Pi (mathematical constant)/Proofs/Student level proof that 22 over 7 exceeds π

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We work out the following integral:

$I\equiv \int _{0}^{1}{\frac {t^{4}(1-t)^{4}}{1+t^{2}}}\,\mathrm {d} t\qquad \qquad \qquad (1)$

It is possible to divide polynomials in a manner that is analogous to long division of decimal numbers. By doing this it, can be shown that

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

where −4 is the remainder of the polynomial division.

Using:

$\int _{0}^{1}t^{n}\,\mathrm {d} t={\frac {1}{n+1}}$

for n=6, 5, 4, 2, and 0, respectively, one obtains

$\int _{0}^{1}(t^{6}-4t^{5}+5t^{4}-4t^{2}+4)\,\mathrm {d} t={\frac {1}{7}}-{\frac {4}{6}}+{\frac {5}{5}}-{\frac {4}{3}}+{\frac {4}{1}}={\frac {22}{7}}$

The following holds

$-4\int _{0}^{1}{\frac {1}{1+t^{2}}}\,\mathrm {d} t=-4\left[\arctan(t)\right]_{0}^{1}=-4{\frac {\pi }{4}}=-\pi$

The latter integral is easily evaluated by making the substitution

$t=\tan(x)\,$

Hence

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

The integrand (expression under the integral) of the integral in equation (1) is everywhere positive on the integration interval [0, 1] and, remembering that an integral can be defined as a sum of integrand values, it follows that $I\,$ is positive. Finally,

${\frac {22}{7}}-\pi >0\quad \Longrightarrow \quad {\frac {22}{7}}>\pi$

which was to be proved.