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 $h / 3$.