Question

The area enclosed between the curves $y={\log }_e(x+e),x={\log }_e\left(\frac{1}{y}\right)$ and the $x$-axis is

• A. $2\text{sq. units}$
• B. $1\text{sq. units}$
• C. $4\text{sq. units}$
• D. None of these

Answer 1

The correct option is A $2\text{sq. units}$

$y={\log }_e(x+e)$
$x={\log }_e\left(\frac{1}{y}\right)\Rightarrow y=e^{-x}$

For $y={\log }_e(x+e),$ shift the graph of $y={\log }_{e}x,e$ units to the left hand side.
Required area $\left(A\right)$ is,
$A=\underset{1-e}{\overset{0}{\int }}{\log }_e(x+e)dx+\underset{0}{\overset{\infty }{\int }}{e}^{-x}dx$
$=[x{\log }_e(x+e){]}_{1-e}^0-\underset{1-e}{\overset{0}{\int }}\frac{x}{x+e}dx-[e^{-x}{]}_0^{\infty }$
$=\underset{0}{\overset{1-e}{\int }}(1-\frac{e}{x+e})dx-e^{-\infty }+e^0$
$=[x-e\log (x+e){]}_0^{1-e}-0+1$
$=1-e+e\log e+1$
$=2\text{sq. units}$