Simpson's Rule for Arc Length

Why Simpson’s Rule Applies to Arc Length

Arc length is often expressed as an integral whose integrand is a square root involving derivatives. Even when the curve is smooth, that integral may not have a convenient elementary antiderivative. Simpson’s rule gives a systematic way to approximate it using function values at evenly spaced points.

The rule is useful only when its mesh and weights are set up correctly. In particular, the standard composite formula uses an even number of subintervals. The function values at the endpoints, odd-indexed interior points, and even-indexed interior points receive different weights.

Definition: For a graph \(y=f(x)\) on \([a,b]\), its arc-length integrand is \(g(x)=\sqrt{1+(f'(x))^2}\), and its length is \(\int_a^b g(x)\,dx\), when the derivative is continuous on the interval. For a parametrized curve \(x=x(t), y=y(t)\) on \([a,b]\), the speed integrand is \(g(t)=\sqrt{(x'(t))^2+(y'(t))^2}\), and its length is \(\int_a^b g(t)\,dt\), when the coordinate functions are continuously differentiable.

Once the appropriate integrand has been formed, the numerical procedure is the same for either kind of curve. Simpson’s rule approximates the integral of that integrand; it does not change the underlying arc-length formula.

The Composite Simpson Formula

Divide \([a,b]\) into \(n\) equal subintervals, where \(n\) is a positive even integer. Let \(h=(b-a)/n\) be the step size, and let \(x_i=a+ih\) for \(i=0,1,\ldots,n\). For an integrand \(g\), write \(g_i=g(x_i)\). The composite Simpson estimate is

S_n=\frac{h}{3}\left[g_0+g_n +4\sum_{\substack{i=1\\ i\ \mathrm{odd}}}^{n-1}g_i +2\sum_{\substack{i=2\\ i\ \mathrm{even}}}^{n-2}g_i\right].

The endpoint values have weight \(1\). Interior values with odd indices have weight \(4\), while interior values with even indices have weight \(2\). The sums are omitted if their index ranges are empty. For \(n=2\), for example, there is one odd-indexed point and no even-indexed interior point.

The even-subinterval condition matters: Simpson’s rule fits a quadratic through three consecutive points, so one application covers two subintervals. The composite formula joins such pairs across the entire interval. Thus \(n=4\) uses two panels, \(n=6\) uses three, and so on.

Why the Weights Are \(1,4,1\)

Consider one panel from \(u\) to \(u+2h\), with equally spaced points \(u\), \(u+h\), and \(u+2h\). Set \(z=(x-u)/h\), so these points correspond to \(z=0,1,2\). The quadratic interpolating the three function values \(g_0,g_1,g_2\) is

p(z)=g_0\frac{(z-1)(z-2)}{2} -g_1z(z-2) +g_2\frac{z(z-1)}{2}.

Since \(dx=h\,dz\), integrate this polynomial from \(z=0\) to \(z=2\). The integral of the first basis polynomial is \(1/3\); that of the middle basis polynomial is \(4/3\); and that of the last basis polynomial is \(1/3\). Therefore, the integral of the interpolating polynomial is \(h(g_0+4g_1+g_2)/3\). This gives the Simpson formula on one panel.

For the composite rule, apply that panel formula to each pair of subintervals and add the results. Each point between panels is an endpoint of one panel and appears in the neighboring panel as well. Combining its contributions gives weight \(2\) for even-indexed interior points; odd-indexed points occur only as panel midpoints and have weight \(4\). The two outer endpoints occur once each, with weight \(1\). This establishes the displayed composite formula.

Why the method works: Simpson’s rule integrates the quadratic interpolant through the values on each two-subinterval panel. It is exact whenever the integrand itself is a polynomial of degree at most two on each panel, because then that interpolant equals the integrand.

Worked Example: A Graph with Two Mesh Sizes

Worked Example: Comparing Two Simpson Estimates for a Graph

Approximate the length of \(y=x^2/2\) on \([0,1]\) first with \(n=2\), then with \(n=4\), and compare the estimates.

Differentiate the graph: \(y'=x\). Its arc-length integrand is \(g(x)=\sqrt{1+x^2}\).

For \(n=2\), \(h=(1-0)/2=1/2\), so the points are \(x_0=0\), \(x_1=1/2\), and \(x_2=1\). The function values are

g_0=1,\qquad g_1=\sqrt{1+\frac14}=\frac{\sqrt5}{2},\qquad g_2=\sqrt2.

Substitute these into the one-panel formula:

S_2=\frac{1/2}{3} \left(1+4\frac{\sqrt5}{2}+\sqrt2\right) =\frac{1+2\sqrt5+\sqrt2}{6} \approx 1.147724920.

For \(n=4\), \(h=1/4\), and the points are \(0,1/4,1/2,3/4,1\). Their corresponding values, in order, are

1,\quad \frac{\sqrt{17}}{4},\quad \frac{\sqrt5}{2},\quad \frac54,\quad \sqrt2.

