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

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 &minus;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]^1_0 = -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.