Question
$$\int\limits_0^\infty {\left[ {\frac{2}{{{e^x}}}} \right]} dx$$ is equal to ( $$\left[ x \right] = $$ greatest integer $$ \leqslant x$$ )
A.
$${\log _e}2$$
B.
$${e^2}$$
C.
$$0$$
D.
$$\frac{2}{e}$$
Answer :
$${\log _e}2$$
Solution :
$$\eqalign{
& {\text{We have, if }}{e^x} > 2,\,\frac{2}{{{e^x}}} < 1 \cr
& {\text{Also, }}\frac{2}{{{e^x}}} > 0 \Rightarrow 0 < \frac{2}{{{e^x}}} < 1 \cr
& \therefore \,{\text{If }}x > {\log _e}2,\,\left[ {\frac{2}{{{e^x}}}} \right] = 0 \cr
& {\text{Again if }}0 < x < {\log _e}2{\text{ then }}1 < {e^x} < 2 \cr
& \Rightarrow 1 > \frac{1}{{{e^x}}} > \frac{1}{2} \cr
& \Rightarrow 2 > \frac{2}{{{e^x}}} > 1{\text{ or }}1 < \frac{2}{{{e^x}}} < 2 \cr
& \therefore \,\left[ {\frac{2}{{{e^x}}}} \right] = 1 \cr
& \therefore \,I = \int\limits_0^\infty {\left[ {\frac{2}{{{e^x}}}} \right]dx} \cr
& = \int\limits_0^\infty {\left[ {2{e^{ - x}}} \right]} dx \cr
& = \int\limits_0^{\log \,2} {\left[ {2{e^{ - x}}} \right]dx} + \int\limits_{\log \,2}^\infty {\left[ {2{e^{ - x}}} \right]} dx \cr
& = \int\limits_0^{\log \,2} {\left( 1 \right)dx} + \int\limits_{\log \,2}^\infty {\left( 0 \right)} dx \cr
& = {\log _e}2 \cr} $$