Answer
The area under the graph of $f(x)=\frac{1}{1+x}$ for $0\leq x\leq 2$
Work Step by Step
Consider $x_i=\frac{2i}{n}$.
The lower bound: $x_0=\frac{2\cdot 0}{n}=0$
The upper bound: $x_n=\frac{2\cdot n}{n}=2$
Then,
$\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{2}{n}\frac{1}{1+(2i/n)}=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\Delta x$
$\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{2}{n}\frac{1}{1+(2i/n)}=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\cdot \frac{\text{upper bound - lower bound}}{n}$
$\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{2}{n}\frac{1}{1+x_i}=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\cdot \frac{2-0}{n}$
$\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{1}{1+x_i}\cdot \frac{2}{n}=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\cdot \frac{2}{n}$
Comparing both sides, we get the function $f(x)=\frac{1}{1+x}$.
Thus, the limit represents the area under the graph of $f(x)=\frac{1}{1+x}$ for $0\leq x\leq 2$.
