t) = \langle t^3 + C_1, \cos t + C_2, 2t + C_3\rangle$. At $t = 0$: $\langle C_1, 1 + C_2, C_3\rangle = \langle 0, 1, -1\rangle$, so $C_1 = 0$, $C_2 = 0$, $C_3 = -1$.
$\mathbf{r}(t) = \langle t^3, \cos t, 2t - 1\rangle$.