A) \[\frac{3}{2}\]
B) \[\frac{5}{2}\]
C) 3
D) 5
Correct Answer: B
Solution :
\[\int_{{{e}^{-1}}}^{{{e}^{2}}}{\left| \frac{{{\log }_{e}}x}{x} \right|dx=\int_{{{e}^{-1}}}^{1}{\left| \frac{{{\log }_{e}}x}{x} \right|\,dx+\int_{1}^{{{e}^{2}}}{\left| \frac{{{\log }_{e}}x}{x} \right|\,dx}}}\] \[=\int_{{{e}^{-1}}}^{1}{-\frac{\log x}{x}dx+\int_{1}^{{{e}^{2}}}{\frac{\log x}{x}dx}}\]\[=\int_{-1}^{0}{-zdz+\int_{0}^{2}{zdz}}\], (Putting \[{{\log }_{e}}x=z\] Þ \[(1/x)\,dx=dz)\] \[=\left[ -\frac{{{z}^{2}}}{2} \right]_{-1}^{0}+\left[ \frac{{{z}^{2}}}{2} \right]_{0}^{2}=\frac{1}{2}+2=\frac{5}{2}\].F
You need to login to perform this action.
You will be redirected in
3 sec
