Definite Integration Question 50
Question: The value of $ \int _{{e^{-1}}}^{e^{2}}{| \frac{{\log _{e}}x}{x} |dx} $ is
[IIT Screening 2000]
Options:
A) $ \frac{3}{2} $
B) $ \frac{5}{2} $
C) 3
D) 5
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int _{{e^{-1}}}^{e^{2}}{| \frac{{\log _{e}}x}{x} |dx=\int _{{e^{-1}}}^{1}{| \frac{{\log _{e}}x}{x} |dx+\int_1^{e^{2}}{| \frac{{\log _{e}}x}{x} |dx}}} $
$ =\int _{{e^{-1}}}^{1}{-\frac{\log x}{x}dx+\int_1^{e^{2}}{\frac{\log x}{x}dx}} $
$ =\int _{-1}^{0}{-zdz+\int_0^{2}{zdz}} $ , (Putting $ {\log _{e}}x=z $
therefore $ (1/x)dx=dz) $
$ =[ -\frac{z^{2}}{2} ] _{-1}^{0}+[ \frac{z^{2}}{2} ]_0^{2}=\frac{1}{2}+2=\frac{5}{2} $ .F
BETA
