Integrating
Vector-Valued Functions
Learn how to compute antiderivatives component by component, evaluate definite integrals of vector functions, and connect integration to displacement and area in multidimensional motion.
AP Calculus BC
What Is a Vector-Valued Function?
A vector-valued function assigns a vector to each input value \( t \). In 2D, it takes the form:
In 3D, we add a third component: \( \mathbf{r}(t) = \langle f(t),\, g(t),\, h(t) \rangle \). The parameter \( t \) often represents time.
Tracing a Parametric Curve
As \( t \) varies, the tip of \( \mathbf{r}(t) \) traces out a parametric curve in the plane. Integration lets us recover the curve from its rate of change.
Why Integrate Vector Functions?
Displacement
If \( \mathbf{r}'(t) = \mathbf{v}(t) \), then \( \int_a^b \mathbf{v}(t)\,dt \) gives the net displacement vector from \( t=a \) to \( t=b \).
Position Recovery
Given a velocity function and an initial position, integration lets you reconstruct the complete position function \( \mathbf{r}(t) \).
Accumulation
Just like scalar integrals represent accumulated area, vector integrals represent accumulated change in each coordinate direction.
The Component-Wise Integration Rule
The constant of integration \( \mathbf{C} = \langle C_1, C_2 \rangle \) is a constant vector, not a scalar. Each component gets its own constant.
Visualizing Component Integration
Extension to Three Dimensions
The same principle applies: integrate each component function as a separate scalar integral. Nothing changes except the number of components.
\( x \)-component
\( \int f(t)\,dt \)
\( y \)-component
\( \int g(t)\,dt \)
\( z \)-component
\( \int h(t)\,dt \)
Indefinite vs. Definite Integrals
| Indefinite | Definite | |
|---|---|---|
| Result type | Vector function of \( t \) | Constant vector |
| Includes \( +\mathbf{C} \)? | Yes: vector constant | No |
| Notation | \( \int \mathbf{r}(t)\,dt \) | \( \int_a^b \mathbf{r}(t)\,dt \) |
| Physical meaning | Family of antiderivatives | Net displacement vector |
| FTC applies? | Yes, to find \( \mathbf{R}(t) \) | Yes: \( \mathbf{R}(b) - \mathbf{R}(a) \) |
Step-by-Step Integration Process
1
Identify each component. Write out \( f(t) \), \( g(t) \), and \( h(t) \) separately.
2
Integrate each component independently. Apply all standard scalar integration techniques (power rule, substitution, etc.).
3
Reassemble. Combine the results back into angle-bracket notation \( \langle \ldots \rangle \).
4
Add the constant vector \( \mathbf{C} \) for indefinite integrals, or evaluate bounds for definite integrals.
5
Use initial conditions if given (e.g., \( \mathbf{r}(0) = \langle 1, 3 \rangle \)) to solve for \( \mathbf{C} \).
Example 1: Indefinite Integral
1
\( x \)-component: \( \int 3t^2\,dt = t^3 + C_1 \)
2
\( y \)-component: \( \int \cos t\,dt = \sin t + C_2 \)
3
Result: \( \langle t^3,\; \sin t \rangle + \mathbf{C} \) where \( \mathbf{C} = \langle C_1, C_2 \rangle \).
Example 2: Definite Integral
1
\( x \)-component: \( \int_0^{\pi} \sin t\,dt = [-\cos t]_0^{\pi} = -\cos\pi - (-\cos 0) = 1 + 1 = \) \( 2 \)
2
\( y \)-component: \( \int_0^{\pi} 2t\,dt = [t^2]_0^{\pi} = \pi^2 - 0 = \) \( \pi^2 \)
3
Result: \( \langle 2,\; \pi^2 \rangle \). This is a constant vector: no \( +\mathbf{C} \).
Example 3: Using an Initial Condition
1
Integrate: \( \mathbf{r}(t) = \int \langle e^t, 4t \rangle\,dt = \langle e^t + C_1,\; 2t^2 + C_2 \rangle \)
2
Apply initial condition: \( \mathbf{r}(0) = \langle e^0 + C_1,\; 0 + C_2 \rangle = \langle 1 + C_1,\; C_2 \rangle = \langle 2,\; -1 \rangle \)
3
Solve: \( C_1 = 1 \), \( C_2 = -1 \)
4
Final answer: \( \mathbf{r}(t) = \langle e^t + 1,\; 2t^2 - 1 \rangle \)
Displacement vs. Distance
Displacement
\( \int_a^b \mathbf{v}(t)\,dt \): a vector from start to end. Accounts for direction.
Total Distance
\( \int_a^b \|\mathbf{v}(t)\|\,dt \): a scalar. Always \( \geq \) displacement magnitude.
