6.2: Composite Rules

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 \(h / 3\).