$\int_0^{100}\frac{e^{-x}}{x+100}dx>0.005$? $\int_0^{100}\frac{e^{-x}}{x+100}dx>0.005$?
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?
 A: 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 
$$
I \ge \frac{1}{121} \approx 0.008264462809917356 \,.
$$
A: Use $e^u\ge 1+u$ to conclude 
$$
\frac1{1+u}\ge e^{-u}\implies \frac1{100+x}\ge\frac1{100}e^{-x/100}.
$$
Then, again using $1-e^{-u}\ge1-\frac1{1+u}$,
$$
\int_0^{100}\frac{e^{-x}}{100+x}dx
\ge\frac1{100} \int_0^{100} e^{-\frac{101}{100}x} dx
=\frac1{100} \frac{100}{101} \left( 1-e^{-101} \right)
\ge\frac1{101}\left(1-\frac1{102}\right)=\frac1{102}
$$
which is decidedly larger than $\frac1{200}$.
A: Using the MatDeck function for numerical integration (Gaussian rule of integration in this case) you can get approximate value of 0.00990194
MatDeck numerical integration - Gaussian rule of integration
