Section 8.5 Simpson rule (\(n=2\))
Take \(x_0=a\text{,}\) \(x_1=(a+b)/2\) and \(x_2=b\text{.}\) Then
\begin{equation*}
w_2 = \int_a^b\frac{x-\frac{a+b}{2}}{b-\frac{a+b}{2}}\frac{x-a}{b-a}dx = \frac{b-a}{6} = w_0
\end{equation*}
and
\begin{equation*}
w_1 = \int_a^b\frac{x-a}{\frac{a+b}{2}-a}\frac{x-b}{\frac{a+b}{2}-b}dx = 2\frac{b-a}{3}.
\end{equation*}
Hence the method formula is
\begin{equation*}
Q_2(f,[a,b]) = \frac{b-a}{6}\left[f(a)+4f(\frac{a+b}{2})+f(b)\right].
\end{equation*}
Simpson rule error. From the Newton-Cotes error formula (recall that \(n\) is even and so we need to use the remainder for \(n=3\)), the error bound is given by
\begin{equation*}
\frac{(b-a)^5}{24}\displaystyle\max_{[a,b]}|f^{(4)}(x)|\int_0^2s(s-1)(s-2)(s-3)ds.
\end{equation*}
Hence,
\begin{equation*}
\left|\int_a^bf(x)dx-Q_2(f,[a,b])\right|\leq\frac{(b-a)^5}{90}\displaystyle\max_{[a,b]}|f^{(4)}(x)|.
\end{equation*}
After the usual decomposition procedure, we get finally that
\begin{equation*}
\left|\int_a^bf(x)dx-Q_{2,h}(f,[a,b])\right|\leq h^4\frac{b-a}{90}\max_{[a,b]}|f^{(4)}(x)|.
\end{equation*}
Convergence Speed. The formula above shows that the truncation error of the Riemann sum is \(O(h^4)\text{.}\) Below we visualize this scaling law by evaluating the error in the computation of \(\int_0^2\sin(x)dx\) with \(h=2^{-j}, 1\leq j\leq10\text{.}\) Due to the high convergence speed of this method, the round-off error shows already before \(h^{-3}\) but the error, before the round-off error kicks-off, goes down to about \(10^{-15}\) (for a much larger \(h\) than it was possible with the previous methods).