Definite Integrals of Vector-Valued Functions

From Scalar Accumulation to Vector Accumulation

A scalar definite integral records accumulated signed contributions across an interval. When the quantity being accumulated has several components, each component accumulates separately, and the results combine into a vector.

For instance, suppose a vector-valued function assigns a vector to each parameter value. Its first component may describe one changing quantity, while its second and third components describe two others. Integrating the function over an interval gives the total signed accumulation in each component. The result is a vector, not a single scalar.

This componentwise viewpoint also gives a precise definition. It specifies exactly what the vector integral means, when it exists, and how to interpret its components.

Definition of a Definite Vector Integral

Let \(\mathbf{F}\) be a function from a closed interval \([a,b]\) into \(\mathbb{R}^n\), where \(a<b\). Write its components as scalar functions \(f_1,\ldots,f_n\):

\mathbf{F}(t) = \bigl(f_1(t),f_2(t),\ldots,f_n(t)\bigr).

If each component is integrable on \([a,b]\), the definite vector integral of \(\mathbf{F}\) is defined by integrating each component over the same interval.

Definition: If every \(f_j\) is integrable on \([a,b]\), then

\int_a^b \mathbf{F}(t)\,dt = \left( \int_a^b f_1(t)\,dt,\, \int_a^b f_2(t)\,dt,\, \ldots,\, \int_a^b f_n(t)\,dt \right).

The integral is a vector in \(\mathbb{R}^n\), with one scalar integral in each component.

In three dimensions, for example, if \(\mathbf{F}(t)=\langle f(t),g(t),h(t)\rangle\), then

\int_a^b \mathbf{F}(t)\,dt = \left\langle \int_a^b f(t)\,dt,\, \int_a^b g(t)\,dt,\, \int_a^b h(t)\,dt \right\rangle.

The integral exists whenever all of its component integrals exist. In particular, if \(\mathbf{F}\) is continuous on \([a,b]\), then each component is continuous there, so each scalar component is integrable and the vector integral exists. Continuity is a useful sufficient condition; it is not required by the definition if the components are integrable for some other reason.

Why the Components Can Be Integrated Separately

The componentwise definition agrees with the usual idea of adding up small vector contributions. Divide \([a,b]\) into subintervals with widths \(\Delta t_i\), and choose a sample point \(t_i^*\) in each subinterval. A vector Riemann sum is the sum of the sampled vectors multiplied by their subinterval widths.

\sum_{i=1}^{m}\mathbf{F}(t_i^*)\,\Delta t_i.

For a three-component function, write \(\mathbf{F}(t)=\langle f(t),g(t),h(t)\rangle\). Each sampled vector has components \(f(t_i^*)\), \(g(t_i^*)\), and \(h(t_i^*)\). Adding the vectors therefore adds their first components, their second components, and their third components separately:

\sum_{i=1}^{m}\mathbf{F}(t_i^*)\,\Delta t_i = \left\langle \sum_{i=1}^{m}f(t_i^*)\,\Delta t_i,\, \sum_{i=1}^{m}g(t_i^*)\,\Delta t_i,\, \sum_{i=1}^{m}h(t_i^*)\,\Delta t_i \right\rangle.

If the component functions are integrable, the scalar sums in the right-hand side approach their respective definite integrals as the subdivisions become finer. Limits of vector-valued functions are determined componentwise, so the vector sums approach the vector whose components are those scalar integral limits. This establishes why the Riemann-sum meaning of a vector integral gives precisely the componentwise definition.

Key idea: A vector Riemann sum accumulates each coordinate independently. Its limit therefore has, in each coordinate, the same value as the definite integral of that scalar component.

Worked Examples

Worked Example: Integrating a Constant Vector

Find the definite integral of \(\mathbf{F}(t)=\langle 2,-1,3\rangle\) on \([1,4]\).

The interval has width \(4-1=3\). By the definition, integrate each constant component. Each integral is the signed area of a rectangle with width \(3\) and height equal to that component.

\begin{aligned} \int_1^4 \mathbf{F}(t)\,dt &= \left\langle \int_1^4 2\,dt,\, \int_1^4 (-1)\,dt,\, \int_1^4 3\,dt \right\rangle\\ &= \left\langle 2(4-1),\, (-1)(4-1),\, 3(4-1) \right\rangle\\ &= \langle 6,-3,9\rangle. \end{aligned}

Check: The three component accumulations are \(2\cdot3=6\), \((-1)\cdot3=-3\), and \(3\cdot3=9\). They match the three coordinates of the result.

Worked Example: Accumulating Three Changing Components

Find the integral of \(\mathbf{F}(t)=\langle t,2-t,1\rangle\) on \([0,2]\), using signed areas to evaluate the scalar component integrals.

For the first component, \(y=t\) forms a triangle above the horizontal axis, with base \(2\) and height \(2\). Its area is \(\frac12(2)(2)=2\).

