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?

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