Unit 9.5 – Integrating Vector-Valued Functions BC ONLY

AP® Calculus BC | Antiderivatives and Initial Value Problems

Why This Matters: Integration of vector-valued functions reverses differentiation! Just as we integrate to find position from velocity, we integrate vector functions component-wise. This is crucial for solving motion problems where you know acceleration or velocity and need to find the position. This topic appears on every BC exam!

📐 The Antiderivative (Indefinite Integral)

Indefinite Integral Formula

THE FORMULA:

For \(\vec{r}(t) = \langle x(t), y(t) \rangle\):

\[ \int \vec{r}(t) \, dt = \left\langle \int x(t) \, dt, \int y(t) \, dt \right\rangle + \vec{C} \]

where \(\vec{C} = \langle C_1, C_2 \rangle\) is a constant vector

Key Insight:

Integrate component-wise! Just like differentiation, integration of vectors is done separately for each component. Don't forget the constant vector!

🔢 The Definite Integral

Definite Integral Formula

THE FORMULA:

\[ \int_a^b \vec{r}(t) \, dt = \left\langle \int_a^b x(t) \, dt, \int_a^b y(t) \, dt \right\rangle \]

Using the Fundamental Theorem:

If \(\vec{R}(t)\) is an antiderivative of \(\vec{r}(t)\), then:

\[ \int_a^b \vec{r}(t) \, dt = \vec{R}(b) - \vec{R}(a) \]

📝 Note: For definite integrals, there's no constant vector because it cancels out when evaluating at the bounds!

🎯 Initial Value Problems

SOLVING FOR POSITION FROM VELOCITY

Given Velocity, Find Position:

If \(\vec{v}(t)\) is velocity and \(\vec{r}(t_0) = \vec{r}_0\) is initial position:

\[ \vec{r}(t) = \int \vec{v}(t) \, dt = \vec{R}(t) + \vec{C} \]

Use initial condition to find \(\vec{C}\):

\[ \vec{r}_0 = \vec{R}(t_0) + \vec{C} \quad \Rightarrow \quad \vec{C} = \vec{r}_0 - \vec{R}(t_0) \]

Given Acceleration, Find Velocity:

If \(\vec{a}(t)\) is acceleration and \(\vec{v}(t_0) = \vec{v}_0\) is initial velocity:

\[ \vec{v}(t) = \int \vec{a}(t) \, dt + \vec{C} \]

Use initial velocity to find \(\vec{C}\)

🔗 The Integration Chain

From Acceleration to Position:

Step 1: Given \(\vec{a}(t)\) and \(\vec{v}(t_0) = \vec{v}_0\)

\[ \vec{v}(t) = \int \vec{a}(t) \, dt + \vec{C}_1 \]

Find \(\vec{C}_1\) using initial velocity

Step 2: Given \(\vec{v}(t)\) and \(\vec{r}(t_0) = \vec{r}_0\)

\[ \vec{r}(t) = \int \vec{v}(t) \, dt + \vec{C}_2 \]

Find \(\vec{C}_2\) using initial position

📖 Comprehensive Worked Examples

Example 1: Basic Indefinite Integral

Problem: Find \(\int \langle 2t, 3t^2 \rangle \, dt\).

Solution:

Integrate each component:

\[ \int \langle 2t, 3t^2 \rangle \, dt = \left\langle \int 2t \, dt, \int 3t^2 \, dt \right\rangle \]

\[ = \langle t^2 + C_1, t^3 + C_2 \rangle = \langle t^2, t^3 \rangle + \langle C_1, C_2 \rangle \]

ANSWER: \(\langle t^2, t^3 \rangle + \vec{C}\)

Example 2: Definite Integral

Problem: Evaluate \(\int_0^1 \langle 6t, 4t^3 \rangle \, dt\).

Find antiderivative:

\[ \vec{R}(t) = \langle 3t^2, t^4 \rangle \]

Evaluate at bounds:

\[ \vec{R}(1) - \vec{R}(0) = \langle 3, 1 \rangle - \langle 0, 0 \rangle = \langle 3, 1 \rangle \]

Example 3: Finding Position from Velocity

Problem: A particle has velocity \(\vec{v}(t) = \langle \cos t, \sin t \rangle\) and initial position \(\vec{r}(0) = \langle 1, 0 \rangle\). Find \(\vec{r}(t)\).

Step 1: Integrate velocity

\[ \vec{r}(t) = \int \langle \cos t, \sin t \rangle \, dt = \langle \sin t, -\cos t \rangle + \vec{C} \]

Step 2: Use initial condition

\[ \vec{r}(0) = \langle 0, -1 \rangle + \langle C_1, C_2 \rangle = \langle 1, 0 \rangle \]

\[ \langle C_1, C_2 \rangle = \langle 1, 1 \rangle \]

Step 3: Final answer

\[ \vec{r}(t) = \langle \sin t + 1, -\cos t + 1 \rangle \]

Example 4: From Acceleration to Position

Problem: Given \(\vec{a}(t) = \langle 0, -g \rangle\) (gravity), \(\vec{v}(0) = \langle v_0, 0 \rangle\), and \(\vec{r}(0) = \langle 0, h_0 \rangle\), find position.

Step 1: Find velocity

\[ \vec{v}(t) = \int \langle 0, -g \rangle \, dt = \langle 0, -gt \rangle + \vec{C}_1 \]

Using \(\vec{v}(0) = \langle v_0, 0 \rangle\):

\[ \langle 0, 0 \rangle + \vec{C}_1 = \langle v_0, 0 \rangle \quad \Rightarrow \quad \vec{C}_1 = \langle v_0, 0 \rangle \]

\[ \vec{v}(t) = \langle v_0, -gt \rangle \]

Step 2: Find position

\[ \vec{r}(t) = \int \langle v_0, -gt \rangle \, dt = \left\langle v_0t, -\frac{gt^2}{2} \right\rangle + \vec{C}_2 \]

Using \(\vec{r}(0) = \langle 0, h_0 \rangle\):

\[ \vec{C}_2 = \langle 0, h_0 \rangle \]

\[ \vec{r}(t) = \left\langle v_0t, h_0 - \frac{1}{2}gt^2 \right\rangle \]

This is projectile motion! (horizontal and vertical components)

📋 Properties of Vector Integration

1. Constant Multiple:

\[ \int c\vec{r}(t) \, dt = c \int \vec{r}(t) \, dt \]

2. Sum/Difference:

\[ \int [\vec{u}(t) \pm \vec{v}(t)] \, dt = \int \vec{u}(t) \, dt \pm \int \vec{v}(t) \, dt \]

3. Linearity:

\[ \int [a\vec{u}(t) + b\vec{v}(t)] \, dt = a\int \vec{u}(t) \, dt + b\int \vec{v}(t) \, dt \]

📊 Complete Formula Reference

Vector Integration Formulas
Type	Formula	Notes
Indefinite Integral	\(\int \vec{r}(t)\,dt = \langle \int x(t)\,dt, \int y(t)\,dt \rangle + \vec{C}\)	Include constant vector
Definite Integral	\(\int_a^b \vec{r}(t)\,dt = \vec{R}(b) - \vec{R}(a)\)	No constant needed
From Velocity	\(\vec{r}(t) = \int \vec{v}(t)\,dt + \vec{C}\)	Use initial position for \(\vec{C}\)
From Acceleration	\(\vec{v}(t) = \int \vec{a}(t)\,dt + \vec{C}\)	Use initial velocity for \(\vec{C}\)

💡 Essential Tips & Strategies

✅ Success Strategies:

Integrate component-wise: Each component separately
Don't forget constant vector: \(\vec{C} = \langle C_1, C_2 \rangle\) for indefinite
Use initial conditions: To find constant vector
Two-step process: Acceleration → velocity → position
Check your work: Differentiate to verify
Write as vector: Use \(\langle \, \rangle\) notation
Each component independent: Integrate separately
Units matter: Position, velocity, acceleration have different units

🔥 Common Scenarios:

