## Calculus: Early Transcendentals 8th Edition

$A = \lim\limits_{n \to \infty} \sum_{i=1}^{n}[\frac{2(1+\frac{2i}{n})}{(1+\frac{2i}{n})^2+1}(\frac{2}{n})]$
For each $x_i$, such that $1 \leq i \leq n$, note that $x_i = (1+i\frac{2}{n}) = (1+\frac{2i}{n})$ $\Delta x = \frac{2}{n}$ We can express the area under the curve as a limit: $A = \lim\limits_{n \to \infty} [f(x_1)\Delta x+f(x_2)\Delta x+...+f(x_n)\Delta x]$ $A = \lim\limits_{n \to \infty} [f(1+\frac{2}{n})(\frac{2}{n})+f(1+\frac{4}{n})(\frac{2}{n})+...+f(1+\frac{2n}{n})(\frac{2}{n})]$ $A = \lim\limits_{n \to \infty} [\frac{2(1+\frac{2}{n})}{(1+\frac{2}{n})^2+1}(\frac{2}{n}) + \frac{2(1+\frac{4}{n})}{(1+\frac{4}{n})^2+1}(\frac{2}{n})+...+\frac{2(1+\frac{2n}{n})}{(1+\frac{2n}{n})^2+1}(\frac{2}{n})]$ $A = \lim\limits_{n \to \infty} \sum_{i=1}^{n}[\frac{2(1+\frac{2i}{n})}{(1+\frac{2i}{n})^2+1}(\frac{2}{n})]$