3 The composite trapezium rule

The general idea here is to split the interval $\left[a,b\right]$ into a sequence of $N$ smaller subintervals of equal width $h=\left(b-a\right)∕N$ . Then we apply the simple trapezium rule to each of the subintervals.

Figure 7 below shows the case where $N=2$ (and $\therefore$ $h=\frac{1}{2}\left(b-a\right)$ ). To simplify notation later on we let $f_0=f\left(a\right)$ , $f_1=f\left(a+h\right)$ and $f_2=f\left(a+2h\right)=f\left(b\right)$ .

Applying the simple trapezium rule to each subinterval we get

where we remember that the width of each of the subintervals is $h$ , rather than the $b-a$ we had in the simple trapezium rule.

The next improvement will come from taking $N=3$ subintervals (Figure 8). Here $h=\frac{1}{3}\left(b-a\right)$ is smaller than in Figure 7 above and we denote $f_0=f\left(a\right)$ , $f_1=f\left(a+h\right)$ , $f_2=f\left(a+2h\right)$ and $f_3=f\left(a+3h\right)=f\left(b\right)$ . (Notice that $f_1$ and $f_2$ mean something different from what they did in the $N=2$ case.)

As Figure 8 shows, the approximation is getting closer to the grey shaded area and in this case we have

The pattern is probably becoming clear by now, but here is one more improvement. In Figure 9 $N=4$ , $h=\frac{1}{4}\left(b-a\right)$ and we denote $f_0=f\left(a\right)$ , $f_1=f\left(a+h\right)$ , $f_2=f\left(a+2h\right)$ , $f_3=f\left(a+3h\right)$ and $f_4=f\left(a+4h\right)=f\left(b\right)$ .

This leads to

We generalise this idea into the following Key Point.

Example 12

Using 4 subintervals in the composite trapezium rule, and working to 6 decimal

 places, approximate

${\int \; <!--nolimits\; -->}_0^{2}cosh\left(x\right)dx$

Solution

In this case $h=\left(2-0\right)∕4=0.5$ .

We require $cosh\left(x\right)$ evaluated at five $x$ -values and the results are tabulated below to 6 d.p.

$\begin{array}{cc}\hfill x_n\hfill & \hfill f_n=cosh\left(x_n\right)\hfill \\ ̲& ̲\\ \hfill 0\hfill & \hfill 1.000000\hfill \\ \hfill 0.5\hfill & \hfill 1.127626\hfill \\ \hfill 1\hfill & \hfill 1.543081\hfill \\ \hfill 1.5\hfill & \hfill 2.352410\hfill \\ \hfill 2\hfill & \hfill 3.762196\hfill \end{array}$

It follows that

Task!

Using 4 subintervals in the composite trapezium rule approximate

${\int \; <!--nolimits\; -->}_1^{2}ln\left(x\right)dx$

Answer

3.1 How good is the composite trapezium rule?

We can work out an upper bound on the error incurred by the composite trapezium method. Fortunately, all we have to do here is apply the method for the error in the simple rule over and over again. Let $e_T^N$ denote the error in the composite trapezium rule with $N$ subintervals. Then

This is all very well as a piece of theory, but it is awkward to use in practice. The process of working out the maximum value of $\left|f^{″}\right|$ separately in each subinterval is very time-consuming. We can obtain a more user-friendly, if less accurate, error bound by replacing each term in the last bracket above with the biggest one. Hence we obtain

$\left|e_T^N\right|\le \frac{h^3}{12}\left(N\underset{a\le x\le b}{max}\left|f^{″}\left(x\right)\right|\right)$

This upper bound can be rewritten by recalling that $Nh=b-a$ , and we now summarise the result in a Key Point.

The formula in Key Point 8 can be used to decide how many subintervals to use to guarantee a specific accuracy.

Example 13

The function $f$ is known to have a second derivative with the property that

$\left|f^{″}\left(x\right)\right|<12$

for $x$ between 0 and 4.

Using the error bound given in Key Point 8 determine how many subintervals are required so that the composite trapezium rule used to approximate

${\int \; <!--nolimits\; -->}_0^{4}f\left(x\right)dx$

can be guaranteed to be in error by less than $\frac{1}{2}\times 10^{-3}$ .

Solution

We require that

$12\times \frac{\left(b-a\right)h^2}{12}<0.0005$

that is

$4h^2<0.0005.$

This implies that $h^2<0.000125$ and therefore $h<0.0111803$ .

Now $N=\left(b-a\right)∕h=4∕h$ and it follows that

$N>357.7708$

Clearly, $N$ must be a whole number and we conclude that the smallest number of subintervals which guarantees an error smaller than 0.0005 is $N=358$ .

It is worth remembering that the error bound we are using here is a pessimistic one. We effectively use the same (worst case) value for $f^{″}\left(x\right)$ all the way through the integration interval. Odds are that fewer subintervals will give the required accuracy, but the value for $N$ we found here will guarantee a good enough approximation.

Next are two Tasks for you to try.

Task!

The function $f$ is known to have a second derivative with the property that

$\left|f^{″}\left(x\right)\right|<14$

for $x$ between $-1$ and 4.

Using the error bound given in Key Point 8 determine how many subintervals are required so that the composite trapezium rule used to approximate

${\int \; <!--nolimits\; -->}_{-1}^{4}f\left(x\right)dx$

can be guaranteed to have an error less than 0.0001.

Answer

Task!

It is given that the function $e^{-x^2∕2}$ has a second derivative that is never greater than 1 in absolute value.

1. Use this fact to determine how many subintervals are required for the composite trapezium method to deliver an approximation to

   ${\int \; <!--nolimits\; -->}_0^1\frac{1}{\sqrt{2\pi }}{e}^{-x^2∕2}dx$

   that is guaranteed to have an error less than $\frac{1}{2}\times 10^{-2}$ .

2. Find an approximation to the integral that is in error by less than $\frac{1}{2}\times 10^{-2}$ .

Answer

Answer

Example 14

Determine the minimum number of steps needed to guarantee an error not

exceeding $\pm 0.001$ , when evaluating

${\int \; <!--nolimits\; -->}_0^{1}cosh\left(x^2\right)dx$

using the trapezium rule.

Solution

$f\left(x\right)=cosh\left(x^2\right)f^{\prime }\left(x\right)=2xsinh\left(x^2\right)f^{″}\left(x\right)=2sinh\left(x^2\right)+4x^{2}cosh\left(x^2\right)$

Using the error formula in Key Point 8

$E=\left|-\frac{1}{12}{h}^2\left\{2sinh\left(x^2\right)+4x^{2}cosh\left(x^2\right)\right\}\right|x\in \left[0,1\right]$

${\left|E\right|}_{\text{max}}$ occurs when $x=1$

Task!

Determine the minimum of strips, $n$ , needed to evaluate by the trapezium rule:

${\int \; <!--nolimits\; -->}_0^{\pi ∕4}\left\{3x^2-1.5sin\left(2x\right)\right\}dx$

such that the error is guaranteed not to exceed $\pm 0.005$ .

Answer