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Integrals of Vector-Valued Functions

Formulas & Definitions
Core Definition

If r⃗(t)=⟨f(t),g(t),h(t)⟩=f(t)i^+g(t)j^+h(t)k^\vec{r}(t) = \langle f(t), g(t), h(t) \rangle = f(t)\hat{i} + g(t)\hat{j} + h(t)\hat{k} is a continuous vector-valued function on an interval [a,b][a, b], its integral is evaluated component-wise:

  1. Indefinite Integral:
    ∫r⃗(t) dt=(∫f(t) dt)i^+(∫g(t) dt)j^+(∫h(t) dt)k^=R⃗(t)+C⃗\int \vec{r}(t) \, dt = \left( \int f(t) \, dt \right)\hat{i} + \left( \int g(t) \, dt \right)\hat{j} + \left( \int h(t) \, dt \right)\hat{k} = \vec{R}(t) + \vec{C}
    where R⃗(t)=⟨F(t),G(t),H(t)⟩\vec{R}(t) = \langle F(t), G(t), H(t) \rangle is an antiderivative vector and C⃗=⟨C1,C2,C3⟩\vec{C} = \langle C_1, C_2, C_3 \rangle is an arbitrary constant vector of integration.
  2. Definite Integral (Fundamental Theorem of Calculus in R3\mathbb{R}^3):
    ∫abr⃗(t) dt=⟨∫abf(t) dt,∫abg(t) dt,∫abh(t) dt⟩=R⃗(b)−R⃗(a)\int_{a}^{b} \vec{r}(t) \, dt = \left\langle \int_{a}^{b} f(t) \, dt, \int_{a}^{b} g(t) \, dt, \int_{a}^{b} h(t) \, dt \right\rangle = \vec{R}(b) - \vec{R}(a)

Concept Explanation

Indefinite Integrals & The Constant Vector C⃗\vec{C}

When integrating scalar functions, a single integration constant +C+ C is produced. Because a 3D vector-valued function consists of three independent component integrations, each component produces its own independent constant (C1,C2,C3C_1, C_2, C_3):

∫r⃗(t) dt=⟨F(t)+C1,G(t)+C2,H(t)+C3⟩=⟨F(t),G(t),H(t)⟩+⟨C1,C2,C3⟩=R⃗(t)+C⃗\int \vec{r}(t) \, dt = \langle F(t) + C_1, G(t) + C_2, H(t) + C_3 \rangle = \langle F(t), G(t), H(t) \rangle + \langle C_1, C_2, C_3 \rangle = \vec{R}(t) + \vec{C}

Never Leave C⃗\vec{C} as a Scalar

In vector calculus, the constant of integration C⃗\vec{C} must be a vector ⟨C1,C2,C3⟩\langle C_1, C_2, C_3 \rangle. Writing + C+\, C (a scalar added to a vector) is algebraically invalid.

Physical Kinematics & Initial Value Problems (IVPs)

Integrating vector-valued functions allows engineers to recover position and velocity profiles from acceleration data:

  1. From Acceleration to Velocity:
    v⃗(t)=∫a⃗(t) dt+C⃗1  ⟹  Given v⃗(0)=v⃗0  ⟹  C⃗1=v⃗0−V⃗(0)\vec{v}(t) = \int \vec{a}(t) \, dt + \vec{C}_1 \implies \text{Given } \vec{v}(0) = \vec{v}_0 \implies \vec{C}_1 = \vec{v}_0 - \vec{V}(0)
  2. From Velocity to Position:
    r⃗(t)=∫v⃗(t) dt+C⃗2  ⟹  Given r⃗(0)=r⃗0  ⟹  C⃗2=r⃗0−R⃗(0)\vec{r}(t) = \int \vec{v}(t) \, dt + \vec{C}_2 \implies \text{Given } \vec{r}(0) = \vec{r}_0 \implies \vec{C}_2 = \vec{r}_0 - \vec{R}(0)

Definite Integrals: Net Displacement

Evaluating the definite integral of velocity v⃗(t)\vec{v}(t) over a time interval a≤t≤ba \le t \le b yields the net displacement vector Δr⃗\Delta \vec{r}:

Δr⃗=r⃗(b)−r⃗(a)=∫abv⃗(t) dt\Delta \vec{r} = \vec{r}(b) - \vec{r}(a) = \int_{a}^{b} \vec{v}(t) \, dt
The displacement vector Δr⃗\Delta \vec{r} represents the straight-line directed vector connecting the starting point r⃗(a)\vec{r}(a) directly to the ending point r⃗(b)\vec{r}(b) in R3\mathbb{R}^3.

