Question

how to calculate average of e^-x over 0 to infinity?

Answer 1

e^(-x) is the same as
\[1 \div e ^{x}\]
as x approaches infinity, e^x gets really huge really fast. As a result, 1/(e^x) gets really small very fast. The graph looks like:


Now, those y values get as close to 0 as you'd like, but they never get negative.