Question

$\int\limits^{\infty}_{{0}}$ $\left[\frac{2}{e^{x}}\right]dx$ is equal to $\left(\left[x\right] = greatest\, integer \le x\right)$

• A. $log_e \,2$
• 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 $\frac{2}{e^{x}}>0$
$\Rightarrow 0<\frac{2}{e^{x}}<1\,\therefore If x>log_{e}\,2, \left[\frac{2}{e^{x}}\right]=0$
Again if $0 < x< log_{e}\, 2$ 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 $ $I$ = $\int\limits^{\infty}_{{0}}$ $\left[\frac{2}{e^{x}}\right]dx+\int\limits^{\infty}_{{0}}[2e^{-x}]dx$
=$\int\limits_{0}^{log 2}\left[2e^{-x}\right] dx +\int\limits_{log 2}^{\infty}\left[2e^{-x}\right] dx$
=$\int\limits_{0}^{log 2}\left(1\right) dx +\int\limits_{log 2}^{\infty}\left(0\right) dx = log_{e} 2$