Proof that 22/7 exceeds π


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

That the integral is positive follows from the fact that the integrand is a quotient whose numerator and denominator are both nonnegative, being sums or products of even powers of real numbers. So the integral from 0 to 1 is positive.

So therefore, to prove that 22/7 is greater than pi, it must be proven that 22/7 - pi is a positive number.

It remains to show that the integral in fact evaluates to the desired quantity:


 * $$0\,$$
 * $$<\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx$$
 * $$=\int_0^1\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}\,dx$$
 * (expanded terms in numerator)
 * $$=\int_0^1 \left(x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}\right) \,dx$$
 * (performed polynomial long division, an important aspect of formulating algebraic geometry)
 * $$=\left.\frac{x^7}{7}-\frac{2x^6}{3}+ x^5- \frac{4x^3}{3}+4x-4\arctan{x}\,\right|_0^1$$
 * (definite integration)
 * $$=\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\pi\ $$
 * (substitute one for x, then zero for x, and subtract them—arctan(1) = &pi;/4)
 * $$=\frac{22}{7}-\pi.$$
 * (addition)
 * }
 * $$=\left.\frac{x^7}{7}-\frac{2x^6}{3}+ x^5- \frac{4x^3}{3}+4x-4\arctan{x}\,\right|_0^1$$
 * (definite integration)
 * $$=\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\pi\ $$
 * (substitute one for x, then zero for x, and subtract them—arctan(1) = &pi;/4)
 * $$=\frac{22}{7}-\pi.$$
 * (addition)
 * }
 * $$=\frac{22}{7}-\pi.$$
 * (addition)
 * }
 * }