The area of the figure bounded by the curves $y = ln x$ & $y = {(ln x)}^{2}$ is

e + 1

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e - 1

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3 - e

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1

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Solution

The correct option is B $3 - e$
Given curves, $y = ln x$ & $y = {(ln x)}^{2}$
Points of intersection of given curves are, $(1, 0)$ and $(e, 1)$
Hence required area is,
$A = \int_{1}^{e} (ln x - (ln x)^{2}) d x$
On solving it by parts we get
$A = [x ln x - x]_{1}^{e} - [x (ln x)^{2} - \int_{1}^{e} 2 ln x d x]$
$A = [x ln x - x]_{1}^{e} - {[x (ln x)^{2} - 2 x ln x + 2 x]}_{1}^{2}$
$A = {[x ln x - x - x (ln x)^{2} + 2 x ln x - 2 x]}_{1}^{2}$
$A = {| 3 x (ln x - 1) |}_{1}^{e} - | x (ln x)^{2} |_{1}^{e} = 3 - e$

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