Prove that
$$
\lim _{x \rightarrow \infty} \int_{1}^{x} \frac{1}{1+t^{2}} d t=\int_{0}^{1} \frac{1}{1+t^{2}} d t=\frac{\pi}{4}
$$
that is, $\lim _{x \rightarrow \infty} \arctan x=\arctan 1=\pi / 4$. Deduce that $2.88<\pi<3.39$.
(Hint: Substitute $t=1 / s$ and use Proposition 6.20. Divide $[0,1]$ into subintervals of length $\frac{1}{4}$.)