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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 $

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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 $\int_0^1{\frac{e^{-x}}{1+x}} \leq \int_0^1{\frac{1}{1+x}} = \log(2)$.

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  • $\begingroup$ thanks but still if $\ x = 0 $ then $\ e^{-x} = 1 $ and $\ \frac{1}{1+x} = 1 $ and then it is greater than $\ \ln 2 $ $\endgroup$
    – bm1125
    Jan 6 '19 at 11:01
  • $\begingroup$ For $x\in[0,1]$,$$\frac{e^{-x}}2\le\frac{e^{-x}}{1+x}\le\frac1{1+x}$$ $\endgroup$ Jan 6 '19 at 11:10

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