Question

If $\underset{0}{\overset{\infty }{\int }}\left[3e^{-x}\right]dx=l$ where $[.]$ denotes the greatest integer function, then the value of $l$ is

• A. $0$
• B. $\ln 3$
• C. $e^3$
• D. $3e^{-1}$

Answer 1

Answer: (B)

Let $I=\underset{0}{\overset{\infty }{\int }}\left[3e^{-x}\right]dx$
$3e^{-x}$ is decreasing function for $x\in [0,\infty )$.
When $x\in (0,\ln 3)$, then $\left[3e^{-x}\right]=1$ And
when $x\in (\ln 3,\infty )$, then $\left[3e^{-x}\right]=0$
$\therefore I=\underset{0}{\overset{\ln 3}{\int }}\left[3e^{-x}\right]dx+\underset{\ln 3}{\overset{\infty }{\int }}\left[3e^{-x}\right]dx$
$\Rightarrow I=\underset{0}{\overset{\ln 3}{\int }}dx+0\Rightarrow I=\ln 3$