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. $2$
• B. $1$
• C. $4$
• D. $3$

Answer 1

The correct option is A $2$

$y={\log }_e(x+e)$

$x={\log }_e\left[1/y\right]$

$x={\log }_{e}1-{\log }_{e}y$

$x=-{\log }_{e}y$

${\log }_{e}y=-x$

$y=e^{-x}$

$y={\log }_e(x+e)$

$0={\log }_e(x+c)$

$e^0=x+e$

$1=2+e$

$x=1-c$

Area $\to {\int }_{1-e}^0{\log }_e(x+e)dx+{\int }_0^{\infty }{e}^{-x}dx$

$\Rightarrow x{\log }_e(x+e)-\int \frac{1}{(x+e)}xdx+{(-e^{-x}]}_0^{\infty }$

$\Rightarrow x\cdot {\log }_e(x+e)-\int \left(\frac{x+e-e}{x+e}\right)dx-[e^{-\infty }-e^{-0}]$

$\Rightarrow x{\log }_e(x+e)-\int (1-\frac{e}{x+e})dx-[\frac{1}{\infty }-\frac{1}{1}]$

$\Rightarrow x{\log }_e(x+e)-x+{\mathrm{elog}}_1(x+e)-[0-1]$

${\Rightarrow [x+e){\log }_e(x+e)-x]}_{1-e}^0+1$

$\Rightarrow \left[e{\log }_e\right(e)-0]-\left[\right(1-\mathrm{e/}+\mathrm{e/}\left){\log }_e\right(1-\mathrm{e/}+\mathrm{e/})-(1-e\left)\right]+1$

$\Rightarrow e+1-e+1$

$\Rightarrow 2sq\cdot unit$ Answer

Correct option is (A).