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

\frac{2}{e}

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\frac{4}{e}

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2 (e + \frac{1}{e})

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4 (e - \frac{1}{e})

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Solution

The correct option is B $\frac{4}{e}$
$y = {ln}^{2} x - 1$
$A = | \int_{1 / e}^{e} ({ln}^{2} x - 1) d x |$
$\int ({ln}^{2} x - 1) d x = ({ln}^{2} x) x - \int 2 ln x \frac{1}{x} . x d x - x$
$= x ({ln}^{2} x) - x - 2 [x ln x - x]$
$= x {ln}^{2} x - x - 2 x ln x + 2 x$
$= x {ln}^{2} x + x - 2 x l n x$
$A = | [x {ln}^{2} x + x - 2 x ln x]^{e} 1 / e | = | e - \frac{1}{e} + e - 1 / e - 2 [e + 1 / e] | = + 4 / e$

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