Simpson's rule

Remarks

$S$ is the closed Newton–Cotes formula with $n = 2$, and its degree of exactness is $3$, one more than the degree of the interpolating parabola, by symmetry. Example: for $f(x) = x^3$ on $[0, 1]$, $S(f) = \frac{1}{6}\big(0 + \frac{1}{2} + 1\big) = \frac{1}{4} = \int_0^1 x^3\, dx$. Non-example of exactness: for $f(x) = x^4$ on $[0, 1]$, $S(f) = \frac{1}{6}\big(0 + \frac{1}{4} + 1\big) = \frac{5}{24}$, while the integral is $\frac{1}{5}$; so the degree is not $4$. If $f$ has a continuous fourth derivative, then $\int_a^b f - S_N(f) = -\frac{(b - a) h^4}{180} f^{(4)}(\xi)$ for some $\xi \in [a, b]$. For $x^4$ on $[0, 1]$ with $N = 2$ this gives $-\frac{1}{16 \cdot 180} \cdot 24 = -\frac{1}{120}$, matching $\frac{1}{5} - \frac{5}{24}$.