Thoughts on Numerical Integration (Part 9): Midpoint rule and exploration of error analysis

Numerical integration is a standard topic in first-semester calculus. From time to time, I have received questions from students on various aspects of this topic, including:

  • Why is numerical integration necessary in the first place?
  • Where do these formulas come from (especially Simpson’s Rule)?
  • How can I do all of these formulas quickly?
  • Is there a reason why the Midpoint Rule is better than the Trapezoid Rule?
  • Is there a reason why both the Midpoint Rule and the Trapezoid Rule converge quadratically?
  • Is there a reason why Simpson’s Rule converges like the fourth power of the number of subintervals?

In this series, I hope to answer these questions. While these are standard questions in a introductory college course in numerical analysis, and full and rigorous proofs can be found on Wikipedia and Mathworld, I will approach these questions from the point of view of a bright student who is currently enrolled in calculus and hasn’t yet taken real analysis or numerical analysis.

In the previous post in this series, I discussed three different ways of numerically approximating the definite integral \displaystyle \int_a^b f(x) \, dx, the area under a curve f(x) between x=a and x=b.

In this series, we’ll choose equal-sized subintervals of the interval [a,b]. If h = (b-a)/n is the width of each subinterval so that x_k = x_0 + kh, then the integral may be approximated as

\int_a^b f(x) \, dx \approx h \left[f(x_0) + f(x_1) + \dots + f(x_{n-1}) \right] \equiv L_n

using left endpoints,

\int_a^b f(x) \, dx \approx h \left[f(x_1) + f(x_2) + \dots + f(x_n) \right] \equiv R_n

using right endpoints, and

\int_a^b f(x) \, dx \approx h \left[f(c_1) + f(c_2) + \dots + f(c_n) \right] \equiv M_n

using the midpoints of the subintervals. We have also derived the Trapezoid Rule

\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{2} [f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)] \equiv T_n

and Simpson’s Rule (if n is even)

\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{3} \left[y_0 + 4 y_1 + 2 y_2 + 4 y_3 + \dots + 2y_{n-2} + 4 y_{n-1} +  y_{n} \right] \equiv S_n.

In the previous post in this series, we saw that both the left-endpoint and right-endpoint rules have a linear rate of convergence: if twice as many subintervals are taken, then the error appears to go down by a factor of 2. If ten times as many subintervals are used, then the error should go down by a factor of 10. Let’s now explore the results of the midpoint rule applied to \displaystyle \int_1^2 x^9 \, dx = 102.3 using different numbers of subintervals. The results are summarized in the table below. The first immediate observation is that these approximations are far better than the left- and right-endpoint rule approximations! Indeed, we see that M_{10} \approx 101.3, using only ten subintervals, is a far better approximation than (from the previous post) either L_{100} = 99.8 or R_{100} \approx 104.9 using 100 subintervals! The lesson to learn: choosing a good algorithm is often far better than simply doing lots of computations.

There’s a second observation: the rate of convergence appears to be much, much faster. Indeed, the data points appear to fit a parabola very well instead of a straight line. Said another way, if twice as many subintervals are taken, then the error appears to go down by a factor of 4. If ten times as many subintervals are used, then the error should go down by a factor of 100. This illustrates quadratic convergence, as opposed to the mere linear convergence of the left- and right-endpoint rules.

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