Question

Area enclosed by the graph of the function $y={\ln }^{2}x-1$ lying in the $4^{\text{th }}$ quadrant is

• A. $\frac{2}{e}$
• B. $\frac{4}{e}$
• C. $2\left(e+\frac{1}{e}\right)$
• D. $4\left(e-\frac{1}{e}\right)$

Answer 1

Answer: (B)

$y={\ln }^{2}x-1$
$y^{\prime }=\frac{2\ln x}{x}=0\Rightarrow x=1$
$x>1$, $y\uparrow$ and $0<x<1$, $y$ is $\downarrow$
$A=\left|\underset{1/e}{\overset{e}{\int }}\left({\ln }^{2}x-1\right)dx\right|$
${=\mid x{\ln }^{2}x]}_{1/e}^e-2\underset{1/e}{\overset{e}{\int }}\left(\frac{\ln x}{x}\right)\cdot xdx-\left(e-\frac{1}{e}\right)\mid$
$=\left|\left(e-\frac{1}{e}\right)-2\underset{1/e}{\overset{e}{\int }}\left(\frac{\ln x}{x}\right)\cdot xdx-\left(e-\frac{1}{e}\right)\right|$
$=\mid -2[x\ln x{]}_{1/e}^e-\underset{1/e}{\overset{e}{\int }}dx]|=|-2\left[\left(e+\frac{1}{e}\right)-\left(e-\frac{1}{e}\right)\right]|=|\frac{4}{e}\mid =\frac{4}{e}$