9.5 Integrating Vector-Valued Functions

Integrating componentwise

To integrate a vector-valued function r(t) = ⟨x(t), y(t)⟩, integrate each component separately: ∫r(t) dt = ⟨∫x(t) dt, ∫y(t) dt⟩. Each integral gets its own constant of integration, so the antiderivative is ⟨X(t)+C₁, Y(t)+C₂⟩. This mirrors how derivatives of vector functions are found componentwise.

Finding position from velocity

If v(t) is a velocity vector function, then position r(t) = ∫v(t) dt. Given an initial position r(t₀), solve for the constants C₁ and C₂ so the antiderivative matches that initial condition. This is the vector analog of using initial conditions to find a particular antiderivative in single-variable calculus.

Definite integrals give displacement

The definite integral ∫ₐᵇ v(t) dt gives the displacement vector over [a,b], found by integrating each component from a to b. Displacement differs from total distance traveled, which requires integrating the speed |v(t)| = √((x'(t))² + (y'(t))²) — the arc length formula.

Key terms

How do you integrate a vector-valued function r(t) = ⟨x(t), y(t)⟩?

Integrate each component separately: ∫r(t) dt = ⟨∫x(t) dt, ∫y(t) dt⟩, each with its own constant of integration.

How do you find position from a velocity vector function?

Integrate v(t) componentwise, then use an initial position condition to solve for the constants of integration.

What does ∫ₐᵇ v(t) dt represent?

The displacement vector of the particle from t=a to t=b.

How do you find total distance traveled by a particle with vector velocity?

Integrate the speed |v(t)| = √((x'(t))² + (y'(t))²) over the interval (the arc length formula).

Displacement vs. distance for vector motion

Displacement is the net change in position (a vector, from the definite integral of velocity); distance is total path length traveled (a scalar, from integrating speed).

Try an AP-style question

∫ ⟨2t, 3t²⟩ dt = ?

1. ⟨t², t³⟩
2. ⟨t², t³⟩ + C
3. t² + t³ + C
4. ⟨2, 6t⟩ + C