Prove $ \frac{1}{201} < \int^{100}_{0} \frac{e^{-x}}{100+x} dx < \frac{1}{100}$ Prove: $\displaystyle \frac{1}{201} < \int^{100}_{0} \frac{e^{-x}}{100+x} dx < \frac{1}{100}$
What i was trying to do:
On the right side: 
$$
\int^{100}_{0} \frac{e^{-x}}{100+x} dx < \int^{100}_{0} \frac{1}{100+x} dx = ln(100+x) |^{100}_0 = ln(200) - ln(100) = ln(2)
$$
But 
$$\frac{1}{100} < ln(2)$$
$$
\int^{100}_{0} \frac{e^{-x}}{100+x} dx < \int^{100}_{0} \frac{1}{e^x} dx = -\frac{1}{(e^x)^2} |^{100}_0 = 1 - \frac{1}{e^{99}}
$$
But as the previous: 
$$
\frac{1}{100} < 1 - \frac{1}{e^{99}}
$$
For the left side i dont realy have an idea. 
So i thought maybe my way of proving is not the right direction, can someone give me a hint? 

Using the estimation as already given in @BrianMoehring's comment and noting that 


*

*$\int_0^{100}e^{-x}dx = 1- \frac{1}{e^{100}}$ you have


$$\frac 1{200}\left(1- \frac{1}{e^{100}}\right) < \int^{100}_{0} \frac{e^{-x}}{100+x} dx < \frac 1{100}\left(1- \frac{1}{e^{100}}\right) < \frac 1{100}$$
The lower bound seems a bit artificial but can be obtained by a further rough estimate using $e^x > 1+x+ \frac{x^2}{2}$ for $x>0$. It is enough to show that
$$\frac 1{201}< \frac 1{200}\left(1- \frac{1}{e^{100}}\right)$$
$$\Leftrightarrow 1-\frac{1}{201}< 1- \frac{1}{e^{100}}$$
$$\Leftrightarrow \frac{1}{201} >\frac{1}{e^{100}}$$
which is true since
$$\frac{1}{e^{100}} < \frac{1}{1+100+\frac{100^2}{2}}< \frac 1{201}$$
