Question

Let ${\int }_0^{\infty }\left[2e^{-x}\right]dx$, where $\left[x\right]$ represents the greatest integer function be equal to $\ln k$.Find $k$ ?

Answer 1

Since, $2e^{-x}$ is decreasing for $xϵ[0,\infty )$.
$\therefore 0\le 2e^{-x}\le 2,\forall xϵ[0,\infty )$
For $x>\ln 2,\left[2e^{-x}\right]=0$
$⟹{\int }_0^{\infty }\left[2e^{-x}\right]dx={\int }_0^{\ln 2}\left[2e^{-x}\right]dx+{\int }_{\ln 2}^{\infty }\left[2e^{-x}\right]dx$
$={\int }_0^{\ln 2}1dx+{\int }_{\ln 2}^{\infty }0dx=(x{)}_0^{\ln 2}=\ln 2$
$\therefore {\int }_0^{\infty }\left[2e^{-x}\right]dx=\ln 2$