The odd-indexed values are those at \(1/4\) and \(3/4\); the even-indexed interior value is at \(1/2\). Thus

\begin{aligned} S_4 &=\frac{1/4}{3}\left[ 1+\sqrt2 +4\left(\frac{\sqrt{17}}{4}+\frac54\right) +2\left(\frac{\sqrt5}{2}\right)\right]\\ &=\frac{6+\sqrt2+\sqrt5+\sqrt{17}}{12} \approx 1.147782264. \end{aligned}

The finer mesh estimate is about \(0.000057344\) greater than the \(n=2\) estimate. Both values are approximations to the same arc length; the comparison alone does not certify the error in either value. The mesh and weights can be checked directly: there are four subintervals, the endpoints have weight \(1\), the two odd-indexed interior points have weight \(4\), and the one even-indexed interior point has weight \(2\).

Worked Example: A Parametric Curve

Worked Example: Simpson’s Rule for a Parametric Length

Approximate the length of the curve \(x=t-t^3/3,\ y=t^2\) for \(0\le t\le1\), using \(n=2\).

Differentiate both coordinates: \(x'(t)=1-t^2\) and \(y'(t)=2t\). The speed is

\begin{aligned} g(t) &=\sqrt{(1-t^2)^2+(2t)^2}\\ &=\sqrt{1-2t^2+t^4+4t^2}\\ &=\sqrt{1+2t^2+t^4} =\sqrt{(1+t^2)^2} =1+t^2. \end{aligned}

The last equality holds because \(1+t^2\) is positive. With \(n=2\), the step is \(h=1/2\), and the parameter points are \(0,1/2,1\). The speed values are \(g(0)=1\), \(g(1/2)=1+1/4=5/4\), and \(g(1)=2\). Therefore,

S_2=\frac{1/2}{3}\left(1+4\cdot\frac54+2\right) =\frac16(1+5+2) =\frac86 =\frac43.

This estimate is exact: the speed \(1+t^2\) is a quadratic, so its interpolating polynomial on the panel is the speed itself. As a direct check, an antiderivative is \(t+t^3/3\), whose endpoint difference is \((1+1/3)-0=4/3\), matching the Simpson calculation.

Worked Example: Organizing a Graph Mesh

Worked Example: A Cubic Graph with Two Subintervals

Approximate the length of \(y=x^3\) on \([0,1]\) with \(n=2\).

Here \(y'=3x^2\), so the integrand is \(g(x)=\sqrt{1+9x^4}\). The step is \(h=1/2\), and the three points are \(0,1/2,1\). Evaluate each value:

g(0)=1,\qquad g\left(\frac12\right) =\sqrt{1+\frac9{16}}=\frac54,\qquad g(1)=\sqrt{10}.

There is one midpoint, with weight \(4\), and the two endpoints have weight \(1\). Hence

S_2=\frac{1/2}{3}\left(1+4\cdot\frac54+\sqrt{10}\right) =\frac{6+\sqrt{10}}{6} \approx 1.527046276.

The arithmetic can be checked from the weighted sum: \(1+4(5/4)+\sqrt{10}=6+\sqrt{10}\), and the prefactor is \(1/6\). This is a numerical estimate, not an exact evaluation of the arc-length integral.

A Common Panel-Count Mistake

A frequent error is to apply a Simpson-like weighting pattern when the number of subintervals is odd. For example, take three subintervals on \([0,1]\), with step \(h=1/3\) and values \(g_0,g_1,g_2,g_3\). The expression

\frac{h}{3}(g_0+4g_1+2g_2+g_3)

is not the composite Simpson formula for those three subintervals. Its weights are \(1,4,2,1\), which sum to \(8\), not \(9\). For the constant function \(g(x)=1\), it gives \(8/9\), rather than the integral \(1\), so it is not a valid composite Simpson formula. Three subintervals cannot be grouped entirely into two-subinterval panels.

The correction is to choose an even number of equal subintervals, such as \(n=2\) or \(n=4\), and then use the corresponding composite weights. Do not try to repair an odd-panel calculation by changing a single weight or retaining the same prefactor.

Using the Estimates Carefully

For either a graph or a parametrized curve, the central practical task is to evaluate the correct integrand before making the table. A graph uses \(\sqrt{1+(f')^2}\); a parametrized curve uses \(\sqrt{(x')^2+(y')^2}\). Confusing these formulas or omitting one squared derivative changes the length being approximated.

After forming the integrand, list the mesh points in order and calculate every value at those points. Mark the index of each value before applying the weights. This is especially helpful when \(n\) is larger: endpoints receive \(1\), odd interior indices receive \(4\), and even interior indices receive \(2\). Then multiply the complete weighted sum by \(h/3\).

Comparing two mesh sizes, as in the first example, is a useful way to see how an estimate changes when the spacing is reduced. It is not by itself an error bound or a proof that either estimate is closer to the true length. The estimates should be reported with sensible rounding, and the number of digits shown should not be mistaken for a guarantee of accuracy.

Check Your Understanding

Use the definitions and formula above to answer these questions.

What integrand should be used to approximate the length of a graph \(y=f(x)\)?
Why must the number of subintervals in the standard composite Simpson formula be even?
For \(n=6\), which interior indices receive weight \(4\), and which receive weight \(2\)?
For a parametrized curve, what quantities are squared and added inside the speed integrand?
Why does a comparison between two Simpson estimates not, by itself, certify the error in either estimate?

What You'll Learn