# How to prove that $\frac{e-1}{2e} \le \int_0^1 \frac{e^{-x}}{1+x}dx \le \ln 2$ [closed]

Prove: $$\ \frac{e-1}{2e} \le \int_0^1 \frac{e^{-x}}{1+x}dx \le \ln 2$$

Attempt:

I can clearly see that $$\ 2e \ge 1+x \ge 1$$ for every $$\ 0 \le x \le 1$$ but $$\ \frac{e^{-x}}{1+x} \ge \ln 2$$ for $$\ x = 1$$

First inequality: in the domain, $$\frac{1}{1+x} \geq \frac{1}{2}$$. Thus $$\int_0^1{\frac{e^{-x}}{1+x}} \geq \frac{1}{2}\int_0^1{e^{-x}}=\frac{e-1}{2e}$$.
Second inequality: in the domain, $$e^{-x} \leq 1$$. Thus $$\int_0^1{\frac{e^{-x}}{1+x}} \leq \int_0^1{\frac{1}{1+x}} = \log(2)$$.
• thanks but still if $\ x = 0$ then $\ e^{-x} = 1$ and $\ \frac{1}{1+x} = 1$ and then it is greater than $\ \ln 2$ Jan 6 '19 at 11:01
• For $x\in[0,1]$,$$\frac{e^{-x}}2\le\frac{e^{-x}}{1+x}\le\frac1{1+x}$$ Jan 6 '19 at 11:10