6.1: Elementary Formulas
We first consider integration from 0 to \(h\) , with \(h\) small, to serve as the building blocks for integration over larger domains. We here define \(I_{h}\) as the following integral:
\[I_{h}=\int_{0}^{h} f(x) d x . \nonumber \]
To perform this integral, we consider a Taylor series expansion of \(f(x)\) about the value \(x=h / 2\) :
\[\begin{aligned} f(x)=f(h / 2)+(x-h / 2) f^{\prime} &(h / 2)+\frac{(x-h / 2)^{2}}{2} f^{\prime \prime}(h / 2) \\ &+\frac{(x-h / 2)^{3}}{6} f^{\prime \prime \prime}(h / 2)+\frac{(x-h / 2)^{4}}{24} f^{\prime \prime \prime \prime}(h / 2)+\ldots \end{aligned} \nonumber \]
6.1.1. Midpoint rule
The midpoint rule makes use of only the first term in the Taylor series expansion. Here, we will determine the error in this approximation. Integrating,
\[\begin{aligned} I_{h}=h f(h / 2)+\int_{0}^{h}(&(x-h / 2) f^{\prime}(h / 2)+\frac{(x-h / 2)^{2}}{2} f^{\prime \prime}(h / 2) \\ &\left.+\frac{(x-h / 2)^{3}}{6} f^{\prime \prime \prime}(h / 2)+\frac{(x-h / 2)^{4}}{24} f^{\prime \prime \prime \prime}(h / 2)+\ldots\right) d x . \end{aligned} \nonumber \]
Changing variables by letting \(y=x-h / 2\) and \(d y=d x\) , and simplifying the integral depending on whether the integrand is even or odd, we have
\[\begin{aligned} I_{h}=& h f(h / 2) \\ &+\int_{-h / 2}^{h / 2}\left(y f^{\prime}(h / 2)+\frac{y^{2}}{2} f^{\prime \prime}(h / 2)+\frac{y^{3}}{6} f^{\prime \prime \prime}(h / 2)+\frac{y^{4}}{24} f^{\prime \prime \prime \prime}(h / 2)+\ldots\right) d y \\ =& h f(h / 2)+\int_{0}^{h / 2}\left(y^{2} f^{\prime \prime}(h / 2)+\frac{y^{4}}{12} f^{\prime \prime \prime \prime}(h / 2)+\ldots\right) d y \end{aligned} \nonumber \]
The integrals that we need here are
\[\int_{0}^{\frac{h}{2}} y^{2} d y=\frac{h^{3}}{24}, \quad \int_{0}^{\frac{h}{2}} y^{4} d y=\frac{h^{5}}{160} \nonumber \]
Therefore,
\[I_{h}=h f(h / 2)+\frac{h^{3}}{24} f^{\prime \prime}(h / 2)+\frac{h^{5}}{1920} f^{\prime \prime \prime \prime}(h / 2)+\ldots \nonumber \]
6.1.2. Trapezoidal rule
From the Taylor series expansion of \(f(x)\) about \(x=h / 2\) , we have
\[f(0)=f(h / 2)-\frac{h}{2} f^{\prime}(h / 2)+\frac{h^{2}}{8} f^{\prime \prime}(h / 2)-\frac{h^{3}}{48} f^{\prime \prime \prime}(h / 2)+\frac{h^{4}}{384} f^{\prime \prime \prime \prime}(h / 2)+\ldots, \nonumber \]
and
\[f(h)=f(h / 2)+\frac{h}{2} f^{\prime}(h / 2)+\frac{h^{2}}{8} f^{\prime \prime}(h / 2)+\frac{h^{3}}{48} f^{\prime \prime \prime}(h / 2)+\frac{h^{4}}{384} f^{\prime \prime \prime \prime}(h / 2)+\ldots \ldots \nonumber \]
Adding and multiplying by \(h / 2\) we obtain
\[\frac{h}{2}(f(0)+f(h))=h f(h / 2)+\frac{h^{3}}{8} f^{\prime \prime}(h / 2)+\frac{h^{5}}{384} f^{\prime \prime \prime \prime}(h / 2)+\ldots . \nonumber \]
We now substitute for the first term on the right-hand-side using the midpoint rule formula:
\[\begin{array}{r} \frac{h}{2}(f(0)+f(h))=\left(I_{h}-\frac{h^{3}}{24} f^{\prime \prime}(h / 2)-\frac{h^{5}}{1920} f^{\prime \prime \prime \prime}(h / 2)\right) \\ +\frac{h^{3}}{8} f^{\prime \prime}(h / 2)+\frac{h^{5}}{384} f^{\prime \prime \prime \prime}(h / 2)+\ldots, \end{array} \nonumber \]
and solving for \(I_{h}\) , we find
\[I_{h}=\frac{h}{2}(f(0)+f(h))-\frac{h^{3}}{12} f^{\prime \prime}(h / 2)-\frac{h^{5}}{480} f^{\prime \prime \prime \prime}(h / 2)+\ldots \nonumber \]
6.1.3. Simpson’s rule
To obtain Simpson’s rule, we combine the midpoint and trapezoidal rule to eliminate the error term proportional to \(h^{3}\) . Multiplying (6.3) by two and adding to (6.4), we obtain
\[3 I_{h}=h\left(2 f(h / 2)+\frac{1}{2}(f(0)+f(h))\right)+h^{5}\left(\frac{2}{1920}-\frac{1}{480}\right) f^{\prime \prime \prime \prime}(h / 2)+\ldots, \nonumber \]
or
\[I_{h}=\frac{h}{6}(f(0)+4 f(h / 2)+f(h))-\frac{h^{5}}{2880} f^{\prime \prime \prime \prime}(h / 2)+\ldots . \nonumber \]
Usually, Simpson’s rule is written by considering the three consecutive points \(0, h\) and \(2 h\) . Substituting \(h \rightarrow 2 h\) , we obtain the standard result
\[I_{2 h}=\frac{h}{3}(f(0)+4 f(h)+f(2 h))-\frac{h^{5}}{90} f^{\prime \prime \prime \prime}(h)+\ldots \nonumber \]