# Prove using integration: π is not 22/7.

Consider the infamous definite integral

\begin{equation*} I = {\int_{0}^{1}}{\frac{x^4(1-x)^4}{x^2+1}}{\,dx}. \end{equation*}

To solve this integral, we long divide the integrand.

Hence,

\begin{equation*} \begin{aligned} I &= {\int_{0}^{1}}x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{x^2+1} {\,dx}\\[3ex] &= \left [ \frac{1}{7}x^7 - \frac{2}{3}x^6 + x^5 - \frac{4}{3}x^3 + 4x - 4\arctan(x) \right ]_{0}^{1} \\[3ex] &= \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 - 4\arctan(1) \\[3ex] &= \frac{22}{7} - \pi.\end{aligned} \end{equation*}

Since the integrand is continuous everywhere and greater than zero for $x \in (0, 1)$, therefore $I > 0$. Thus, $\pi < \frac{22}{7}$.

## Error bounds

Now that we've established that $\pi \ne 22/7$, we can try to figure out how accurate this approximation really is. We do this by considering the numerator of the integrand. Since $1 \leq x^2 + 1 \leq 2$ in the interval $[0,1]$, it follows that

\begin{equation*} \frac{{x^4(1-x)^4}}{2} \leq f(x) \leq {x^4(1-x)^4}, \end{equation*}

so

\begin{equation*} {\int_{0}^{1}}\frac{{x^4(1-x)^4}}{2} {\,dx}< \frac{22}{7} - \pi < {\int_{0}^{1}}{x^4(1-x)^4}{\,dx}, \end{equation*}

and once evaluated,

\begin{equation*} \frac{1}{1260} < \frac{22}{7} - \pi < \frac{1}{630}. \end{equation*}

This tells us that 22/7 approximates π correctly to at least two digits and at most three digits. If we look at the decimal expansion of π and 22/7,

\begin{equation*} \begin{aligned} \pi &= \color{green} {3.14}1592653589 \ldots \\ \frac{22}{7} &= \color{green} {3.14}2857142857 \ldots \end{aligned} \end{equation*}

indeed we see three correct digits.

Can we in this way find similar integrals that produce even closer approximations to π?