Midpoint Rule

Summary

The composite midpoint rule approximates $\int_{a}^{b} f (x) d x$ by summing rectangle areas with heights evaluated at subinterval midpoints. It is a second-order method for smooth integrands.

Prerequisites

Definite integrals and partitions of $[a, b]$
Boundedness of $f^{″}$ for the error formula

Problem Type

Approximate $I = \int_{a}^{b} f (x) d x$ from samples of $f$ .

Method Definition

Partition $[a, b]$ into $n$ equal parts of width $h = (b - a) / n$ . On $[x_{i - 1}, x_{i}]$ with midpoint $m_{i} = x_{i - 1} + h / 2$ ,

\int_{x_{i - 1}}^{x_{i}} f (x) d x \approx h f (m_{i}) .

Composite rule:^[1]

\int_{a}^{b} f (x) d x \approx h \sum_{i = 1}^{n} f (a + (i - \frac{1}{2}) h) .

Assumptions / Requirements

$f$ continuous on $[a, b]$ (integrable)
For the standard error bound, $f \in C^{2} [a, b]$

Algorithm

Choose $n$ , set $h = (b - a) / n$ .
Sum $h f (a + (i - 1 / 2) h)$ for $i = 1, \dots, n$ .

Error / Accuracy

If $| f^{″} (x) | \leq M$ on $[a, b]$ , the composite midpoint error satisfies

| E | \leq \frac{(b - a)}{24} h^{2} M = \frac{(b - a)^{3}}{24 n^{2}} M .

Order $O (h^{2})$ , same order as the composite trapezoidal rule (with a smaller constant for many $f$ ).

Worked Example

Approximate $\int_{0}^{1} x^{2} d x = 1 / 3$ with $n = 2$ ( $h = 1 / 2$ ).

Midpoints: $1 / 4$ , $3 / 4$ .

I \approx \frac{1}{2} ((1 / 4)^{2} + (3 / 4)^{2}) = \frac{1}{2} (\frac{1}{16} + \frac{9}{16}) = \frac{5}{16} = 0.3125 .

Error $| 1 / 3 - 5 / 16 | = 1 / 48 \approx 0.0208$ . Bound with $f^{″} = 2$ , $M = 2$ :

\frac{(1)^{3}}{24 \cdot 4} \cdot 2 = \frac{1}{48},

sharp on this example.

Common Failure Modes

Using endpoint samples instead of midpoints
Applying the $O (h^{2})$ bound when $f^{″}$ is unbounded near an endpoint

Connections

References

Burden & Faires, Numerical Analysis, midpoint / rectangle rules; NIST DLMF Ch. 3, https://dlmf.nist.gov/3 ↩︎