Composite Rules

divide the interval of integration into n subintervals of equal length $h=\frac{b-a}{n}$. Then

$$\int _a^bf(x)dx=\left\{ \int _a^{a+h}+\int _{a+h}^{a+2h}+\cdots +\int _{a+(n-1)h}^b \right\} f(x)dx.$$

We now apply our quadrature rule to each subinterval.

The trapezium rule gives

$$\begin{array} {lcl}\int _a^bf(x)dx & = & \frac{h}{2}\left\{ ... ...a+(n-1)h)+f(b) \right\} -\frac{h^3}{12}f''(\xi _n),\end{array}$$

with the $\xi$'s in the usual intervals. Collecting these terms, we find

$$\int _a^bf(x)dx=\frac{h}{2}\left\{ f(a)+f(b) \right\} +h\sum _{k=1}^{n-1}f(a+kh)-\frac{h^3}{12}\sum _{k=1}^nf''(\xi _k).$$

The error term can be put in a neater form as follows:

$$\begin{array} {lcl}E_n(f) & = & -\frac{h^3}{12}\sum _{k=1}^n... ...(b-a)}{12}f''(\eta ).\quad a<\xi _1<\eta <\xi _n<b,\end{array}$$

where we have used the intermediate value theorem to replace the mean value $\bar{f}''$ in the last line.

In the composite rule we have lost a power of h, and we see that doubling the number of subintervals halves h and therefore reduces the eror by a factor of 4, assuming that $f''(\eta )$ is not altered.

To compute an integral we keep doubling the number of subintervals until successive answers differ by less than a prescribed tolerance.