Question

The value of the integral $ \int\limits^{e^2}_{e^{-1}} \bigg | \frac{ \log_e \, x }{ x } \bigg | \, dx $ is

• A. $\frac{3}{2} $
• B. $\frac{5}{2} $
• C. 3
• D. 5

Answer 1

Answer: (B)

$ \int\limits_{e^{-1}}^{e^{2}}\left|\frac{\log _{e} x}{x}\right| d x =\int\limits_{e^{-1}}^{1}\left|\frac{\log _{e} x}{x}\right| d x-\int\limits_{1}^{\epsilon^{2}}\left|\frac{\log _{e} x}{x}\right| d x $
$ {\left[\text{since, 1 is turning point for}\left|\frac{\log _{e} x}{x}\right| \text { for }+\text { ve and - ve values }\right] }$
$=-\int\limits_{e^{-1}}^{1} \frac{\log _{e} x}{x} d x+\int\limits_{1}^{e^{2}}\left|\frac{\log _{e} x}{x}\right| d x$
$=-\frac{1}{2}\left[\left(\log _{e} x\right)^{2}\right]_{e^{-1}}^{1}+\frac{1}{2}\left[\left(\log _{e} x\right)^{2}\right]_{1}^{e^{2}} $
$=-\frac{1}{2}\left\{0-(-1)^{2}\right\}+\frac{1}{2}\left(2^{2}-0\right)=\frac{5}{2} $