Question

The area(in $\text{sq.units}$) of the region bounded by the curves $y=ex\ln x$ and $y=\frac{\ln x}{ex}$ is

• A. $\frac{e^2-5}{4e}$
• B. $\frac{e^2+4}{5e}$
• C. $\frac{e^2-4}{4e}$
• D. $\frac{e^2-5}{5e}$

Answer 1

The correct option is A $\frac{e^2-5}{4e}$

The bounded region shown in the figure :
The required area
$=\underset{1/e}{\overset{1}{\int }}(\frac{\ln x}{ex}-ex\ln x)dx=\frac{1}{e}{\left[\frac{{\ln }^{2}x}{2}\right]}_{1/e}^1-e{\left[\frac{x^2}{4}\right(2\ln x-1\left)\right]}_{1/e}^1=\frac{e^2-5}{4e}$