Review question

Ref: R5642

## Solution

The graph $y=f(x)$ of a function is drawn below for $0 \le x \le 1$.

The trapezium rule is then used to estimate

$\displaystyle\int_0^1 \! f(x) \, \mathrm{d}x$

by dividing $0 \le x \le 1$ into $n$ equal intervals. The estimate calculated will equal the actual integral when

1. $n$ is a multiple of $4$;

2. $n$ is a multiple of $6$;

3. $n$ is a multiple of $8$;

4. $n$ is a multiple of $12$.

The function $f(x)$ is linear on the intervals

$0 \le x \le \frac{1}{3},$

$\frac{1}{3} \le x \le \frac{1}{2} ,$

$\frac{1}{2} \le x \le \frac{3}{4},$

$\frac{3}{4} \le x \le 1,$

and in order to use the trapezium rule, we need to divide the interval $0 \le x \le 1$ into $n$ equal intervals.

However, if the values $x = \dfrac{1}{3}, \dfrac{1}{2}, \dfrac{3}{4}$ aren’t at the end of an interval, then the trapezium rule will overestimate the value of the integral.

For example, if we took $\dfrac{1}{4} \le x \le \dfrac{1}{2}$ as one of the intervals ($n=4$), then we would be overestimating the value of the integral, because we would also be including the area of triangle A, as shown below.

Therefore we need to choose a value of $n$ such that $x = \dfrac{1}{3}, \dfrac{1}{2}, \dfrac{3}{4}$ lie at the end of intervals.

The largest such interval would be of width $\dfrac{1}{12}$, that is, $n=12$, and similarly, $n$ can be any multiple of $12$.