Answer
The integral can not be zero, as the graph of $e^{-x}$ is above the x-axis, and the area below it is not 0.
This is an improper integral, and it converges to 1.
CAS confirms this result.
Work Step by Step
From the
Definition of Improper Integrals with Infinite Integration Limits
1. If $f$ is continuous on the interval $[a, \infty)$, then $\displaystyle \int_{a}^{\infty}f(x)d\mathrm{x}=\lim_{b\rightarrow\infty}\int_{a}^{b}f(x)dx$.
This integral has an infinite bound and belongs to the type of improper integrals defined above.
$\displaystyle \int_{0}^{\infty}e^{-x}dx= \displaystyle \lim_{b\rightarrow\infty}\int_{0}^{b}e^{-x}dx$
$=\displaystyle \lim_{b\rightarrow\infty}[-e^{-x}]_{0}^{b}$
$=\displaystyle \lim_{b\rightarrow\infty}[-e^{-b}+1]$
$=1$
The integral can not be zero, as the graph of $e^{-x}$ is above the x-axis, and the area below it is not 0.
This is an improper integral, and it converges to 1.
CAS confirms this result.
Geogebra CAS input:
Integral(exp(-x),0,infinity)
