Question

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

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

Answer 1

Answer: (B)

$\underset{e^{-1}}{\overset{e^2}{\int }}\left|\frac{{\log }_{e}x}{x}\right|dx=\underset{e^{-1}}{\overset{1}{\int }}\left|\frac{{\log }_{e}x}{x}\right|dx-\underset{1}{\overset{{ϵ}^2}{\int }}\left|\frac{{\log }_{e}x}{x}\right|dx$
$\left[\text{since, 1 is turning point for}\left|\frac{{\log }_{e}x}{x}\right|\text{ for }+\text{ ve and - ve values }\right]$
$=-\underset{e^{-1}}{\overset{1}{\int }}\frac{{\log }_{e}x}{x}dx+\underset{1}{\overset{e^2}{\int }}\left|\frac{{\log }_{e}x}{x}\right|dx$
$=-\frac{1}{2}{\left[{\left({\log }_{e}x\right)}^2\right]}_{e^{-1}}^1+\frac{1}{2}{\left[{\left({\log }_{e}x\right)}^2\right]}_1^{e^2}$
$=-\frac{1}{2}\left\{0-(-1{)}^2\right\}+\frac{1}{2}\left(2^2-0\right)=\frac{5}{2}$