Question

The value of the integral ${\int }_{e^{-1}}^{e^2}\left|\frac{lo{g}_{e}x}{x}\right|dx$ is

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

Answer 1

Answer: (B)

$\frac{5}{2}$

We have,
$I={\int }_{e^{-1}}^{e^2}\left|\frac{lo{g}_{e}x}{x}\right|dx$
$\Rightarrow I={\int }_{1/e}^{e^2}\frac{|lo{g}_{e}x|}{x}dx$
$\Rightarrow I={\int }_{1/e}^1\frac{|lo{g}_{e}x|}{x}dx+{\int }_1^{e^2}\frac{|lo{g}_{e}x|}{x}dx$
$\Rightarrow I=-{\int }_{1/e}^1\frac{lo{g}_{e}x}{x}dx+{\int }_1^{e^2}\frac{lo{g}_{e}x}{x}dx\left[\begin{array}{c}\because lo{g}_{e}x<0for\frac{1}{e}<x<1\\ >0forx>1\end{array}\right]$
$\Rightarrow I=-{\int }_{1/e}^{1}lo{g}_{e}xd\left(lo{g}_{e}x\right)+{\int }_1^{e^2}lo{g}_{e}xd\left(lo{g}_{e}x\right)$
$\Rightarrow I=-{\left[\frac{1}{2}\right(lo{g}_{e}x{)}^2]}_{1/e}^1+{\left[\frac{1}{2}\right(lo{g}_{e}x{)}^2]}_1^{e^2}$
$\Rightarrow I=-[\frac{1}{2}\times 0-\frac{1}{2}]+[\frac{1}{2}\times 4-0]$
$=\frac{1}{2}+2=\frac{5}{2}$
Hence, option 'B' is correct.