Integrating Vector-Valued Functions: Study Guide (Calculus BC 9.5)

BC-only material. Vector-valued functions are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

This topic joins two ideas you already have: vectors from Topic 9.4, and “integrate a rate, then add a starting value” from earlier units. If any row is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Vector-valued functions and their derivatives	9.4	r′(t) = ⟨x′(t), y′(t)⟩ is what you are undoing
Fundamental Theorem of Calculus	6.7	∫ from a to b of r′(t) dt = r(b) − r(a)
Basic antiderivatives and substitution	6.8, 6.9	Integrating each component
Particular solutions from an initial condition	7.7	Finding the constants
Position from velocity along a line	8.2	The same idea, now in two dimensions

Notation on this page. A vector-valued function is written r(t) = ⟨x(t), y(t)⟩. Some books write x(t) i + y(t) j; it means the same thing. ∫ from a to b of f(t) dt is a definite integral with lower limit a and upper limit b.

Integrate one component at a time

In Topic 9.4 you differentiated a vector by differentiating each component:

if r(t) = ⟨x(t), y(t)⟩, then r′(t) = ⟨x′(t), y′(t)⟩.

Integration is the reverse process, so it also works one component at a time. To find an antiderivative of ⟨f(t), g(t)⟩, find an antiderivative of f and an antiderivative of g separately:

∫ ⟨f(t), g(t)⟩ dt = ⟨∫ f(t) dt, ∫ g(t) dt⟩ = ⟨F(t), G(t)⟩ + C

Here F′ = f, G′ = g, and C = ⟨C₁, C₂⟩ is a constant vector. You can check the answer by differentiating: the derivative of ⟨F(t) + C₁, G(t) + C₂⟩ is ⟨f(t), g(t)⟩, because the derivative of each constant is 0.

Two constants, not one. The x-component and the y-component are separate functions. Each has its own constant, and the two constants are usually different. Writing “+ C” after the vector is fine, as long as you remember that C is a vector.

Definite integrals give a vector. For a definite integral, integrate each component between the same limits. For example,

∫ from 0 to 1 of ⟨6t², eᵗ⟩ dt = ⟨[2t³] from 0 to 1, [eᵗ] from 0 to 1⟩ = ⟨2, e − 1⟩.

The answer is a vector with two components, not a single number. Do not add the components together.

Parametric form is the same problem. You may be given dx/dt = f(t) and dy/dt = g(t) instead of a vector. That is exactly the rate vector ⟨f(t), g(t)⟩ written as two separate equations. Integrate each one and use the given starting point.

Net change and the position formula

Apply the Fundamental Theorem of Calculus to each component of r′(t):

∫ from a to b of r′(t) dt = ⟨x(b) − x(a), y(b) − y(a)⟩ = r(b) − r(a)

So the definite integral of a rate vector is the net change in the vector over [a, b]. Rearranging gives the formula you will use most:

r(b) = r(a) + ∫ from a to b of r′(t) dt

In words: new value = known value + accumulated change. This is the same idea as x(b) = x(a) + ∫ v dt for motion along a line, applied to each coordinate. When r′ is a velocity, the net change is the particle’s displacement; Topic 9.6 builds on that.

The same formula works one step down. If a(t) is the acceleration vector, then

v(b) = v(a) + ∫ from a to b of a(t) dt.

Finding a particular solution: two methods

A problem that gives you a rate vector and the value of the vector at one time is an initial value problem. You can solve it in either of two ways.

Method 1: antiderivative, then constants.

1. Integrate each component and attach a constant: r(t) = ⟨F(t) + C₁, G(t) + C₂⟩.
2. Substitute the known time t₀ into each component and set it equal to the known value.
3. Solve for C₁ and C₂ separately.

Method 2: accumulation form. Write the answer directly as

r(t) = r(t₀) + ∫ from t₀ to t of r′(s) ds

The letter s is a dummy variable, used so that t can be the upper limit. This form is the one to use when an antiderivative cannot be found by hand.

