Answer
$\int_{0}^{1}\frac{1}{1+x^{2}}dx$
Work Step by Step
$\displaystyle \lim_{n \to \infty} \frac{1}{n} \displaystyle \sum_{i=1}^{n}\frac{1}{1+(\frac{i}{n})^{2}}$
Since $\frac{1}{n} $ does not depend on $i$, move $\frac{1}{n} $ inside the sum because it is a constant:
$\displaystyle \lim_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{1}{n}\cdot\frac{1}{1+(\frac{i}{n})^{2}}$
$\displaystyle \lim_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{1}{n}\cdot\frac{1}{1+(0+i\cdot \frac{1}{n})^{2}}$
$\displaystyle \lim_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{1-0}{n}\cdot\frac{1}{1+(0+i\cdot \frac{1-0}{n})^{2}}$
Put $\Delta x=\frac{1-0}{n}$ and $x_{i}=0+i\cdot \frac{1-0}{n}$
Using theorem $4$ it follows that:
$\displaystyle \lim_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{1-0}{n}\cdot\frac{1}{1+(0+i\cdot \frac{1-0}{n})^{2}}
=\int_{0}^{1}\frac{1}{1+x^{2}}dx$
