0
1
0.5
2
Answer is Right!
Answer is Wrong!
The correct answer is $\boxed{\text{A) }0}$.
The improper integral $\int_0^\infty {e^{-2t}} dt$ converges because the function $e^{-2t}$ is continuous and decreasing on $[0, \infty)$, and the limit of the integral as $a \to \infty$ is $0$.
To see this, note that for any $a > 0$, we have
$$\int_a^\infty {e^{-2t}} dt \leq \int_a^\infty dt = a.$$
Taking the limit as $a \to \infty$, we get
$$\lim_{a \to \infty} \int_a^\infty {e^{-2t}} dt \leq \lim_{a \to \infty} a = 0.$$
Since the limit of the integral is less than or equal to $0$, the integral must converge to $0$.