The Bounds on an Arc-Length Average

Why an Average Has Bounds

A scalar field may vary along a curve, but its average value with respect to arc length cannot be smaller than every value it takes or larger than every value it takes. This simple fact gives a useful check on calculations: if a proposed average falls outside the field’s range on the curve, something has gone wrong.

The relevant average is weighted by distance along the curve. The parameter used to describe the curve can affect how quickly the curve is traversed, so averaging the field over a parameter interval is not generally the same as averaging it over arc length.

Definition and Assumptions

Let \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) be a piecewise continuously differentiable parametrization of a curve \(C\). Its length is \(L=\int_a^b \|\mathbf{r}'(t)\|\,dt\). Assume \(L>0\), and let \(f\) be a scalar field continuous on the image of the curve. The scalar line integral uses the arc-length element \(ds=\|\mathbf{r}'(t)\|\,dt\).

Definition: The arc-length average of \(f\) along \(C\) is the scalar line integral of \(f\) divided by the curve’s length:

f_{\mathrm{avg}} = \frac{1}{L}\int_C f\,ds = \frac{\displaystyle\int_a^b f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt} {\displaystyle\int_a^b \|\mathbf{r}'(t)\|\,dt}.

Because the parameter interval is closed and bounded, its image under the continuous parametrization is compact. The composition \(f(\mathbf{r}(t))\) is continuous, so it attains a minimum and a maximum on the curve. Write these values as \(m\) and \(M\). They describe the field’s range on the curve, not necessarily its range throughout the surrounding region.

The Bounds Theorem

Theorem: If \(f\) is continuous on a curve \(C\) of positive finite length, and \(m\) and \(M\) are the minimum and maximum values of \(f\) on \(C\), then its arc-length average satisfies

m\leq f_{\mathrm{avg}}\leq M.

The key reason is that the arc-length element is nonnegative. Every value of \(f\) along the curve lies between \(m\) and \(M\), and integrating those inequalities with respect to arc length preserves their order.

Use the pointwise bounds.
For every point on \(C\), the definitions of the minimum and maximum give \(m\leq f\leq M\).

Integrate along the curve.
Since \(ds\) is nonnegative, integrating the inequalities gives \(m\int_C ds\leq\int_C f\,ds\leq M\int_C ds\).

Divide by the length.
Since \(\int_C ds=L>0\), division by \(L\) preserves the inequalities and gives \(m\leq f_{\mathrm{avg}}\leq M\).

This argument proves the theorem directly. In particular, it requires a finite, positive curve length so that the average is defined and division by \(L\) is valid. Continuity ensures that the minimum and maximum on the curve exist; the inequality argument itself uses their pointwise bounds.

Worked Examples

Worked Example: A Field on a Line Segment

Find the arc-length average of \(f(x,y)=x^2+1\) along the segment parametrized by \(\mathbf{r}(t)=(t,0)\), \(0\leq t\leq 2\), and check it against the bounds.

Along the segment, \(f(\mathbf{r}(t))=t^2+1\). The speed is \(\|\mathbf{r}'(t)\|=\|(1,0)\|=1\), so the length is

L=\int_0^2 1\,dt=2.

On this segment, \(0\leq x\leq 2\), so \(1\leq x^2+1\leq 5\). Thus \(m=1\) and \(M=5\). The scalar line integral is

\int_C f\,ds =\int_0^2(t^2+1)(1)\,dt =\left[\frac{t^3}{3}+t\right]_0^2 =\left(\frac{8}{3}+2\right)-0 =\frac{14}{3}.

Therefore,

f_{\mathrm{avg}}=\frac{14/3}{2}=\frac{7}{3}.

The result passes the bounds check: \(1\leq 7/3\leq 5\). As a further check, multiplying the average by the length recovers the integral: \((7/3)(2)=14/3\).

Worked Example: A Field on a Semicircle

Find the arc-length average of \(f(x,y)=x^2\) along the upper semicircle of radius \(2\), parametrized by \(\mathbf{r}(t)=(2\cos t,2\sin t)\), \(0\leq t\leq\pi\).

Along the curve, \(f(\mathbf{r}(t))=4\cos^2t\). The derivative is \(\mathbf{r}'(t)=(-2\sin t,2\cos t)\), so the speed is

\|\mathbf{r}'(t)\| =\sqrt{4\sin^2t+4\cos^2t} =2.

