Currently I'm seeing a numerical methods course, I have been trying to understand the topics with the following book:
G. Phillips M. and P. Taylor J., Theory and Applications of Numerical Analysis. Elsevier, 1996, pp. 167.
In this book, the authors show the derivations for Trapezoidal and Midpoint rules and with those they compose the following quadrature rule: $\newcommand{\pth}[1]{\left( #1 \right)}$ \begin{align} \int_{a}^{b} f(x) dx &\simeq \frac{1}{3}\pth{2M(h) + T(h)}\\ &\begin{split} &\simeq\frac{2}{3}\pth{h\sum_{i=0}^{N-1}f\pth{x_i + \frac{1}{2}h} + \pth{x_N - x_0}\frac{h^2}{24} f''(\xi)}\\ &~~~+\frac{1}{3}\left( \frac{h}{2}\pth{\pth{\sum_{i=0}^{N-1}f(x_{i})} + \pth{f(x_N) - f(x_0)}} - \pth{x_N - x_0} \frac{h^2}{12} f''(\xi) \right) \end{split} \end{align} They are clever explainig why the usual quadrature rule using weights is not possible to apply for some cases and they show how the most common rules (Midpoint, Trapezoidal and Simpson's) differ its results, hence they derive this quadrature rule.
I have read the book quite the time and couldn't find the error for such rule.
Thank you beforehand
Edit:
The authors also show that $$ (2M(h) + T(h))/3 = S\pth{\frac{1}{2}h} $$ So that the Simpson's rule can be obtained from the initial expression. Fact that makes us think about the answer given by Carl Christian. The consequence is that we can apply the same error estimate to that of the Simpson's rule for this quadrature rule having some considerations.
Let $h>0$ be given. We begin by consider the subinterval $I_h = [-h, h]$. Here the midpoint method takes the form $$M_h(f) = 2h f(0)$$ and the trapezoidal rule takes the form $$T_h(f) = h (f(-h) + f(h)).$$ The new method is given by $$S_h(f) = \frac{1}{3} \Big[ 2 M_h(f) + T_h(f) \Big] = \frac{h}{3}\Big[f(-h) + 4 f(0) + f(h)\Big].$$ At this point we recognize that the new rule is the celebrated Simpson rule and no further analysis is required.