Properties of Vector Integration
1
Constant scalar multiple: \( \int c\,\mathbf{r}(t)\,dt = c\int \mathbf{r}(t)\,dt \)
2
Sum / difference: \( \int [\mathbf{r}(t) \pm \mathbf{s}(t)]\,dt = \int \mathbf{r}(t)\,dt \pm \int \mathbf{s}(t)\,dt \)
3
FTC for vectors: If \( \mathbf{R}'(t) = \mathbf{r}(t) \), then \( \int_a^b \mathbf{r}(t)\,dt = \mathbf{R}(b) - \mathbf{R}(a) \)
4
Additivity over intervals: \( \int_a^b \mathbf{r}(t)\,dt + \int_b^c \mathbf{r}(t)\,dt = \int_a^c \mathbf{r}(t)\,dt \) (each component satisfies this)
Common Mistakes to Avoid
✕ Scalar Constant
Writing \( +C \) (a single number) instead of \( +\mathbf{C} = \langle C_1, C_2 \rangle \). Each component gets its own constant.
✕ Mixing Up Displacement
Confusing \( \int \mathbf{v}(t)\,dt \) (vector displacement) with \( \int \|\mathbf{v}(t)\|\,dt \) (scalar arc length/distance).
✕ Forgetting Initial Conditions
Solving for \( \mathbf{r}(t) \) but not using \( \mathbf{r}(t_0) \) to determine the specific constants \( C_1, C_2 \).
Quick Reference Summary
| Operation | Formula | Result Type |
|---|---|---|
| Indefinite integral | \( \langle \int f(t)\,dt,\; \int g(t)\,dt \rangle + \mathbf{C} \) | Vector function |
| Definite integral | \( \langle \int_a^b f(t)\,dt,\; \int_a^b g(t)\,dt \rangle \) | Constant vector |
| Displacement | \( \int_a^b \mathbf{v}(t)\,dt \) | Vector |
| Arc length | \( \int_a^b \|\mathbf{v}(t)\|\,dt \) | Scalar |
| Position from velocity | \( \mathbf{r}(t) = \int \mathbf{v}(t)\,dt + \mathbf{C} \) | Vector function |
Question 1 of 5
Evaluate \( \displaystyle\int \langle 6t,\; e^{2t} \rangle\, dt \).
The \( x \)-component: \( \int 6t\,dt = 3t^2 \). The \( y \)-component: \( \int e^{2t}\,dt = \frac{1}{2}e^{2t} \). The constant must be a vector \( \mathbf{C} \), not a scalar \( C \). Choice (c) uses a scalar constant.
Question 2 of 5
Evaluate \( \displaystyle\int_0^1 \langle 2t,\; 3t^2 \rangle\, dt \).
The \( x \)-component: \( [t^2]_0^1 = 1 \). The \( y \)-component: \( [t^3]_0^1 = 1 \). Result: \( \langle 1, 1 \rangle \). Definite integrals produce a constant vector with no \( +\mathbf{C} \), which eliminates (d).
Question 3 of 5
A particle has velocity \( \mathbf{v}(t) = \langle 4,\; -2t \rangle \) and position \( \mathbf{r}(0) = \langle 1,\; 5 \rangle \). What is \( \mathbf{r}(t) \)?
Integrate: \( \int 4\,dt = 4t + C_1 \) and \( \int -2t\,dt = -t^2 + C_2 \). Apply \( \mathbf{r}(0) = \langle 1, 5 \rangle \): \( C_1 = 1 \), \( C_2 = 5 \). So \( \mathbf{r}(t) = \langle 4t + 1,\; -t^2 + 5 \rangle \). Choice (c) incorrectly uses \( -2t^2 \) (forgot the \( \frac{1}{2} \) from the power rule).
Question 4 of 5
Which expression gives the total distance traveled by a particle with velocity \( \mathbf{v}(t) \) from \( t = a \) to \( t = b \)?
Total distance requires integrating the speed (magnitude of velocity): \( \int_a^b \|\mathbf{v}(t)\|\,dt \). Choice (a) gives the magnitude of displacement, not distance. Choice (b) gives the displacement vector. Choice (c) uses single bars (absolute value), not double bars (norm).
Question 5 of 5
Evaluate \( \displaystyle\int_0^{\pi/2} \langle \cos t,\; \sin t,\; 1 \rangle\, dt \).
Component by component: \( [\sin t]_0^{\pi/2} = 1 - 0 = 1 \), \( [-\cos t]_0^{\pi/2} = 0 - (-1) = 1 \), \( [t]_0^{\pi/2} = \pi/2 \). Result: \( \langle 1, 1, \pi/2 \rangle \). Choice (a) incorrectly evaluates \( \int_0^{\pi/2} \cos t\,dt \) as 0.