Both methods give the same function. Method 1 is often quicker for a formula; Method 2 is quicker for a single value at one time.

Worked example 1: an initial condition that is not at t = 0

Question. A particle moves in the plane with velocity vector r′(t) = ⟨3t² − 4t, 2e^(t−1)⟩ for t ≥ 0. At t = 1 the particle is at the point (5, −1). Find r(t), then find the position at t = 3. No calculator.

1. Integrate each component.
   • ∫ (3t² − 4t) dt = t³ − 2t² + C₁
   • ∫ 2e^(t−1) dt = 2e^(t−1) + C₂ (the inner function t − 1 has derivative 1, so no extra factor)
2. Use the condition at t = 1, not t = 0.
   • x(1) = 1 − 2 + C₁ = 5, so C₁ = 6.
   • y(1) = 2e⁰ + C₂ = 2 + C₂ = −1, so C₂ = −3.
3. Write the particular solution. r(t) = ⟨t³ − 2t² + 6, 2e^(t−1) − 3⟩
4. Evaluate at t = 3.
   • x(3) = 27 − 18 + 6 = 15
   • y(3) = 2e² − 3 ≈ 11.778

Answer. At t = 3 the particle is at (15, 2e² − 3), about (15, 11.778).

Check with Method 2. r(3) = r(1) + ∫ from 1 to 3 of r′(t) dt.

• ∫ from 1 to 3 of (3t² − 4t) dt = [t³ − 2t²] from 1 to 3 = 9 − (−1) = 10, so x(3) = 5 + 10 = 15. ✓
• ∫ from 1 to 3 of 2e^(t−1) dt = [2e^(t−1)] from 1 to 3 = 2e² − 2, so y(3) = −1 + 2e² − 2 = 2e² − 3. ✓

Check by differentiating. d/dt ⟨t³ − 2t² + 6, 2e^(t−1) − 3⟩ = ⟨3t² − 4t, 2e^(t−1)⟩ ✓, and r(1) = ⟨5, −1⟩ ✓.

The trap. If you treat (5, −1) as the position at t = 0, you get C₁ = 5 and C₂ = −1 − 2e⁻¹, and the x-coordinate at t = 3 comes out as 14 instead of 15. Always substitute the time you are actually given.

Worked example 2: from acceleration to position

Context. A radio-controlled model car moves on a flat test area. Its position is measured in metres from a corner post, with x east and y north, and t is in seconds. Its acceleration vector is

a(t) = ⟨−2, 6t − 12⟩ m/s²

At t = 0 its velocity is ⟨8, 9⟩ m/s and its position is (2, 1). Find v(t), r(t), the velocity at t = 2 and the position at t = 4. No calculator.

1. First integration: velocity. v(t) = ⟨−2t + C₁, 3t² − 12t + C₂⟩. At t = 0: C₁ = 8 and C₂ = 9. v(t) = ⟨8 − 2t, 3t² − 12t + 9⟩ m/s
2. Second integration: position. r(t) = ⟨8t − t² + D₁, t³ − 6t² + 9t + D₂⟩. At t = 0: D₁ = 2 and D₂ = 1. r(t) = ⟨8t − t² + 2, t³ − 6t² + 9t + 1⟩ m
3. Velocity at t = 2. v(2) = ⟨8 − 4, 12 − 24 + 9⟩ = ⟨4, −3⟩ m/s: moving east and south.
4. Position at t = 4. r(4) = ⟨32 − 16 + 2, 64 − 96 + 36 + 1⟩ = ⟨18, 5⟩, so the car is at (18, 5).

Checks.

