Integral Calculus Question 248
Question: $ \int\limits_0^{\infty }{[ \frac{2}{e^{x}} ]},dx $ is equal to ([x] = greatest integer $ \le $ x)
Options:
A) $ {\log_{e}}2 $
B) $ e^{2} $
C) 0
D) $ \frac{2}{e} $
Show Answer
Answer:
Correct Answer: A
Solution:
[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>{\log_{e}}2,[ \frac{2}{e^{x}} ]=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 [ \frac{2}{e^{x}} ]=1\therefore I=\int\limits_0^{\infty }{[ \frac{2}{e^{x}} ]dx=\int\limits_0^{\infty }{[ 2{e^{-x}} ]},dx} $ $ =\int\limits_0^{\log ,2}{[ 2{e^{-x}} ]dx+\int\limits_{\log 2}^{\infty }{[ 2{e^{-x}} ]}dx} $ $ =\int\limits_0^{\log 2}{(1)dx+\int\limits_{\log 2}^{\infty }{(0)dx={\log_{e}}2}} $
BETA
