## FANDOM

238,419 Pages

This proof uses integral calculus. If 22/7 > π, then 22/7 - π is a positive number. We claim that this integral comes out to 22/7 - π.

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

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. It remains to be shown that this integral evaluates to 22/7 - π. To wit:

 $0\,$ $<\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx$ $=\int_0^1\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}\,dx$ (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 w:polynomial long division, an important aspect of formulating w: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) = π/4) $=\frac{22}{7}-\pi.$ (addition)

Thus, 22/7 - π > 0 and it follows that 22/7 > π