# 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
$$x^{4}-4x^{5}+6x^{6}-4x^{7}+x^{8}$$

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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