A comparison of area estimates, and Simpson's rule

We have mentioned several different ways of estimating the area under a curve: left-endpoint, right-endpoint, midpoint and trapezoidal estimates. It's interesting to compare them.

Recall that we consider the function \(f(x)\) over the interval \([a,b]\) and divide the interval into \(n\) subintervals \([x_0, x_1], [x_1, x_2], \dots, [x_{n-1}, x_n]\) each of width \(\Delta x = \frac{1}{n} (b-a)\).

All of the estimates discussed give the approximate area under \(y=f(x)\) in the form of a sum of products of values of \(f\) with the width \(\Delta x\). In fact, the left-endpoint, right-endpoint and trapezoidal estimates are all of the form

\[ \big[ c_0 f(x_0) + c_1 f(x_1) + c_2 f(x_2) + \dots + c_{n-1} f(x_{n-1}) + c_n f(x_n) \big] \, \Delta x. \]

(The midpoint rule is slightly different, since it evaluates the function \(f\) at the midpoints of subintervals.) The coefficients \(c_j\) are compared in the following table.

The coefficients in the formulas for area estimates
Estimate \(c_0\) \(c_1\) \(c_2\) \(c_3\) \(c_4\) \(\dots\) \(c_{n-2}\) \(c_{n-1}\) \(c_n\)
left endpoint 1 1 1 1 1 \(\dots\) 1 1 0
right endpoint 0 1 1 1 1 \(\dots\) 1 1 1
trapezoidal \(\frac{1}{2}\) 1 1 1 1 \(\dots\) 1 1 \(\frac{1}{2}\)
Simpson \(\frac{1}{3}\) \(\frac{4}{3}\) \(\frac{2}{3}\) \(\frac{4}{3}\) \(\frac{2}{3}\) \(\dots\) \(\frac{2}{3}\) \(\frac{4}{3}\) \(\frac{1}{3}\)

There is another estimate, called Simpson's rule, which takes a slightly different set of coefficients, as shown in the table. It requires that the number of subintervals \(n\) is even. The coefficients start at \(\frac{1}{3}\), then alternate between \(\frac{4}{3}\) and \(\frac{2}{3}\), before ending again at \(\frac{1}{3}\). (It is as if every second coefficient in the list for the trapezoidal estimate donated \(\frac{1}{6}\) to its~neighbours.)

Explicitly, the estimate from Simpson's rule for the area under \(y=f(x)\) between \(x=a\) and \(x=b\) is

\[ \Big[ \frac{1}{3} f(x_0) + \frac{4}{3} f(x_1) + \frac{2}{3} f(x_2) + \frac{4}{3} f(x_3) + \frac{2}{3} f(x_4) + \dots + \frac{2}{3} f(x_{n-2}) + \frac{4}{3} f(x_{n-1}) + \frac{1}{3} f(x_n) \Big] \Delta x. \]

Although the coefficients may appear somewhat bizarre, Simpson's rule almost always gives a much more accurate answer than any of the other estimates mentioned. We'll see why in the next section.

Next page - Appendix - Exactness of area estimates