A Neat Pi Approximation Day Integral!

Mathematics and dates (not the romantic kind) tend to yield in fun number play. March 14th, written as 3/14, is considered to be “Pi Day”, while July 22nd, written as, 22/7 is “Pi Approximation Day”.

We have this nice integral from @daveinstpaul

In case you can’t see the image:
$$\pi + \int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1 + x^{2}}\ dx = \frac{22}{7}$$

Shall we solve it?

Let’s tackle the integral one step at a time.

$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1 + x^{2}}\ dx$$

First, let’s expand the numerator of the integrand.
$$x^{4}(1-x)^{4} = x^{4}(1-4x + 6x^{2}-4x^{3}+x^{4})$$

and this gives

now this expression is being divided by \(\frac{1}{1+x^{2}}\) which means we have to do get through some polynomial division.

I don’t like polynomial division. I like doing it this way.
$$x^{4} = (x^{2} + 1)Q(x) + R(x)$$ where \(Q(x)\) is constructed in such a way to best match the left hand side and \(R(x)\) is the remainder such that \(\mbox{deg}(R(x)) < \mbox{deg}(x^{2}+1)\). In this case, \(Q(x)\)’s construction works like so: the greatest power function that gets us to \(x^{4}\) is \(x^{2}\), yielding \(x^{4} + x^{2}\). To negate the spurious \(x^{2}\) we need an additional term of \(-1\), giving \(x^{4} + x^{2} – x^{2} – 1\). Thus the whole part (\(Q(x)\)) is \(x^{2} – 1\) and the remainder, \(R(x)\) is \(1\). As such, \(\frac{x^{4}}{x^{2} + 1} = x^{2}-1 + \frac{1}{x^{2} + 1}\).

Following this process (which is just polynomial division without all the long messy steps) and collecting all the whole parts and remainders, we have
$$\int_{0}^{1}x^{6} – 4x^{5} + 5x^{4} – 4x^{2} + 4 \ dx + \int_{0}^{1}\frac{-4}{x^{2} + 1}\ dx$$

The left integral works out to \(\frac{1}{7} – \frac{2}{3} + 1 – \frac{4}{3} + 4\) which is, huzzah! \(\frac{22}{7}\) and the integral on the right is just \(-4\arctan(x) \Big|_{0}^{1}\), which evaluates to \(-\pi\).

Putting it all together from the original statement, we have $$\pi + \underbrace{-\pi + \frac{22}{7}}_{\mbox{the integral}} = \frac{22}{7}$$

Great find by David Radcliffe who also supplies a reference!

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