Constant acceleration: \(\vec{a} = \langle a_x, a_y \rangle\) → parabolic path
Gravity only: \(\vec{a} = \langle 0, -g \rangle\) → projectile motion
Circular motion: Trig functions in velocity
Initial value problems: Always given initial position or velocity

❌ Common Mistakes to Avoid

Mistake 1: Forgetting the constant vector in indefinite integrals
Mistake 2: Not using initial conditions to find constants
Mistake 3: Mixing up which constant goes with which component
Mistake 4: Forgetting to integrate both components
Mistake 5: Wrong integration (power rule errors)
Mistake 6: Not subtracting vectors correctly in definite integrals
Mistake 7: Confusing velocity with position
Mistake 8: Sign errors (especially with gravity)
Mistake 9: Forgetting negative sign in trig integrals
Mistake 10: Not simplifying final answer

📝 Practice Problems

Solve these:

Find \(\int \langle e^t, t^2 \rangle \, dt\).
Evaluate \(\int_0^2 \langle 3t^2, 6t \rangle \, dt\).
Given \(\vec{v}(t) = \langle 2t, 3 \rangle\) and \(\vec{r}(0) = \langle 1, 2 \rangle\), find \(\vec{r}(t)\).
Given \(\vec{a}(t) = \langle 0, -10 \rangle\), \(\vec{v}(0) = \langle 5, 0 \rangle\), \(\vec{r}(0) = \langle 0, 0 \rangle\), find \(\vec{r}(t)\).

Answers:

\(\langle e^t, \frac{t^3}{3} \rangle + \vec{C}\)
\(\langle 8, 12 \rangle\)
\(\vec{r}(t) = \langle t^2 + 1, 3t + 2 \rangle\)
\(\vec{r}(t) = \langle 5t, -5t^2 \rangle\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Show component integration: Integrate each component
Include constant vector: For indefinite integrals
Use initial conditions: Show how you found constants
Set up equation: \(\vec{r}(t_0) = \vec{R}(t_0) + \vec{C}\)
Solve for constant: Show algebra
Write final answer: As a vector
Check dimensions: Each component separately
Simplify: Combine like terms

💯 Exam Strategy:

Identify what you're given (acceleration, velocity, or position)
Identify what you need to find
Integrate component-wise
Include constant vector for indefinite
Use initial condition to find constants
Substitute constants back
Write final answer as vector
Verify by differentiating (if time)

⚡ Quick Reference Guide

VECTOR INTEGRATION ESSENTIALS

Indefinite Integral:

\[ \int \vec{r}(t) \, dt = \langle \int x(t)\,dt, \int y(t)\,dt \rangle + \vec{C} \]

Definite Integral:

\[ \int_a^b \vec{r}(t) \, dt = \vec{R}(b) - \vec{R}(a) \]

Motion Chain:

\(\vec{a}(t)\) → integrate → \(\vec{v}(t)\) (use \(\vec{v}_0\))
\(\vec{v}(t)\) → integrate → \(\vec{r}(t)\) (use \(\vec{r}_0\))

Remember:

Integrate each component separately!
Include \(\vec{C}\) for indefinite
Use initial conditions to find \(\vec{C}\)

Master Vector Integration! Integration of vector-valued functions is done component-wise: \(\int\vec{r}(t)\,dt = \langle\int x(t)\,dt, \int y(t)\,dt\rangle + \vec{C}\) where \(\vec{C} = \langle C_1, C_2\rangle\) is the constant vector. For definite integrals: \(\int_a^b\vec{r}(t)\,dt = \vec{R}(b) - \vec{R}(a)\) (no constant). Motion problems: given acceleration, integrate to get velocity (use initial velocity for constant); given velocity, integrate to get position (use initial position for constant). Two-step process: \(\vec{a} \to \vec{v} \to \vec{r}\). Each integration adds one constant vector—find using initial conditions. Properties: linearity, constant multiple, sum/difference all work component-wise. Common application: projectile motion with \(\vec{a} = \langle 0, -g\rangle\). Verify answers by differentiating. This is guaranteed BC content—appears on every exam! Practice initial value problems until automatic! 🎯✨