The value of pi is not 22/7

Evaluate $ \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \; dx$

Then conclude that $\pi \ne \frac{22}{7}$ and explain why.

Answer:

$\frac{x^4(1-x)^4}{1+x^2} = \frac{x^4(x^4-4x^3+6x^2-4x+1)}{1+x^2} $

$= \frac{x^8 - 4x^7 + 6x^6 - 4x^5 + x^4}{1 + x^2} $

$= (x^6-4x^5+5x^4-4x^2 + 4) -4 \frac{1}{1+x^2}$

So, $ \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \; dx$

$= \int_0^1 (x^6-4x^5+5x^4-4x^2 + 4) \; dx -4 \int_0^1 \frac{1}{1+x^2} \; dx $

$= \frac{1}{7} x^7 - \frac{4}{6}x^6 + x^5 - \frac{4}{3}x^3 + 4x \biggr. \biggr|_0^1 - 4\tan^{-1} x \biggr. \biggr|_0^1 $

$= \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 (\tan^{-1} 1 - \tan^{-1} 0) $

$= \frac{22}{7} - \pi $

Since the integrand is always positive on the interval $(0,1)$, the value of the integral is not negative or zero. So, $\frac{22}{7} \gt \pi $