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) none of these

Answer 1

(a) 2



The point of intersection of the curves $y={\log }_e\left(x+e\right)\mathrm{and}x={\log }_e\left(\frac{1}{y}\right)$ is (0, 1)
$$\begin{gathered} y={\log }_e\left(x+e\right) \\ \Rightarrow x+e=e^y \\ \Rightarrow x=e^y-e \\ \mathrm{Here}\mathrm{taking},x_1=e^y-e \\ \mathrm{and}{x}_2={\log }_e\left(\frac{1}{y}\right) \end{gathered}$$
Therefore, area of the required region,
$$\begin{gathered} A={\int }_0^1\left(x_2-x_1\right)\mathit{d}y\mathit{}\left[\mathrm{Where},x_1=e^y-e\mathrm{and}{x}_2={\log }_e\left(\frac{1}{y}\right)\right] \\ A={\int }_0^1{\log }_e\left(\frac{1}{y}\right)\mathit{d}y-{\int }_0^1\left(e^y-e\right)\mathit{d}y \\ A={\int }_0^1{\log }_e\left(\frac{1}{y}\right)dy-{\left[e^y-ey\right]}_0^1\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(\mathit{1}\right) \\ \mathrm{Let}\mathit{}I=\int {\log }_e\left(\frac{1}{y}\right)\mathit{d}y \\ \mathrm{Putting}\frac{1}{y}=t \\ \Rightarrow -\frac{1}{y^2}dy=dt \\ \Rightarrow dy=-y^{2}dt \\ \Rightarrow dy=-\frac{1}{t^2}dt \\ \mathrm{Therefore},\mathrm{integral}\mathrm{becomes} \\ \mathrm{I}=\int -\frac{1}{t^2}{\log }_{\mathrm{e}}t\mathit{}dt \\ =-{\log }_{\mathrm{e}}\mathit{}t\mathit{}\int \frac{1}{t^2}dt\mathit{-}\int \frac{1}{t}\times \frac{\mathit{1}}{\mathit{t}}dt \\ =\frac{1}{t}{\log }_e\mathit{}t+\frac{1}{t} \\ =y{\log }_e\frac{1}{y}+y \\ \mathrm{Now},\left(1\right)\mathrm{becomes} \\ A={\left[y{\log }_e\frac{1}{y}+y\right]}_0^1-{\left[e^y-ey\right]}_0^1\mathit{} \\ ={\left[y{\log }_e\left(\frac{1}{y}\right)+y-e^y+ey\right]}_0^1 \\ =\left[{\log }_e\left(1\right)+1-e^1+e\left(1\right)\right]-\left[0+0-e^{\left(0\right)}+e\left(0\right)\right] \\ =2 \end{gathered}$$