Integrating vector-valued functions is an AP Calculus BC-only skill. If you are given a rate vector, such as velocity or acceleration, you integrate each component separately to recover a position or velocity function. Initial conditions then let you find the constants and write a particular solution.

This guide builds on 9.4 Defining and Differentiating Vector-Valued Functions. The key idea is that vector integration works like regular antidifferentiation, but component by component.

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Integrating Vector-Valued Functions

It’s good to first recognize the format of your function. It is either written in two or three dimensions; we want to know which one so that we are not confused about which terms should be paid attention to when integrating.

A two-dimension vector function might look like this:

r(t)=f(t)i+g(t)j

or like this:

r(t)=⟨f(t),g(t)⟩

These both mean the same thing.

How do we integrate?

Luckily, we integrate each component of a vector function separately. No need to make things more complicated than they need to be! You look at each component as if it was its own problem. Let’s go through an example just to see how the College Board might ask a question like this.

Integrating Vector-Valued Functions Walkthrough

We are given:

r(t)=⟨t^2,\sin(t),e^t⟩

We know we are going to need to integrate this vector-valued function, so let’s re-write this including integrals on each component.

\int{r(t)}dt=⟨∫t^2dt,∫\sin(t)dt,∫e^tdt⟩

You can solve each of these integrals separately. If you need help with integrals, check out our guide 6.14 Selecting Techniques for Antidifferentiation. Although it says for AB, it will apply to this.

Our final answer will be:

R(t)=⟨\frac13t^3+C1,−\cos(t)+C2,e^t+C3⟩

Please note that C1, C2, and C3 just represent different possible constants and I added them for clarity. For these types of problems, you will typically get definite integrals so that there is less ambiguity. Follow your typical steps for solving these problems!

AP Exam Relevance

For AP Calculus BC, Topic 9.5 focuses on determining a particular solution from a rate vector and initial conditions. On the exam, that usually means you are given velocity and an initial position, or acceleration and an initial velocity, and you need to integrate to find the requested vector-valued function.

Watch for units and interpretation. Integrating velocity gives displacement, not total distance traveled. If the question asks for distance from the origin at a time, first find the position vector at that time, then use the distance formula.

Additional Practice Problem

Let’s try a word problem now. This would be a calculator-allowed style question.

At time 𝑡 ≥ 0, a particle moving in the 𝑥𝑦-plane has a velocity vector given by.

v(t)=⟨3t^2,3⟩

If the particle is at point (1,2) at time 𝑡 = 0, how far is the particle from the origin at time 𝑡 = 2?

Try the problem on your own first! If you are stuck read through our explanation below!

Solution to Practice Problem

Step one is integrating the function. As we’ve discussed, you will integrate each of the components of the vector function separately. Before applying conditions, this is what you should have for your integrated vector function:

x(t)=∫3t^2dt=t^3+Cx

y(t)=∫3dt=3t+Cy

The Cx and Cy constants are just placeholders. We are going to figure out those numbers now:

Since we are given the position for t=0, we can use that information to work backward and find our constants.

x(t):1=0^3+Cx\therefore Cx=1

y(t):2=3(0)+Cy\therefore Cy=2

Now that we know the values for our constants let's put those into our integrated vector functions:

⟨t^3+1,3t+2 ⟩

Now we’re done with the hard parts! From here on it’s just simple math to finish off the problem. We will plug in t=2 to our equation to find the position of the particle at that point. This gives us

(9,8)

Finally, we’ll use the distance formula to find how far this point is from the origin. Since this would be a calculator-allowed question, plug this into your calculator and give a number rounded to three decimal places rather than leaving it in radical form.

\sqrt{(9-0)^2+(8-0)^2}

You should get 12.042 units as your final answer.

Why integrate vector-valued functions?

With these types of problems, you want to understand what each vector function represents. Vector functions often describe position, velocity, or acceleration. Integrating a velocity vector gives displacement and helps you find position when paired with an initial condition. Integrating an acceleration vector gives change in velocity and helps you find velocity when paired with an initial condition.

You can get more practice applying these ideas in 9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
initial condition	Specified values of a function at particular points that determine which particular solution to a differential equation is selected.
parametric function	Functions where x and y coordinates are each expressed as separate functions of a third variable, typically time (t), rather than y as a function of x.
rate vector	A vector-valued function that describes the rate of change of position with respect to time, representing velocity or acceleration.
vector-valued function	Functions that output vectors rather than scalar values, where each component is a function of the same independent variable.

Frequently Asked Questions

How do you integrate a vector-valued function?

Integrate each component separately. For example, if r(t) = , then an antiderivative is R(t) = , plus constants for each component.

What does integrating a velocity vector give you?

Integrating a velocity vector gives displacement. If you also know an initial position, you can add the constants and find the particle's position function.

How do initial conditions work with vector-valued functions?

After integrating each component, plug the initial time and initial position or velocity into the vector equation. Solve each component equation for its constant.

Is AP Calculus Topic 9.5 AB or BC?

Topic 9.5, Integrating Vector-Valued Functions, is BC only. It belongs to the Parametric Equations, Polar Coordinates, and Vector-Valued Functions unit.

What is the difference between displacement and distance in vector motion?

Displacement comes from integrating velocity component by component. Total distance traveled requires integrating speed, which is the magnitude of the velocity vector.

How do you find distance from the origin after integrating?

First integrate and use initial conditions to find the position vector at the given time. Then use the distance formula from the origin, usually sqrt(x^2 + y^2).