Question

The value of the integral $\int_{e^{- 1}}^{e^{2}}\left| \frac{\log_{e}x}{x} \right|dx$ is

• A. 3/2
• B. 5/2
• C. 3
• D. 5

Answer 1

Answer: (B)

5/2

Answer 2

Put $\log_{e}x = t \Rightarrow e^{t} = x$

∴ $dx = e^{t}dt$

and limits are adjusted as –1 to 2

∴ $I = \int_{- 1}^{2}{\left| \frac{t}{e^{t}} \right|e^{t}dt} = \int_{- 1}^{2}{|t|dt}$ ⇒ $I = \int_{- 1}^{0}{- tdt + \int_{0}^{2}{tdt}}$ ⇒ $I = \left\lbrack \frac{- t^{2}}{2} \right\rbrack_{- 1}^{0} + \left\lbrack \frac{t^{2}}{2} \right\rbrack_{0}^{2}$ ⇒ $I = 5 ⥂ / ⥂ 2$