Question

${\int }_0^{\infty }\left[\frac{2}{e^x}\right]dx$ (where [.] denotes the greatest integer function) equals

• A. $lo{g}_{e}2$
• B. $e^2$
• C. $0$
• D. $\frac{2}{e}$

Answer 1

The correct option is A $lo{g}_{e}2$

We can see that:

$e^xϵ(1,2)$ for $xϵ(0,{\log }_{e}2)$ $($as $e^{{\log }_{e}2}=2)$

$\Rightarrow \frac{2}{e^x}ϵ(1,2)$ for $xϵ(0,{\log }_{e}2)$

$\Rightarrow \left[\frac{2}{e^x}\right]=1$ for $xϵ(0,{\log }_{e}2)$

Similarly,

$e^xϵ(2,\infty )$ for $xϵ({\log }_{e}2,\infty )$

$\Rightarrow \frac{2}{e^x}ϵ(0,1)$ for $xϵ({\log }_{e}2,\infty )$

$\Rightarrow \left[\frac{2}{e^x}\right]=0$ for $xϵ({\log }_{e}2,\infty )$

Now,

${\int }_0^{\infty }\left[\frac{2}{e^x}\right]dx={\int }_0^{{\log }_{e}2}\left[\frac{2}{e^x}\right]dx+{\int }_{{\log }_{e}2}^{\infty }\left[\frac{2}{e^x}\right]dx$

$={\int }_0^{{\log }_{e}2}1.dx+{\int }_{{\log }_{e}2}^{\infty }0.dx$

$={\int }_0^{{\log }_{e}2}1.dx$

$={\left[x\right]}_0^{{\log }_{e}2}$

$={\log }_{e}2$.

Hence, Option $A$ is correct.