Engineering Applications

Example Problems

Example 1: Indefinite Vector Integration

Evaluate the indefinite integral:

∫⟨5cos⁡(7t),3cos⁡(4t),−9t⟩dt\int \left\langle 5\cos(7t), 3\cos(4t), \frac{-9}{t} \right\rangle dt

Solution
  1. Integrate Component 1 (xx-component):
    ∫5cos⁡(7t) dt=57sin⁡(7t)+C1\int 5\cos(7t) \, dt = \frac{5}{7}\sin(7t) + C_1
  2. Integrate Component 2 (yy-component):
    ∫3cos⁡(4t) dt=34sin⁡(4t)+C2\int 3\cos(4t) \, dt = \frac{3}{4}\sin(4t) + C_2
  3. Integrate Component 3 (zz-component):
    ∫−9t dt=−9ln⁡∣t∣+C3\int \frac{-9}{t} \, dt = -9\ln|t| + C_3
  4. Combine into Vector Form with C⃗\vec{C}:
    ∫⟨5cos⁡(7t),3cos⁡(4t),−9t⟩dt=⟨57sin⁡(7t),34sin⁡(4t),−9ln⁡∣t∣⟩+C⃗\int \left\langle 5\cos(7t), 3\cos(4t), \frac{-9}{t} \right\rangle dt = \left\langle \frac{5}{7}\sin(7t), \frac{3}{4}\sin(4t), -9\ln|t| \right\rangle + \vec{C}
    where C⃗=⟨C1,C2,C3⟩\vec{C} = \langle C_1, C_2, C_3 \rangle.

Example 2: Initial Value Problem (Recovering Position from Acceleration)

An autonomous rover has acceleration a⃗(t)=⟨6t,12t2,4⟩ m/s2\vec{a}(t) = \langle 6t, 12t^2, 4 \rangle \text{ m/s}^2 with initial velocity v⃗(0)=⟨2,−1,5⟩ m/s\vec{v}(0) = \langle 2, -1, 5 \rangle \text{ m/s} and initial position r⃗(0)=⟨0,10,−3⟩ m\vec{r}(0) = \langle 0, 10, -3 \rangle \text{ m}. Find its position function r⃗(t)\vec{r}(t).

Solution
  1. Integrate Acceleration to find Velocity v⃗(t)\vec{v}(t):
    v⃗(t)=∫⟨6t,12t2,4⟩ dt=⟨3t2,4t3,4t⟩+C⃗1\vec{v}(t) = \int \langle 6t, 12t^2, 4 \rangle \, dt = \langle 3t^2, 4t^3, 4t \rangle + \vec{C}_1
    Apply initial condition v⃗(0)=⟨2,−1,5⟩\vec{v}(0) = \langle 2, -1, 5 \rangle:
    v⃗(0)=⟨0,0,0⟩+C⃗1=⟨2,−1,5⟩  ⟹  C⃗1=⟨2,−1,5⟩\vec{v}(0) = \langle 0, 0, 0 \rangle + \vec{C}_1 = \langle 2, -1, 5 \rangle \implies \vec{C}_1 = \langle 2, -1, 5 \rangle
    v⃗(t)=⟨3t2+2,4t3−1,4t+5⟩\vec{v}(t) = \langle 3t^2 + 2, 4t^3 - 1, 4t + 5 \rangle
  2. Integrate Velocity to find Position r⃗(t)\vec{r}(t):
    r⃗(t)=∫⟨3t2+2,4t3−1,4t+5⟩ dt=⟨t3+2t,t4−t,2t2+5t⟩+C⃗2\vec{r}(t) = \int \langle 3t^2 + 2, 4t^3 - 1, 4t + 5 \rangle \, dt = \langle t^3 + 2t, t^4 - t, 2t^2 + 5t \rangle + \vec{C}_2
    Apply initial condition r⃗(0)=⟨0,10,−3⟩\vec{r}(0) = \langle 0, 10, -3 \rangle:
    r⃗(0)=⟨0,0,0⟩+C⃗2=⟨0,10,−3⟩  ⟹  C⃗2=⟨0,10,−3⟩\vec{r}(0) = \langle 0, 0, 0 \rangle + \vec{C}_2 = \langle 0, 10, -3 \rangle \implies \vec{C}_2 = \langle 0, 10, -3 \rangle
    r⃗(t)=⟨t3+2t,t4−t+10,2t2+5t−3⟩ m\vec{r}(t) = \langle t^3 + 2t, t^4 - t + 10, 2t^2 + 5t - 3 \rangle \text{ m}

Connections & References