Question

$\underset{0}{\overset{\infty }{\int }}$ $\left[\frac{2}{e^x}\right]dx$ is equal to $\left(\left[x\right]=greatestinteger\le x\right)$

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

Answer 1

Answer: (A)

We have, if $e^x>2,\frac{2}{e^x}<1$. Also $\frac{2}{e^x}>0$
$\Rightarrow 0<\frac{2}{e^x}<1\therefore Ifx>lo{g}_{e}2,\left[\frac{2}{e^x}\right]=0$
Again if $0<x<lo{g}_{e}2$ then $1<e^x<2$
$\Rightarrow 1>\frac{1}{e^x}>\frac{1}{2}\Rightarrow 2>\frac{2}{e^x}>1$ or $1<\frac{2}{e^x}<2$
$\therefore \left[\frac{2}{e^x}\right]=1$
$\therefore$ $I$ = $\underset{0}{\overset{\infty }{\int }}$ $\left[\frac{2}{e^x}\right]dx+\underset{0}{\overset{\infty }{\int }}[2e^{-x}]dx$
=$\underset{0}{\overset{log2}{\int }}\left[2e^{-x}\right]dx+\underset{log2}{\overset{\infty }{\int }}\left[2e^{-x}\right]dx$
=$\underset{0}{\overset{log2}{\int }}\left(1\right)dx+\underset{log2}{\overset{\infty }{\int }}\left(0\right)dx=lo{g}_{e}2$