Question

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

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

Answer 1

Answer: (B)

$\frac{5}{2}$

Answer 2

$\int^{e^2}_{e^{-1}} \bigg | \frac{ \log_e \, x }{ x } \bigg | \, dx \, = \int^{1}_{e^{-1}} \bigg | \frac{ \log_e \, x }{ x } \bigg | \, dx \, - \int^{e^2}_{{1}} \bigg | \frac{ log_e \, x }{ x } \bigg | \, dx \,$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Bigg [$ since , 1 is turning point for $| \frac{ \log _e \, x }{x}|$ for $+ ve$ and $\, - ve$ values $\Bigg ]$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = - \int^1_{e^{-1} }\frac{ \log _e \, x }{x} \, dx + \int^{e^2}_1 | \frac{ \log _e \, x }{x}| \, dx$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = -\frac{1}{2} [ ( \log_e \, x )^2 ]^1_{e^{-1}} + \frac{ 1 }{2} [ ( \log _ e \, x )^2]^{e^2 } _ 1$
\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = - \frac{1}{2} \big \\{ 0 - ( - 1 )^2 \big \\} + \frac{1}{2} ( 2^2 - 0 ) = \frac{5}{2}