22/7, an overestimate

Mathematics NuclearPlane787 6 min 11.2
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The first approximation anyone learns is $22/7 = 3.142857\ldots$ It errs high, and the proof of that fact fits in one display line:

$$0 \;<\; \int_0^1 \frac{x^4(1-x)^4}{1+x^2}\,dx \;=\; \frac{22}{7}-\pi.$$

Positivity is immediate: the integrand is continuous, nonnegative on $[0,1]$, and not identically zero, so the integral is strictly positive.1 The evaluation is nothing deeper than polynomial division:

1
D. P. Dalzell (1944). The integral was later set as a Putnam problem, where it acquired its fame.
$$\frac{x^4(1-x)^4}{1+x^2} \;=\; x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2},$$

and integrating term by term over $[0,1]$ gives $\tfrac17-\tfrac23+1-\tfrac43+4-\pi=\tfrac{22}{7}-\pi$. That the answer lands exactly on the schoolroom fraction is the charm of the thing.

How wrong is 22/7

The same integrand measures its own error. On $[0,1]$ we have $1\le 1+x^2\le 2$, so the integral is squeezed between half and all of

$$\int_0^1 x^4(1-x)^4\,dx \;=\; B(5,5) \;=\; \frac{4!\,4!}{9!} \;=\; \frac{1}{630},$$

which yields

$$\frac{1}{1260} \;<\; \frac{22}{7}-\pi \;<\; \frac{1}{630}.$$

So $22/7$ overshoots by about $0.00126$ — wrong in the fourth decimal, exactly as observed. Archimedes, squeezing the circle between 96-gons, already had $223/71 < \pi < 22/7$; the integral turns the upper half of his sandwich into a one-line calculus exercise.

The trick generalizes. Heavier integrands of the same family witness the next famous convergent, $355/113$, with its astonishing seven-digit accuracy.2 The digits of $\pi$ start at this article’s own address: the catalog places each constant at its leading digits, and $3.141\ldots$ lives at ℵ3141.

2
S. K. Lucas (2005) exhibits integrals of the same shape for $355/113 - \pi$ and beyond.