Simpson’s Rule (1/3)

Summary

Simpson’s 1/3 rule integrates a quadratic interpolant on two equal subintervals (three nodes). Composite Simpson is fourth-order accurate for smooth $f$ and exact for cubics.

Prerequisites

Trapezoidal Rule
Even number of subintervals for the composite rule

Problem Type

Approximate $\int_{a}^{b} f (x) d x$ with higher accuracy than the trapezoidal rule.

Method Definition

One panel (two subintervals): set $h = (b - a) / 2$ and nodes $a$ , $a + h$ , $b$ .

\int_{a}^{b} f (x) d x \approx \frac{h}{3} (f (a) + 4 f (a + h) + f (b)) .

Composite rule: $n$ even, $h = (b - a) / n$ , $x_{i} = a + i h$ :

\int_{a}^{b} f \approx \frac{h}{3} (f (x_{0}) + f (x_{n}) + 4 \sum_{i odd} f (x_{i}) + 2 \sum_{i even, 0 < i < n} f (x_{i})) .

Weights pattern: $1, 4, 2, 4, 2, \dots, 4, 1$ .^[1]

Assumptions / Requirements

$f$ continuous; for the classical error, $f \in C^{4} [a, b]$
Composite: $n$ even

Error / Accuracy

Single panel error involves $f^{(4)} (ξ)$ and is $O (h^{5})$ locally; composite error is $O (h^{4})$ :

E = - \frac{(b - a)}{180} h^{4} f^{(4)} (ξ)

for some $ξ \in (a, b)$ .

Worked Example

Compute $\int_{0}^{1} x^{2} d x$ with one Simpson panel.

Here $h = (1 - 0) / 2 = 1 / 2$ , nodes $0$ , $1 / 2$ , $1$ :

\int_{0}^{1} x^{2} d x \approx \frac{1 / 2}{3} (0 + 4 \cdot (\frac{1}{2})^{2} + 1) = \frac{1}{6} (0 + 4 \cdot \frac{1}{4} + 1) = \frac{1}{6} \cdot 2 = \frac{1}{3} .

Exact value is $1 / 3$ (Simpson is exact for quadratics). Using the wrong step $h = (b - a) / 3$ would incorrectly produce $2 / 9$ .

Common Failure Modes

Setting $h = (b - a) / 3$ for the three-point formula (incorrect)
Composite rule with odd $n$
Mixing 1/3 and 3/8 weight patterns

Connections

References

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