Integrating Vector-Valued Functions — AP® Calculus BC Notes

In 9.4 you went from position to velocity by differentiating each component. This topic runs the other way: from a velocity vector, and a starting point, back to the position. It is the planar version of what you did for straight-line motion in 8.2, done twice, once for $x$ and once for $y$ .

Antiderivatives one component at a time

To integrate a vector-valued function, integrate each component:

$\int \left\langle f(t), g(t) \right\rangle\,dt = \left\langle \int f(t)\,dt, \int g(t)\,dt \right\rangle = \left\langle F(t), G(t) \right\rangle + \mathbf{C}.$

Each component gets its own constant, so the constant is a vector, $\mathbf{C} = \left\langle C_1, C_2 \right\rangle$ . Definite integrals work the same way and give a vector of numbers:

$\int_0^2 \left\langle 3t^2, 2t + 1 \right\rangle\,dt = \left\langle t^3\Big|_0^2, (t^2 + t)\Big|_0^2 \right\rangle = \left\langle 8, 6 \right\rangle.$

One velocity, many paths

Two paths with the same velocity v(t) = 1, 2t - 2 for t from 0 to 2.8. The emerald path starts at (0, 0) and the blue path starts at (2, 1); the blue path is the emerald one shifted 2 right and 1 up. At t = 2 both carry the same vermillion velocity vector. The antiderivative of v is t, t squared minus 2t, plus a constant vector C: C = 0, 0 for the emerald path and C = 2, 1 for the blue one. The initial position picks C: r(t) = r(0) plus the integral of v. — **Figure 1.** The same velocity from two starting points: two shifted paths. LevlPrep original diagram.

Knowing the velocity tells you how the particle moves, not where it is. The figure shows two paths with the same velocity $\mathbf{v}(t) = \left\langle 1, 2t - 2 \right\rangle$ . Both are antiderivatives, $\left\langle t, t^2 - 2t \right\rangle + \mathbf{C}$ ; they differ by the constant vector, so one is the other shifted. An initial condition picks the path:

$\mathbf{r}(t) = \mathbf{r}(0) + \int_0^t \mathbf{v}(s)\,ds, \qquad \text{or, between two times,} \qquad \mathbf{r}(b) = \mathbf{r}(a) + \int_a^b \mathbf{v}(t)\,dt.$

In components: $x(b)$ $= x(a)$ $+ \int_a^b x'(t)\,dt$ and $y(b)$ $= y(a)$ $+ \int_a^b y'(t)\,dt$ . The integral is the change in position over the interval; the start is added to it.

Worked example 1. A particle has velocity $\mathbf{v}(t) = \left\langle 4t^3, e^t \right\rangle$ and starts at $\mathbf{r}(0) = \left\langle -1, 3 \right\rangle$ . Find $\mathbf{r}(t)$ and $\mathbf{r}(1)$ .

Antiderivatives: $\left\langle t^4, e^t \right\rangle + \mathbf{C}$ .

At $t = 0$ : $\left\langle 0, 1 \right\rangle + \mathbf{C} = \left\langle -1, 3 \right\rangle$ , so $\mathbf{C} = \left\langle -1, 2 \right\rangle$ . Notice $e^0 = 1$ , so $C_2$ is $2$ , not $3$ .

$\mathbf{r}(t) = \left\langle t^4 - 1, e^t + 2 \right\rangle$ and $\mathbf{r}(1) = \left\langle 0, e + 2 \right\rangle$ .

From acceleration: two constants

Given the acceleration vector, integrate twice. The initial velocity fixes the first constant vector and the initial position the second.

Worked example 2. $\mathbf{a}(t) = \left\langle 6t, -4 \right\rangle$ , $\mathbf{v}(0) = \left\langle 0, 8 \right\rangle$ , $\mathbf{r}(0) = \left\langle 2, 0 \right\rangle$ . Find $\mathbf{r}(2)$ .

Velocity: $\mathbf{v}(t)$ $= \left\langle 3t^2, -4t \right\rangle$ $+ \left\langle 0, 8 \right\rangle$ $= \left\langle 3t^2, 8 - 4t \right\rangle$ .

Position: $\mathbf{r}(t)$ $= \left\langle t^3, 8t - 2t^2 \right\rangle$ $+ \left\langle 2, 0 \right\rangle$ $= \left\langle t^3 + 2, 8t - 2t^2 \right\rangle$ .

$\mathbf{r}(2)$ $= \left\langle 8 + 2, 16 - 8 \right\rangle$ $= \left\langle 10, 8 \right\rangle$ .

Velocity from a table

When $x'(t)$ and $y'(t)$ are known only at some times, approximate each integral with a sum, using the table's own widths, and then add the starting coordinate.

Worked example 3. A cart starts at $(0, 2)$ , in feet. Its velocity components, in feet per second, are $x' = 2, 5, 1$ and $y' = 3, -1, 4$ at $t = 0, 1, 4$ seconds. Approximate the cart's position at $t = 4$ with a trapezoidal sum.

$\int_0^4 x'(t)\,dt$ $\approx \frac{2 + 5}{2}(1)$ $+ \frac{5 + 1}{2}(3)$ $= 3.5$ $+ 9$ $= 12.5$ , so $x(4) \approx 0 + 12.5 = 12.5$ feet.

$\int_0^4 y'(t)\,dt$ $\approx \frac{3 + (-1)}{2}(1)$ $+ \frac{-1 + 4}{2}(3)$ $= 1$ $+ 4.5$ $= 5.5$ , so $y(4) \approx 2 + 5.5 = 7.5$ feet.

The cart is near $(12.5, 7.5)$ .

What each integral means

Expression	Meaning
$\int_a^b x'(t)\,dt$	Change in $x$ -coordinate from $t = a$ to $t = b$
$\int_a^b \mathbf{v}(t)\,dt$	Change in position, as a vector
$x(a) + \int_a^b x'(t)\,dt$	The $x$ -coordinate at $t = b$
$\frac{1}{b - a}\int_a^b y'(t)\,dt$	Average vertical velocity on $[a, b]$

In context, give each with its units: the integral of a velocity in meters per minute is in meters. And keep the $dt$ in every integral you write: the scoring guidelines read an integral without its differential as incomplete.