You have used the trapezoid and Simpson's rules with their error bounds in Trapezoid and Simpson's Rules. Here we read those bounds as statements about order: how fast does the error shrink as the panels get thinner, and for which functions is a rule exact?
Key idea
With step $h = \frac{b - a}{n}$, the composite trapezoid rule has error $-\frac{(b - a)h^{2}}{12}f''(\xi)$: order $h^{2}$. Composite Simpson's rule has error $-\frac{(b - a)h^{4}}{180}f^{(4)}(\xi)$: order $h^{4}$. Doubling $n$ cuts the errors by about 4 and 16. Simpson's rule is exact for every cubic.
Why it works
On one panel the trapezoid rule integrates the line through the endpoints; by the interpolation error formula the gap is $\frac{f''(\xi)}{2}(x - x_i)(x - x_{i+1})$, whose integral over the panel is $-\frac{h^{3}}{12}f''$. There are $n = \frac{b - a}{h}$ panels, so the total is $-\frac{(b - a)h^{2}}{12}f''$ for some average point: one power of $h$ is lost in the sum. Simpson's rule integrates a parabola through three points, which should make it exact only for quadratics. But it is symmetric about the middle point, so it also integrates the odd function $(x - m)^{3}$ exactly (both sides give 0). Hence it is exact for cubics, and its error involves $f^{(4)}$.
Worked example: Measuring the orders
For $\int_0^1 e^{x}\,dx = e - 1$, compare errors with $n = 4, 8, 16$.
- Trapezoid: $8.94 \times 10^{-3}$, $2.24 \times 10^{-3}$, $5.59 \times 10^{-4}$: each doubling divides by $4.0$.
- Simpson: $3.70 \times 10^{-5}$, $2.33 \times 10^{-6}$, $1.46 \times 10^{-7}$: each doubling divides by $16$.
- $\log_2 4 = 2$ and $\log_2 16 = 4$: the orders, read straight from the data.
Answer: orders $2$ and $4$
Worked example: Simpson is exact for cubics
Apply Simpson's rule with $n = 2$ to $\int_0^2 x^{3}\,dx$.
- $h = 1$: $S = \frac{1}{3}[f(0) + 4f(1) + f(2)] = \frac{1}{3}[0 + 4 + 8] = 4$.
- Exact: $\frac{2^{4}}{4} = 4$. No error at all, with a parabola fitted to a cubic.
Answer: $4$
Worked example: Error for a quadratic, exactly
Find the trapezoid error for $\int_0^1 x^{2}\,dx$ with $n = 2$.
- $T_2 = \frac{1}{2}\left[\frac{0 + 1}{2} + 0.25\right] = 0.375$; the integral is $\frac{1}{3}$.
- Error $0.375 - \frac{1}{3} = \frac{1}{24}$, exactly $\frac{(b - a)h^{2}}{12}f'' = \frac{1 \cdot 0.25}{12} \cdot 2$, because $f''$ is constant.
Answer: $\frac{1}{24}$
For the same number of function values, Simpson's rule is almost always the better choice for smooth integrands. The trapezoid rule has one surprising strength: for smooth periodic functions integrated over a full period, its error falls faster than any power of $h$.
Common mistake
Thinking Simpson's rule is exact only for quadratics. It is exact for cubics too, thanks to its symmetry, and its error depends on the fourth derivative. A quick test: try Simpson on $x^{3}$ over any interval; the answer is exact.