Simpson’s Rule (3/8)

Summary

Simpson’s 3/8 rule integrates a cubic interpolant on three equal subintervals (four nodes). Composite use requires the number of subintervals to be a multiple of 3.

Prerequisites

Simpson 1-3 Rule
Equally spaced nodes

Problem Type

Approximate $\int_{a}^{b} f (x) d x$ , often combining 3/8 panels with 1/3 panels when $n mod 6$ constraints appear.

Method Definition

One 3/8 panel: $n = 3$ subintervals, $h = (b - a) / 3$ , nodes $x_{0}, x_{1}, x_{2}, x_{3}$ .

\int_{a}^{b} f (x) d x \approx \frac{3 h}{8} (f (x_{0}) + 3 f (x_{1}) + 3 f (x_{2}) + f (x_{3})) .

Composite ( $n$ multiple of 3): weights follow the repeating pattern

1, 3, 3, 2, 3, 3, 2, \dots, 3, 3, 1

scaled by $3 h / 8$ .^[1]

Assumptions / Requirements

$f$ continuous; classical error analysis assumes sufficient smoothness
$n$ divisible by 3 for a pure composite 3/8 rule

Error / Accuracy

Single 3/8 panel error is $O (h^{5})$ involving $f^{(4)}$ , same order family as Simpson 1/3; constants differ. For smooth $f$ , 1/3 is usually preferred panel-for-panel, while 3/8 is useful for fitting $n mod 3$ constraints.

Worked Example

Integrate $f (x) = x^{2}$ on $[0, 3]$ with one 3/8 panel ( $h = 1$ ).

Nodes: $0, 1, 2, 3$ with values $0, 1, 4, 9$ .

\int_{0}^{3} x^{2} d x \approx \frac{3 \cdot 1}{8} (0 + 3 \cdot 1 + 3 \cdot 4 + 9) = \frac{3}{8} (0 + 3 + 12 + 9) = \frac{3}{8} \cdot 24 = 9.

Exact integral: $[x^{3} / 3]_{0}^{3} = 9$ . Exact again for a quadratic.

Counterexample of a wrong formula: applying trapezoidal-style weights $h / 3 (y_{0} + 4 \sum y_{i} + y_{n})$ with $h = 1$ on $[0, 4]$ is not Simpson 3/8 and need not be accurate; always use $3 h / 8$ with pattern $1, 3, 3, 1$ per block of three subintervals.

Common Failure Modes

Using coefficient $h / 3$ instead of $3 h / 8$
$n$ not a multiple of 3
Confusing with composite Simpson 1/3 weights $1, 4, 2, 4, \dots$

Connections

References

Burden & Faires, Numerical Analysis, Simpson’s 3/8 rule; NIST DLMF Ch. 3, https://dlmf.nist.gov/3 ↩︎