Integrating Vector-Valued Functions

AP Course

AP Calculus AB/BC

Study the complete College Board sequence for AP Calculus AB and BC, from limits through infinite series.

Choose an official unit and topic to open its lecture, concept check, or focused practice.

AP Calculus AB/BC · Unit 9 · Topic 9.5 · BC Only

Integrating Vector-Valued Functions

Integrate components and use initial vectors to determine a particular position function.

1. Topic Focus

Represent planar motion parametrically and with vectors, then analyze polar derivatives and areas.

This topic: Integrate components and use initial vectors to determine a particular position function.

2. Key Relationship

\(\int\langle f,g\rangle dt=\left\langle\int fdt,\int gdt\right\rangle\)

Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.

3. Visual Connection

Vector-valued motionDifferentiate or integrate components while keeping position, velocity, acceleration, and their constants distinct.

4. Worked Example

Integrate acceleration to velocity and velocity to position, applying the matching initial condition at each stage.

Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.

5. Concept Development

BC-only topic. Integration of a vector-valued function is performed component by component. If \(\mathbf F'(t)=\mathbf f(t)\), then \(\mathbf F\) is a vector antiderivative of \(\mathbf f\).

Indefinite integrals

\(\int\langle f(t),g(t)\rangle\,dt=\left\langle\int f(t)\,dt,\int g(t)\,dt\right\rangle+\mathbf C,\)

where \(\mathbf C=\langle C_1,C_2\rangle\). Each component has its own integration constant. Writing only one scalar constant would incorrectly force the two components to shift by the same amount.

Definite integrals and the Fundamental Theorem

\(\int_a^b\langle f(t),g(t)\rangle\,dt=\left\langle\int_a^b f(t)\,dt,\int_a^b g(t)\,dt\right\rangle.\)

If \(\mathbf r'(t)=\mathbf v(t)\), then

\(\int_a^b\mathbf v(t)\,dt=\mathbf r(b)-\mathbf r(a).\)

The result is a displacement vector, not total distance. Distance requires the scalar integral \(\int_a^b\|\mathbf v(t)\|dt\), developed further in Topic 9.6.

Use initial conditions directly

The cleanest way to build the particular position function from velocity is

\(\boxed{\mathbf r(t)=\mathbf r(t_0)+\int_{t_0}^{t}\mathbf v(s)\,ds}.\)

The dummy variable \(s\) prevents confusion between the upper limit \(t\) and the integration variable. This form automatically satisfies \(\mathbf r(t_0)\) and avoids solving separately for constants.

From acceleration to position

Acceleration must be integrated twice:

\(\mathbf v(t)=\mathbf v(t_0)+\int_{t_0}^{t}\mathbf a(s)\,ds,\qquad \mathbf r(t)=\mathbf r(t_0)+\int_{t_0}^{t}\mathbf v(s)\,ds.\)

An initial velocity determines the constant introduced by the first integration; an initial position determines the constant introduced by the second. Position data alone cannot determine velocity after integrating acceleration once.

Net change from a table

When velocity components are tabulated, approximate each component integral separately. For trapezoidal integration,

\(\Delta\mathbf r\approx\left\langle T[v_x],T[v_y]\right\rangle.\)

Use the actual time widths, preserve component signs, and add the resulting displacement to the initial position if a final position is requested.

Verification and units

Differentiate the proposed position to recover velocity.
If acceleration was given, differentiate velocity to recover acceleration.
Substitute every initial condition.
Check units: integrating acceleration over time gives velocity; integrating velocity gives position or displacement.

Linearity

For vector functions \(\mathbf f,\mathbf g\) and scalar \(c\), integration preserves sums and constant multiples:

\(\int(\mathbf f+\mathbf g)\,dt=\int\mathbf f\,dt+\int\mathbf g\,dt,\qquad \int c\mathbf f\,dt=c\int\mathbf f\,dt.\)

These properties follow componentwise from ordinary scalar integration.