Hence \(L=\int_0^\pi 2\,dt=2\pi\). On the semicircle, \(0\leq x^2\leq 4\), so the average must lie between \(0\) and \(4\). Compute the scalar line integral:

\int_C f\,ds =\int_0^\pi (4\cos^2t)(2)\,dt =8\int_0^\pi \cos^2t\,dt =8\left(\frac{\pi}{2}\right) =4\pi.

Thus \(f_{\mathrm{avg}}=4\pi/(2\pi)=2\), and \(0\leq 2\leq 4\), as required. The value is also consistent with the symmetry of \(x^2\) along the semicircle: it ranges from \(0\) to \(4\), and its arc-length average is halfway between those bounds.

Worked Example: A Nonuniform Parametrization of a Segment

Find the arc-length average of \(f(x,y)=x\) along \(\mathbf{r}(t)=(t^2,0)\), \(0\leq t\leq 1\). This parametrization moves more slowly near \(t=0\) than near \(t=1\), so the speed must be included.

The curve traces the segment from \((0,0)\) to \((1,0)\). Along it, \(f(\mathbf{r}(t))=t^2\), and \(\mathbf{r}'(t)=(2t,0)\), giving speed \(\|\mathbf{r}'(t)\|=2t\) on this interval. The length is

L=\int_0^1 2t\,dt=[t^2]_0^1=1.

The field ranges from \(0\) to \(1\) on the segment. Its scalar line integral and average are

\int_C f\,ds =\int_0^1(t^2)(2t)\,dt =\int_0^1 2t^3\,dt =\left[\frac{t^4}{2}\right]_0^1 =\frac{1}{2}, \qquad f_{\mathrm{avg}}=\frac{1/2}{1}=\frac{1}{2}.

The bounds check is \(0\leq 1/2\leq 1\). Multiplying by \(L=1\) also recovers the integral, \(f_{\mathrm{avg}}L=1/2\).

A Common Mistake: Averaging Over the Parameter

In the last example, a tempting but incorrect calculation would average \(f(\mathbf{r}(t))=t^2\) over the parameter interval without including the speed:

\int_0^1 t^2\,dt=\frac{1}{3}.

This is an average with respect to \(dt\), not an arc-length average. The curve does not cover equal distances in equal increments of \(t\): its speed is \(2t\). The correct arc-length average includes that speed and divides by the curve length:

\frac{\displaystyle\int_0^1 t^2(2t)\,dt} {\displaystyle\int_0^1 2t\,dt} = \frac{1/2}{1} = \frac{1}{2}.

Both \(1/3\) and \(1/2\) lie between the minimum \(0\) and maximum \(1\), so the bounds alone cannot detect this particular mistake. The bounds are a necessary check, not a substitute for using the correct arc-length element.

Using the Bounds as a Check

After computing an arc-length average, compare it with the minimum and maximum of the field along the actual curve. If a calculation gives a value outside that interval, the result cannot be correct. For instance, in the first example, any proposed average greater than \(5\) would contradict the fact that \(f\leq 5\) everywhere on the segment.

The check concerns the average, not the integral by itself. Since \(\int_C f\,ds=L f_{\mathrm{avg}}\), the corresponding integral bounds are \(mL\leq\int_C f\,ds\leq ML\). Because \(L>0\), dividing these inequalities by \(L\) gives the bounds on the average. A value inside the allowed interval is plausible, but that alone does not establish that the calculation is correct.

Key takeaway: A continuous scalar field’s arc-length average lies between its minimum and maximum on the curve. Use the speed in the arc-length integral, then use the field’s range as a check on the result.

Check Your Understanding

Use the definitions and bounds above to answer these questions.

What assumptions ensure that the arc-length average is defined and that the minimum and maximum on the curve exist?
If the field ranges from \(-2\) to \(6\) on a curve, what must be true of its arc-length average?
Why does multiplying \(m\leq f\leq M\) by \(ds\) preserve the inequality directions?
For the parametrization \(\mathbf{r}(t)=(t^2,0)\), why must the speed appear in the numerator and denominator of the arc-length average?
If a computed average lies between the minimum and maximum, does that prove the computation is correct? Explain.