For the second component, \(y=2-t\) forms a triangle above the axis, with the same base and height. Its area is also \(\frac12(2)(2)=2\). The third component is the constant function \(1\), giving a rectangle of width \(2\) and height \(1\), with area \(2\).

\begin{aligned} \int_0^2 \mathbf{F}(t)\,dt &= \left\langle \int_0^2 t\,dt,\, \int_0^2 (2-t)\,dt,\, \int_0^2 1\,dt \right\rangle\\ &= \left\langle \frac12(2)(2),\, \frac12(2)(2),\, (2)(1) \right\rangle\\ &= \langle 2,2,2\rangle. \end{aligned}

Check: The first two component graphs each enclose a triangle of area \(2\), and the constant third component encloses a rectangle of area \(2\). All three entries in the calculated vector agree with these areas.

Worked Example: Positive and Negative Contributions in One Component

Find the integral of \(\mathbf{F}(t)=\langle 1-t,t,-1\rangle\) on \([0,2]\). In the first component, contributions above and below the axis must be counted with opposite signs.

The graph \(1-t\) crosses the axis at \(t=1\). On \([0,1]\), it forms a positive triangle with base \(1\) and height \(1\), so its signed area is \(\frac12\). On \([1,2]\), it forms a negative triangle with the same base and height, so its signed area is \(-\frac12\). The first component's total is therefore \(\frac12+(-\frac12)=0\).

The second component \(t\) forms a triangle above the axis with base \(2\) and height \(2\), giving area \(\frac12(2)(2)=2\). The third component is constantly \(-1\); over width \(2\), its signed rectangular area is \((-1)(2)=-2\).

\begin{aligned} \int_0^2 \mathbf{F}(t)\,dt &= \left\langle \int_0^2 (1-t)\,dt,\, \int_0^2 t\,dt,\, \int_0^2 (-1)\,dt \right\rangle\\ &= \left\langle \frac12-\frac12,\, \frac12(2)(2),\, (-1)(2) \right\rangle\\ &= \langle 0,2,-2\rangle. \end{aligned}

Check: The positive and negative triangles in the first component cancel, while the other components contribute \(2\) and \(-2\). These signed areas reproduce \(\langle 0,2,-2\rangle\).

Interpreting the Result as Accumulated Change

Each coordinate of \(\int_a^b\mathbf{F}(t)\,dt\) records the signed accumulation of the corresponding coordinate function. Positive values indicate a net contribution in the positive coordinate direction; negative values indicate a net contribution in the negative direction. A zero coordinate means that the positive and negative contributions in that component cancel in total. It does not mean that the component was zero at every parameter value, as the first component in the third example illustrates.

If the components of \(\mathbf{F}\) describe rates of change with respect to \(t\), then their integrals describe the accumulated changes in those respective quantities over the interval. For example, one component could represent accumulated change in one measured quantity and another component the change in a second. The integral collects those component changes into a single vector. The interpretation depends on what \(\mathbf{F}\) represents and on the units of its components.

The parameter interval matters: accumulation is taken over the stated values from \(a\) to \(b\), not over the geometric shape traced by the vector function. The integral records the net componentwise accumulation, so contributions of opposite signs can cancel.

A Common Mistake: Integrating the Magnitude Instead

A definite vector integral is not generally obtained by finding the magnitude of the vector at each parameter and integrating those magnitudes. The componentwise integral produces a vector; the integral of the magnitude produces a scalar. These quantities answer different questions.

For example, let \(\mathbf{F}(t)=\langle 1,1\rangle\) on \([0,1]\). The vector integral is

\int_0^1 \mathbf{F}(t)\,dt = \left\langle\int_0^1 1\,dt,\int_0^1 1\,dt\right\rangle = \langle 1,1\rangle.

A mistaken calculation would replace this vector integral with the integral of the magnitude:

\int_0^1 \|\mathbf{F}(t)\|\,dt = \int_0^1 \sqrt{1^2+1^2}\,dt = \sqrt{2}.

The scalar \(\sqrt{2}\) is not the vector \(\langle 1,1\rangle\). The correction is to integrate the coordinates separately whenever the task asks for the definite integral of a vector-valued function. Integrating magnitude is a different operation and should be used only when that scalar quantity is what the problem asks for.

Check Your Understanding

Use the componentwise definition and signed accumulation to answer these questions.

State the condition on the scalar components that guarantees the definite integral of a vector-valued function is defined by the componentwise definition.
Why does a vector Riemann sum lead to scalar Riemann sums in each coordinate?
What is \(\int_0^3\langle 1,-2\rangle\,dt\)?
For \(\mathbf{G}(t)=\langle 2-t,1\rangle\) on \([0,2]\), does the first component have positive, negative, or zero net accumulation? Explain using signed areas.
In one sentence, explain why \(\int_a^b\|\mathbf{F}(t)\|\,dt\) is not, in general, the definite vector integral of \(\mathbf{F}\).