Answer
$\int_{0}^{1}\frac{1}{1+x^2}~dx$
Work Step by Step
We can use the definition of the integral in theorem 4 to find the definite integral:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
Consider the interval $[0,1]$:
$\Delta x = \frac{b-a}{n} = \frac{(1-0)}{n} = \frac{1}{n}$
$x_i = \frac{i}{n}$
$\lim\limits_{n \to \infty}~\frac{1}{n}~\sum_{i=1}^{n}\frac{1}{1+(i/n)^2}$
$= \lim\limits_{n \to \infty}~\sum_{i=1}^{n}\frac{1}{1+(i/n)^2}\cdot \frac{1}{n}$
$=\lim\limits_{n \to \infty}~\sum_{i=1}^{n}f(x_i)\Delta x$
when $f(x) = \frac{1}{1+x^2}$
Therefore, the definite integral is: $~~\int_{0}^{1}\frac{1}{1+x^2}~dx$
