Question

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

Answer 1

R.E.F image

$y=ln(x+e),x=ln\left(1/y\right)$

$y=ln(x+e),y=e^{-x}$

$ln(x+e)=r^{-x}$

$x=0$

$A=A_1+A_2={\int }_{-e+1}^{0}ln(x+e)dx+{\int }_0^{\infty }{e}^{-n}dx$

$A_1={\int }_0^{-e+1}$ ln $(n+e)dx$

$A_1={\int }_1^{e}10lnudu$

By - Parts

$A_1=\left[lnu{\int }_1^{e}ldu\right]-{\int }_1^{e}du\times \frac{dlnu}{du}du$

[Let $n+e=U,dx=du$

$A_1=(UlnU{)}_1^e-{\int }_1^{e}U\times \frac{1}{u}du$

$A_1=e-(e-1)$

$A_1=1$

$A_2={\int }_0^{\infty }{e}^{-x}dx$

Let $-x=U$

$dx=-du$

$A_2={\int }_0^{-\infty }{e}^{U}du$

$A_2=-[e^u{]}^{\infty }$

$A_2=-[0-1]$

$A_2=1$

$A=A_1+A_2=1+1$

$A=2$

Thus area = 2 sq. units.