Use max-min inequality to find upper and lower bounds for the value of $\int_0^1 \frac{1}{1+x^2}dx$
The answer is $$0.5\leq \int_0^1 \frac{1}{1+x^2}dx \leq 1$$
But when same integral's upper bound and lower bounds were found out using limits as $0 – 0.5$ and $0.5 – 1$ the result is different as shown below
$$0.4\leq \int_0^{0.5} \frac{1}{1+x^2}dx \leq 0.5\hspace{25pt}\text{ and} \hspace{25pt}0.25\leq \int_{0.5}^1 \frac{1}{1+x^2}dx \leq 0.4$$
adding both we get
$$0.65\leq \int_0^1 \frac{1}{1+x^2}dx \leq 0.9$$
Why splitting the integral into two gives better approximation?
Best Answer
On the interval $[0,1]$ the function $f(x)=\frac{1}{x^2+1}$ smoothly (in a Lipschitz-continuous way) goes from the value $1$ to the value $\frac{1}{2}$. If we approximate $\int_{0}^{1}f(x)\,dx$ through Riemann sums, it is not surprising that better bounds come from partitioning the integration range in many parts. We may also notice that:
$$ I = \int_{0}^{1}\frac{dx}{1+x^2} = \int_{0}^{1/2}\left(\frac{1}{1+x^2}+\frac{1}{1+(1-x)^2}\right)\,dx $$ where the new integrand function has a smaller variation on its integration range than the original one, so it is better suited for estimating the integral through Riemann sums. Additionally, it is a concave function, so the approximations coming from the rectangle/trapezoid method are upper/lower bounds for the integral due to the Hermite-Hadamard inequality. If we perform that "symmetrization trick" once more, then consider a Taylor expansion in the origin, we get: $$ I \approx \int_{0}^{1/2}\left(\frac{31}{20}+\frac{x}{8}-\frac{503 x^2}{4000}\right)\,dx = \frac{75397}{96000}$$ that is a pretty accurate approximation of $\frac{\pi}{4}$.