6. Detailed Worked Example and Error Check

Example 1: An indefinite vector integral

Integrate \(\mathbf f(t)=\langle3t^2,2\cos t\rangle\):

\(\int\mathbf f(t)\,dt=\langle t^3,2\sin t\rangle+\langle C_1,C_2\rangle.\)

Differentiating returns \(\langle3t^2,2\cos t\rangle\).

Example 2: Recover position from velocity

Suppose \(\mathbf v(t)=\langle2t,3t^2\rangle\) and \(\mathbf r(1)=\langle4,-2\rangle\). Then

\(\begin{aligned}\mathbf r(t)&=\langle4,-2\rangle+\int_1^t\langle2s,3s^2\rangle ds\\&=\langle4,-2\rangle+\langle t^2-1,t^3-1\rangle\\&=\langle t^2+3,t^3-3\rangle.\end{aligned}\)

Example 3: Integrate acceleration twice

Let \(\mathbf a(t)=\langle6t,-2\rangle\), \(\mathbf v(0)=\langle1,4\rangle\), and \(\mathbf r(0)=\langle2,-1\rangle\). First,

\(\mathbf v(t)=\langle1,4\rangle+\int_0^t\langle6s,-2\rangle ds=\langle3t^2+1,4-2t\rangle.\)

Then

\(\mathbf r(t)=\langle2,-1\rangle+\int_0^t\langle3s^2+1,4-2s\rangle ds=\langle t^3+t+2,-t^2+4t-1\rangle.\)

Example 4: Initial time is not zero

Suppose \(\mathbf a(t)=\langle2,e^t\rangle\), \(\mathbf v(1)=\langle0,e\rangle\), and \(\mathbf r(1)=\langle3,0\rangle\). Then

\(\mathbf v(t)=\langle2t-2,e^t\rangle,\)

\(\mathbf r(t)=\left\langle3+(t-1)^2,e^t-e\right\rangle.\)

Substituting \(t=1\) verifies both initial vectors.

Example 5: A definite integral gives displacement

For \(\mathbf v(t)=\langle t-1,t^2-1\rangle\) on \([0,2]\),

\(\Delta\mathbf r=\int_0^2\mathbf v(t)dt=\left\langle0,\frac23\right\rangle.\)

The horizontal displacement is zero even though the particle may have moved horizontally during the interval.

Example 6: Approximate displacement from a table

At \(t=0,1,3\), suppose velocity is \(\langle2,0\rangle,\langle4,-2\rangle,\langle0,2\rangle\). Applying the trapezoidal rule to each component gives

\(\Delta\mathbf r\approx\left\langle1\left(\frac{2+4}{2}\right)+2\left(\frac{4+0}{2}\right),\;1\left(\frac{0-2}{2}\right)+2\left(\frac{-2+2}{2}\right)\right\rangle=\langle7,-1\rangle.\)

Example 7: Determine the vector constant

Given \(\mathbf r'(t)=\langle\cos t,2t\rangle\) and \(\mathbf r(0)=\langle1,-2\rangle\), a general antiderivative is

\(\mathbf r(t)=\langle\sin t,t^2\rangle+\langle C_1,C_2\rangle.\)

The initial condition gives \(\langle C_1,C_2\rangle=\langle1,-2\rangle\), so \(\mathbf r(t)=\langle\sin t+1,t^2-2\rangle\).

Example 8: Same velocity, translated paths

If \(\mathbf v(t)=\langle1,2t\rangle\), then every possible position function has the form

\(\mathbf r(t)=\langle t,t^2\rangle+\mathbf C.\)

Different initial positions select different constant vectors and translate the entire path without changing its velocity at any time.

Common errors

Using one scalar constant instead of a constant vector.
Forgetting the initial velocity when integrating acceleration.
Using initial position to determine the velocity constant.
Integrating acceleration only once when position is requested.
Confusing \(\int\mathbf v\,dt\) with \(\int\|\mathbf v\|dt\).
Dropping negative signs in a component integral.
Applying a numerical rule to vector magnitudes when displacement components are requested.
Failing to verify all initial conditions.

