9.5 - Integrating Vector-Valued Functions

Overview

Integrating vector-valued functions works component by component. Each component gets integrated separately using standard integration techniques.

General form:

\int \langle f(t), g(t) \rangle \, dt = \left\langle \int f(t) \, dt, \int g(t) \, dt \right\rangle

Indefinite Integrals with Initial Conditions

When you’re finding

\mathbf{r}(t)

from

\mathbf{r}'(t)

, integrate each component and then use the initial conditions to solve for the constants.

Example 1: Exponential Components

Given:

\mathbf{r}'(t) = \langle 6e^{3t}, 4e^{2t} \rangle

and

\mathbf{r}(0) = \langle 3, 2 \rangle

Integrate the $x$ -component:

x(t) = \int 6e^{3t} \, dt

This needs

u

-substitution, since we have

3t

in the exponent. Let

u = 3t

, so

du = 3 \, dt

, meaning

dt = \frac{du}{3}

x(t) = \int 6e^{u} \cdot \frac{du}{3} = \int 2e^{u} \, du = 2e^{u} + c_1 = 2e^{3t} + c_1

Integrate the $y$ -component:

y(t) = \int 4e^{2t} \, dt

Let

u = 2t

, so

du = 2 \, dt

, meaning

dt = \frac{du}{2}

y(t) = \int 4e^{u} \cdot \frac{du}{2} = \int 2e^{u} \, du = 2e^{u} + c_2 = 2e^{2t} + c_2

Apply the initial conditions

\mathbf{r}(0) = \langle 3, 2 \rangle

For the

x

-component:

3 = 2e^{3(0)} + c_1 = 2(1) + c_1 = 2 + c_1

, so

c_1 = 1

For the

y

-component:

2 = 2e^{2(0)} + c_2 = 2(1) + c_2 = 2 + c_2

, so

c_2 = 0

Final answer:

\mathbf{r}(t) = \langle 2e^{3t} + 1, \, 2e^{2t} \rangle

Indefinite Integrals Without Initial Conditions

When no initial condition is given, the constants of integration just stay in the answer.

Example 2: Trig and Inverse Trig Integrals

Find

\int \left\langle \sec^2(t), \frac{1}{1+t^2} \right\rangle \, dt

Integrate the $x$ -component:

x(t) = \int \sec^2(t) \, dt = \tan(t) + c_1

Integrate the $y$ -component, recognizing the inverse tangent derivative:

y(t) = \int \frac{1}{1+t^2} \, dt = \arctan(t) + c_2

Final answer:

\mathbf{r}(t) = \langle \tan(t) + c_1, \, \arctan(t) + c_2 \rangle

Definite Integrals

For definite integrals, evaluate each component separately over the given interval.

General form:

\int_a^b \langle f(t), g(t) \rangle \, dt = \left\langle \int_a^b f(t) \, dt, \int_a^b g(t) \, dt \right\rangle

Example 3: Polynomial Components

Evaluate

\int_{-2}^{2} \langle t^2, t^4 \rangle \, dt

$x$ -component:

\int_{-2}^{2} t^2 \, dt = \left[\frac{t^3}{3}\right]_{-2}^{2} = \frac{8}{3} - \frac{-8}{3} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}

$y$ -component:

\int_{-2}^{2} t^4 \, dt = \left[\frac{t^5}{5}\right]_{-2}^{2} = \frac{32}{5} - \frac{-32}{5} = \frac{32}{5} + \frac{32}{5} = \frac{64}{5}

Final answer:

\left\langle \frac{16}{3}, \frac{64}{5} \right\rangle

Example 4: Fractional Exponents

Evaluate

\int_{-1}^{1} \langle t^3, t^{1/5} \rangle \, dt

$x$ -component:

\int_{-1}^{1} t^3 \, dt = \left[\frac{t^4}{4}\right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0

$y$ -component:

\int_{-1}^{1} t^{1/5} \, dt = \left[\frac{t^{6/5}}{6/5}\right]_{-1}^{1} = \left[\frac{5t^{6/5}}{6}\right]_{-1}^{1} = \frac{5(1)}{6} - \frac{5(1)}{6} = 0

Note that

(-1)^{6/5} = \left[(-1)^{1/5}\right]^6 = (-1)^6 = 1

, since the even power makes it positive.

Final answer:

\langle 0, 0 \rangle

Parametric Form

The same methods work when the rates come as two separate equations,

\frac{dx}{dt}

and

\frac{dy}{dt}

, instead of one vector. Integrate each one, then use the starting point to find each constant.

Example 5: Parametric Initial Value Problem

Given:

\frac{dx}{dt} = 3t^2 - 2

and

\frac{dy}{dt} = \frac{1}{t + 1}

for

t \ge 0

, and the particle is at

(1, 4)

when

t = 0

. Find

x(t)

and

y(t)

Integrate the

x

-rate:

x(t) = \int (3t^2 - 2) \, dt = t^3 - 2t + c_1

Since

x(0) = 1

1 = 0 - 0 + c_1

, so

c_1 = 1

Integrate the

y

-rate:

y(t) = \int \frac{1}{t + 1} \, dt = \ln(t + 1) + c_2

No absolute value is needed, since

t + 1 > 0

for

t \ge 0

. Since

y(0) = 4

4 = \ln(1) + c_2 = 0 + c_2

, so

c_2 = 4

Final answer:

x(t) = t^3 - 2t + 1 \qquad y(t) = \ln(t + 1) + 4

Example 6: No Antiderivative You Can Find

Some rates, like

\sin(t^2)

, don’t have an antiderivative you can write down. Then skip the general solution and use the Fundamental Theorem of Calculus with the starting value:

x(b) = x(a) + \int_a^b \frac{dx}{dt} \, dt \qquad y(b) = y(a) + \int_a^b \frac{dy}{dt} \, dt

Given:

\frac{dy}{dt} = \sin(t^2)

and

y(0) = 1

. Find

y(2)

y(2) = 1 + \int_0^2 \sin(t^2) \, dt \approx 1 + 0.805 = 1.805

The integral is done on a calculator. AP free-response questions about particle motion use this setup a lot (more in Topic 9.6).

Important Notes

Constants of integration: - Each component gets its own constant:

c_1

c_2

, and so on - These constants are generally different values - Use the initial conditions to solve for each one separately

Common integration techniques you’ll need: -

u

-substitution for chain rule cases - Recognizing inverse trig derivatives - The power rule for polynomials - Exponential function integration

Symmetry in definite integrals: - Odd functions over symmetric intervals

[-a, a]

integrate to zero - Even functions over symmetric intervals may not cancel

Key Takeaways

Component-wise integration: $\int \langle f(t), g(t) \rangle \, dt = \left\langle \int f(t) \, dt, \int g(t) \, dt \right\rangle$

Each component integrates independently using standard calculus techniques

The constants of integration are different for each component: $c_1$ , $c_2$ , and so on

Initial conditions $\mathbf{r}(t_0) = \langle x_0, y_0 \rangle$ give you separate equations for each constant

$u$ -substitution is required when there’s a chain rule involved, like with $e^{3t}$

Recognize the inverse trig derivatives: $\int \frac{1}{1+t^2} \, dt = \arctan(t) + C$

Definite integrals: evaluate each component over $[a,b]$ and combine the results

Parametric form works the same way: integrate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ separately. If you can’t find an antiderivative, use $x(b) = x(a) + \int_a^b \frac{dx}{dt} \, dt$ (and the same for $y$ )

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