Difficulty: ** Impressiveness: ****

If you do A-level maths, you do an awful lot of integration. You integrate polynomials, trig functions, partial fractions, exponentials, parametric curves, products of these… and get nice analytical answers.

Here, let me provoke controversy: Unless you’re going to be a pure mathematician or a maths tutor, you will have little or no call for analytical answers outside of exams. Ninety percent of careers - conservatively - will have no use for integration; the other ten, you’ll get away with numerical integration. The odds are you’re going to want a number such as 1.755 rather than $\pi - \ln(4)$, in any case.

Depending on your board, you may learn some techniques for numerical approximation. There’s the trapezium rule, which neatly splits the integral up into… you guessed it, trapeziums, works out the total area under the curve and spits out a number. For many students, that’s the start and end of numerical integration.

Some of you will also pick up on Simpson’s rule, which is a bit neater - it fits piecewise quadratic curves to the points on the curve, and finds the area under those.

There are others - but the best and simplest one I know of is one that I’ve only come across recently: the full-width half-maximum rule. This rule is really designed for functions like the bell curve which have a single, well-defined peak, and which starts at 0 on either side.

The clue on how to work it is in the name: you work out how wide it is halfway up, and multiply that by the height at its highest point. That’s it!

So, if you wanted to estimate the integral of $\cos^2(x)$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, here’s what you’d do:

  • identify the maximum. You know what $\cos^2(x)$ looks like, it has a peak at $(0,1)$, just like $\cos(x)$.
  • Work out where the curve is halfway up: where $\cos^2(x) = \frac{1}{2}$. That gives $\cos(x) = \frac{1}{\sqrt{2}}$, or $x = \pm \frac{\pi}{4}$.
  • Figure out the width from this: it’s $2\times\frac{\pi}{4}$ or $\frac{\pi}{2}$.
  • That means the FWHM estimate is $\frac{\pi}{2} \times 1 = \frac{\pi}{2}$.

What’s the actual integral? Well, that’s a C4 secret. It works out to be… exactly $\frac{\pi}{2}$. As the man says: bazinga.