Review question

# Can we apply the trapezium rule to $f(x) = x/2 - \lfloor x/2 \rfloor$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8140

## Solution

For a real number $x$, we denote by $\lfloor x \rfloor$ the largest integer less than or equal to $x$. Let

$f(x) =\dfrac{x}{2}-\left\lfloor \dfrac{x}{2} \right\rfloor.$

The smallest number of equal width strips for which the trapezium rule produces an overestimate for the integral

$\int_0^5 f(x) \:dx$

is

(a) $2,\quad$ (b) $3,\quad$ (c) $4,\quad$ (d) $5,\quad$ (e) it never produces an overestimate.

If we sketch the graph for $0 \leq x \leq 5$, we find that it takes a saw-tooth shape.

It’s easy to calculate the integral; it is the area of the three triangles shown, which is $\dfrac{9}{4}$.

What happens if we apply the trapezium rule with two strips?

The trapezium rule returns the area of the blue triangle, which is $\dfrac{5}{4} < \dfrac{9}{4}$, an underestimate.

What happens with three strips?

By the trapezium rule, the shaded area now is $\dfrac{1}{2}\times\dfrac{5}{3}\left[0+\dfrac{1}{2}+2\left(\dfrac{5}{6}+ \dfrac{2}{3}\right)\right] = \dfrac{105}{36}> \dfrac{9}{4}$, and so the answer is (b).

If we apply the trapezium rule with $5n$ strips, where $n$ is a positive integer, the result agrees exactly with the given integral.