= Solution
<Richardson extrapolation> cancels a leading discretization error by combining two mesh sizes. If $A(h)=I+ch^p+O(h^q)$ with $q>p$, then
$$
\frac{2^pA(h/2)-A(h)}{2^p-1}=I+O(h^q).
$$
Use a uniform grid $x_i=a+ih$, $0\le i\le n$, with even $n$. The composite trapezium rule is
$$
T_h=h\left[\frac{f_0+f_n}{2}+\sum_{i=1}^{n-1}f_i\right],\qquad f_i=f(x_i).
$$
The composite Simpson rule, formed by integrating a quadratic interpolant on each pair of successive panels, is
$$
S_h=\frac h3\left[f_0+f_n+4\sum_{i\text{ odd}}f_i+2\sum_{\substack{0<i<n\\i\text{ even}}}f_i\right].
$$
Assume an even-power trapezium error expansion $T_h=I+c_2h^2+O(h^4)$; continuous fourth derivative is a standard sufficient smoothness hypothesis. The coarse grid has step $2h$, so one extrapolation step is $(4T_h-T_{2h})/3$. Its endpoint coefficients are $h/3$, its odd-index coefficients are $4h/3$, and its interior even-index coefficients are $2h/3$. Therefore
$$
\boxed{S_h=\frac{4T_h-T_{2h}}3=I+O(h^4)}.
$$
The algebraic identity between the two rules holds for any sampled values; the smoothness and error assumptions are used to infer the improved accuracy.
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