Question

The area bounded by the curves $y={\log }_{e}x$ and $y={\left({\log }_{e}x\right)}^2$ is

• A. $3-e$
• B. $e-3$
• C. $\frac{1}{2}(3-e)$
• D. $\frac{1}{2}(e-3)$

Answer 1

The correct option is A $3-e$

$\begin{array}{c}y=\left({\log }_{e}x\right)----\left(1\right)\\ y={\left({\log }_{e}x\right)}^2-----\left(2\right)\\ Forthepointofinter\sec tion\\ {\left({\log }_{e}x\right)}^2=\left({\log }_{e}x\right)\\ {\left({\log }_{e}x\right)}^2-\left({\log }_{e}x\right)=0\\ \left({\log }_{e}x\right)\left[\left({\log }_{e}x\right)-1\right]\\ \left({\log }_{e}x\right)=0and\left({\log }_{e}x\right)=1\\ x=1andx=e\\ areabounded\\ =\left|{\int }_1^e\left\{{\left({\log }_{e}x\right)}^2-\left({\log }_{e}x\right)\right\}dx\right|\\ consider\\ I=\int \left\{{\left({\log }_{e}x\right)}^2-\left({\log }_{e}x\right)\right\}dx\\ {\log }_{e}x=t\\ x=e^t\\ dx=e^{t.dt}\\ Now,\\ I=\int e^t\left\{t^2-t\right\}dt\\ =e^t\left\{t^2-t\right\}-\left\{\left(2t-1\right)e^t-2e^t\right\}\left[byintergrationbyparts\right]\\ =e^t\left[t^2-t-2t+1+2\right]\\ =e^t\left[t^2-3t+3\right]\\ =x\left[\log {\left(x\right)}^2-3\log \left(x\right)+3\right]\\ Now,\\ \left|{\int }_1^e\left\{{\left({\log }_{e}x\right)}^2-\left({\log }_{e}x\right)\right\}dx\right|\\ =\left|{\left[x\left\{loh{\left(x\right)}^2\right\}-3\log \left(x\right)+3\right]}_1^e\right|\\ =\left|e\left(1-3+3\right)-1\left(0-0+3\right)\right|\\ =\left|e-3\right|\\ =3-e\end{array}$

Hence, the option $B$ is the correct answer.