Thank you for reporting, we will resolve it shortly
Q.
If $ \int\limits_{0}^{\infty}\left[3e^{-x}\right]dx = l $ where $[.]$ denotes the greatest integer function, then the value of $ l $ is
AMUAMU 2018
Solution:
Let $I=\int\limits_{0}^{\infty}\left[3 e^{-x}\right] d x$
$3 e^{-x}$ is decreasing function for $x \in[0, \infty)$.
When $x \in(0, \ln 3)$, then $\left[3 e^{-x}\right]=1$ And
when $x \in(\ln 3, \infty)$, then $\left[3 e^{-x}\right]=0$
$\therefore I=\int\limits_{0}^{\ln 3}\left[3 e^{-x}\right] d x+\int\limits_{\ln 3}^{\infty}\left[3 e^{-x}\right] d x $
$\Rightarrow I=\int\limits_{0}^{\ln 3} d x+0 \Rightarrow I=\ln 3$