• Differentiate twice: r′(t) = ⟨8 − 2t, 3t² − 12t + 9⟩ = v(t) ✓ and r″(t) = ⟨−2, 6t − 12⟩ = a(t) ✓.
• Initial values: v(0) = ⟨8, 9⟩ ✓, r(0) = ⟨2, 1⟩ ✓.
• Net change: r(4) − r(0) = ⟨16, 4⟩ m. The car ends 16 m east and 4 m north of where it started.
• Units: integrating m/s² over seconds gives m/s; integrating m/s gives m. ✓

Interpretation. Integrating velocity tells you only how far the car’s coordinates change. The starting position (2, 1) fixes where the path sits on the test area. Notice in Figure 1 that the dashed net-change arrow is much shorter than the curved path: the net change is not the distance travelled. That distinction is the subject of Topic 9.6.

The trap. If you forget v(0) and write v(t) = ⟨−2t, 3t² − 12t⟩, the position at t = 4 comes out as (−14, −31). That is far from the true (18, 5). Each integration step needs its own initial condition.

Worked example 3: with a graphing calculator

Question. A particle moves so that dx/dt = 1 + cos(t²) and dy/dt = 2e^(−t²/2). At t = 0 the particle is at (−1, 4). Find its position at t = 2. A graphing calculator is allowed.

1. Spot the issue. Neither cos(t²) nor e^(−t²/2) has an antiderivative you can write with elementary functions. So Method 1 is not possible. Use the accumulation form.
2. Set up each coordinate.
   • x(2) = x(0) + ∫ from 0 to 2 of (1 + cos(t²)) dt
   • y(2) = y(0) + ∫ from 0 to 2 of 2e^(−t²/2) dt
3. Evaluate the integrals on the calculator (radian mode).
   • ∫ from 0 to 2 of (1 + cos(t²)) dt ≈ 2.4615
   • ∫ from 0 to 2 of 2e^(−t²/2) dt ≈ 2.3926
4. Add the starting values.
   • x(2) ≈ −1 + 2.4615 = 1.461
   • y(2) ≈ 4 + 2.3926 = 6.393

Answer. At t = 2 the particle is at about (1.461, 6.393).

Checks. Both rates are never negative on [0, 2] (cos(t²) ≥ −1), so both coordinates should increase from their starting values. They do. Keep at least 4 decimal places in the integrals and round only the final answer.

Going backwards in time. The same formula works for an earlier time. If you know r(1) and want r(0), write r(0) = r(1) + ∫ from 1 to 0 of r′(t) dt = r(1) − ∫ from 0 to 1 of r′(t) dt. The minus sign is easy to lose.

Checking your answer

You can check any particular solution in two quick steps:

1. Differentiate it. You should get back the given rate vector.
2. Substitute the given time. You should get back the given position (or velocity).

If both checks pass, the answer is correct. There is only one function that meets both conditions.

Common misconceptions

• “One constant is enough.” ⟨t³ − 2t², 2e^(t−1)⟩ + C needs C = ⟨C₁, C₂⟩. Using the same number for both components is almost always wrong.
• Using t = 0 when the condition is at another time. Substitute the time you are given, such as t = 1.
• Reporting the net change as the position. ∫ from a to b of r′(t) dt is the change. Add r(a) to get the position at b.
• Losing the starting velocity. From acceleration, the first constant comes from v(t₀), the second from r(t₀). Both are needed.
• Adding the components of a definite integral. ∫ from 0 to 1 of ⟨6t², eᵗ⟩ dt is the vector ⟨2, e − 1⟩, not the number e + 1.
• Forgetting a chain-rule factor in one component. ∫ e^(t/2) dt = 2e^(t/2), not ½ e^(t/2).
• Sign errors going backwards in time. r(0) = r(1) − ∫ from 0 to 1 of r′(t) dt.
• Rounding too early in calculator work. Store integral values and round only at the end.

Where this leads

Topic 9.6 uses everything here to solve full motion problems in the plane: velocity, speed and acceleration from derivatives, and displacement and total distance from integrals. Continue with Topic 9.6, Solving Motion Problems Using Parametric and Vector-Valued Functions, or see the order on the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.