7. AP Reasoning Routine

Keep the parameter visible until the requested quantity is formed, track orientation and speed, and choose polar bounds from the traced region.

Identify the representation and requested quantity.
State the rule or theorem and verify its conditions.
Keep exact values until the final requested approximation.
Interpret sign, units, interval, and context.

AP Checkpoint

Integrate each vector-valued function and apply the stated initial conditions.
(a) Find \(\int\langle4t^3,e^t\rangle dt\).
(b) Given \(\mathbf v(t)=\langle2t,\cos t\rangle\) and \(\mathbf r(0)=\langle1,3\rangle\), find \(\mathbf r(t)\).
(c) Given \(\mathbf v(t)=\langle3t^2,-2\rangle\) and \(\mathbf r(1)=\langle0,4\rangle\), find \(\mathbf r(t)\).
(d) Given \(\mathbf a(t)=\langle2,6t\rangle\) and \(\mathbf v(0)=\langle-1,2\rangle\), find \(\mathbf v(t)\).
(e) Add \(\mathbf r(0)=\langle4,-3\rangle\) to part (d) and find \(\mathbf r(t)\).
(f) Given \(\mathbf a(t)=\langle2,e^t\rangle\), \(\mathbf v(1)=\langle0,e\rangle\), and \(\mathbf r(1)=\langle3,0\rangle\), find velocity and position.
(g) Find the displacement generated by \(\mathbf v(t)=\langle t^2,2t-1\rangle\) on \(0\le t\le2\).
(h) At \(t=0,1,3\), velocity is \(\langle2,0\rangle,\langle4,-2\rangle,\langle0,2\rangle\). Use the trapezoidal rule to estimate displacement. If \(\mathbf r(0)=\langle1,2\rangle\), estimate \(\mathbf r(3)\).
(i) Verify that \(\mathbf r(t)=\langle\sin t+1,t^2-2\rangle\) solves \(\mathbf r'(t)=\langle\cos t,2t\rangle\), \(\mathbf r(0)=\langle1,-2\rangle\).
(j) Explain the difference between \(\int_a^b\mathbf v(t)dt\) and \(\int_a^b\|\mathbf v(t)\|dt\).

Check the solution

(a) \(\langle t^4,e^t\rangle+\langle C_1,C_2\rangle\).
(b) \(\mathbf r(t)=\langle1,3\rangle+\int_0^t\langle2s,\cos s\rangle ds=\langle t^2+1,\sin t+3\rangle\).
(c) \(\mathbf r(t)=\langle0,4\rangle+\int_1^t\langle3s^2,-2\rangle ds=\langle t^3-1,6-2t\rangle\).
(d) \(\mathbf v(t)=\langle-1,2\rangle+\int_0^t\langle2,6s\rangle ds=\langle2t-1,3t^2+2\rangle\).
(e) \(\mathbf r(t)=\langle4,-3\rangle+\int_0^t\langle2s-1,3s^2+2\rangle ds=\langle t^2-t+4,t^3+2t-3\rangle\).
(f) \(\mathbf v(t)=\langle2t-2,e^t\rangle\) and \(\mathbf r(t)=\langle3+(t-1)^2,e^t-e\rangle\).
(g) \(\Delta\mathbf r=\int_0^2\langle t^2,2t-1\rangle dt=\langle8/3,2\rangle\).
(h) Componentwise trapezoidal integration gives \(\Delta\mathbf r\approx\langle7,-1\rangle\), so \(\mathbf r(3)\approx\langle8,1\rangle\).
(i) Differentiation gives \(\langle\cos t,2t\rangle\), and substitution at \(t=0\) gives \(\langle1,-2\rangle\), so both requirements hold.
(j) The first integral is the displacement vector and can contain cancellation in each component. The second integrates nonnegative speed and gives scalar total distance traveled.