## Calculus: Early Transcendentals 8th Edition

$\int_{0}^{1} \frac{e^x}{1+x} dx$
We know the following $\int_{a}^{b} f(x)dx=\lim\limits_{n \to \infty}\sum\limits_{i=1}^{n} f(x_{i})\Delta x$ $\Delta x=\frac{a-b}{n}$ $x_{i}=a+i\Delta x$ In the problem we see that $f(x)=\frac{e^x}{1+x}$ and we are given the interval of [0,1] therefore the integral will be from 0 to 1. Therefore $\lim\limits_{n \to \infty}\sum\limits_{n=1}^{n} f(\frac{e^{x_i}}{1+x_i})\Delta x=\int_{0}^{1} \frac{e^x}{1+x} dx$