$\int_0^{100}\frac{e^{-x}}{x+100}dx>0.005$? $\int_0^{100}\frac{e^{-x}}{x+100}dx>0.005$?
My attempt: $$\int_0^{100}\frac{e^{-x}}{x+100}dx>\int_0^{100}\frac{e^{-x}}{200}dx=\frac{1-e^{-100}}{200}$$ A little bit error. How to amend it?

The idea: The integrand decreases rapidly on the given interval, so the idea is to estimate the integral from below by integrating over a shorter interval $[0, a]$, and then continue with your approach, but with a better bound for the denominator: 
For $0 < a < 100$ we can estimate
$$I = \int_0^{100}\frac{e^{-x}}{x+100}dx \ge \int_0^a\frac{e^{-x}}{x+100}dx \\
\ge \int_0^a\frac{e^{-x}}{a+100}dx = \frac{1-e^{-a}}{a+100} \, .
$$
For $a=4$ this gives
$$
I \ge \frac{1-e^{-4}}{104} \approx 0.00943927270299294
$$
which comes fairly close to the result $I \approx 0.009901942286733037$ (obtained by numeric integration with Maxima).
We can also avoid calculating $e^{-a}$ numerically and use $e^a \ge 1+a$ to further estimate
$$
  I \ge \frac{1-e^{-a}}{a+100} \ge \frac{a}{(a+1)(a+100)} \, .
$$
For $a=10$ this gives 
$$
I \ge \frac{1}{121} \approx 0.008264462809917356 \,.
$$
