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Q.
Area enclosed by the graph of the function $y=\ln ^2 x-1$ lying in the $4^{\text {th }}$ quadrant is
Application of Integrals
Solution:
$y=\ln ^2 x-1$
$y^{\prime}=\frac{2 \ln x}{x}=0 \Rightarrow x=1$
$x >1$, $y \uparrow$ and $0< x <1$, $y$ is $\downarrow$
$A=\left|\int\limits_{1 / e }^{ e }\left(\ln ^2 x -1\right) dx \right|$
$\left.=\mid x \ln ^2 x \right]_{1 / e }^{ e }-2 \int\limits_{1 / e }^{ e }\left(\frac{\ln x }{ x }\right) \cdot xdx -\left( e -\frac{1}{ e }\right) \mid $
$=\left|\left( e -\frac{1}{ e }\right)-2 \int\limits_{1 / e }^{ e }\left(\frac{\ln x }{ x }\right) \cdot xdx -\left( e -\frac{1}{ e }\right)\right| $
$\left.=\mid-2[ x \ln x ]_{1 / e }^{ e }-\int\limits_{1 / e }^{ e } dx \right]|=|-2\left[\left( e +\frac{1}{ e }\right)-\left( e -\frac{1}{ e }\right)\right]|=| \frac{4}{ e } \mid=\frac{4}{ e }$
