6.2: Composite Rules

Last updated
Save as PDF

Page ID: 96060

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

We now use our elementary formulas obtained for (6.2) to perform the integral given by \((6.1)\)

6.2.1. Trapezoidal rule

We suppose that the function \(f(x)\) is known at the \(n+1\) points labeled as \(x_{0}, x_{1}, \ldots, x_{n}\), with the endpoints given by \(x_{0}=a\) and \(x_{n}=b\). Define

\[f_{i}=f\left(x_{i}\right), \quad h_{i}=x_{i+1}-x_{i} \nonumber \]

Then the integral of (6.1) may be decomposed as

\[\begin{aligned} \int_{a}^{b} f(x) d x &=\sum_{i=0}^{n-1} \int_{x_{i}}^{x_{i+1}} f(x) d x \\ &=\sum_{i=0}^{n-1} \int_{0}^{h_{i}} f\left(x_{i}+s\right) d s \end{aligned} \nonumber \]

where the last equality arises from the change-of-variables \(s=x-x_{i}\). Applying the trapezoidal rule to the integral, we have

\[\int_{a}^{b} f(x) d x=\sum_{i=0}^{n-1} \frac{h_{i}}{2}\left(f_{i}+f_{i+1}\right) \nonumber \]

If the points are not evenly spaced, say because the data are experimental values, then the \(h_{i}\) may differ for each value of \(i\) and (6.6) is to be used directly.

However, if the points are evenly spaced, say because \(f(x)\) can be computed, we have \(h_{i}=h\), independent of \(i\). We can then define

\[x_{i}=a+i h, \quad i=0,1, \ldots, n \nonumber \]

and since the end point \(b\) satisfies \(b=a+n h\), we have

\[h=\frac{b-a}{n} \nonumber \]

The composite trapezoidal rule for evenly space points then becomes

\[\begin{aligned} \int_{a}^{b} f(x) d x &=\frac{h}{2} \sum_{i=0}^{n-1}\left(f_{i}+f_{i+1}\right) \\ &=\frac{h}{2}\left(f_{0}+2 f_{1}+\cdots+2 f_{n-1}+f_{n}\right) \end{aligned} \nonumber \]

The first and last terms have a multiple of one; all other terms have a multiple of two; and the entire sum is multiplied by \(h / 2\).

6.2.2. Simpson’s rule

We here consider the composite Simpson’s rule for evenly space points. We apply Simpson’s rule over intervals of \(2 h\), starting from \(a\) and ending at \(b\) :

\[\begin{aligned} \int_{a}^{b} f(x) d x=\frac{h}{3}\left(f_{0}+4 f_{1}+f_{2}\right)+\frac{h}{3}\left(f_{2}+4 f_{3}+f_{4}\right)+& \ldots \\ &+\frac{h}{3}\left(f_{n-2}+4 f_{n-1}+f_{n}\right) \end{aligned} \nonumber \]

Note that \(n\) must be even for this scheme to work. Combining terms, we have

\[\int_{a}^{b} f(x) d x=\frac{h}{3}\left(f_{0}+4 f_{1}+2 f_{2}+4 f_{3}+2 f_{4}+\cdots+4 f_{n-1}+f_{n}\right) \nonumber \]

The first and last terms have a multiple of one; the even indexed terms have a multiple of \(2 ;\) the odd indexed terms have a multiple of \(4 ;\) and the entire sum is multiplied by \(h / 3\).