Why is it zero?

In this blog entry, which is more formal than usual, I'm going to look at an interesting integral. I'll be examining a quick way to evaluate it (quicker than the approach the University of Cambridge exam board takes and quicker than Wolfram Alpha) as well as giving an interpretation of, and explanation for, its value.

Question 4 of Paper 2 of the specimen Pre U question paper in Mathematics asks you to evaluate the following integral:

Six marks are offered and the mark scheme suggests the following approach:

First, factorise the denominator:

Then split the fraction into partial fractions:

Now integrate the two fractions:

Finally, substitute in the limits:

Which gives a final answer of 0.

This is not, however, the most efficient way of calculating the integral. You may notice that the derivative of the denominator is 2x – 2, which is 2(x – 1), which is double the numerator. So this integral is of the form f'/f, which integrates to ln(f). This works because if you differentiate ln(f) using the chain rule, you get f'/f. Since integration is the reverse of differentiation, the integral of f'/f is therefore ln(f) and we have reversed the chain rule. In this case we have to divide by 2 because we only have x – 1 and not 2(x – 1), so we only have half of the derivative.

Thus we jump straight from the statement of the question to:

Substituting in the limits gives:

which, once again, is 0.

Wolfram Alpha finds the integral using a substitution, which is always possible in cases of reversing the chain rule, but takes much more effort. (Click on the link to Wolfram Alpha, then click on "Show steps" in orange.)

Having established that the value of the integral is 0, the next question is: why? A look at the graph explains it:

The purple area is equal to the red area, and these cancel out. So this section of the graph is symmetrical. Or is it? Certainly the areas are equal, and they look symmetrical, but that's not enough to satisfy a mathematician. We need to prove it!

The graph appears to have rotational symmetry about the point (1,0). If we translate the graph by 1 unit to the left, it will have rotational symmetry about the origin. This is easier to establish.

A graph has rotational symmetry about the origin if f(-x)=–f(x).

(This is because f(-x) is the reflection of the graph in the y-axis and –f(x) is the reflection of the graph in the x-axis. If these two reflections are the same as each other, then the graph has rotational symmetry.)

Let g(x) be the original function, i.e.

Then g(x+1) translates g(x) by 1 unit to the left. We'll call this f(x), i.e.

To establish rotational symmetry we need to consider f(-x):

Which confirms that f(x) has rotational symmetry about the origin, and so g(x) has rotational symmetry about the point (1,0).

And that explains why the original integral equals 0. In fact, it explains more than that: the integral between any limits centred on 1 will be zero because the whole graph is symmetrical: