The area of the figure bounded by the curves $y = ln x$ and $y = (ln x)^{2}$ 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 C
$3 - e$

$A = \int_{1}^{e} (ln x - (ln x)^{2}) d x \int ln d x = x ln x - x$

$∴ A = (x ln x - x) |_{1}^{e} - [ln x (x ln x - x) |_{1}^{e} - \int_{1}^{e} (ln x - 1)]$

$A = [x ln x - x - x (ln x)^{2} + x ln x + x ln x - x - x]_{1}^{e}$

$A = [- x (ln x)^{2} + 3 x ln x - 3 x]_{1}^{e}$

$A = - e + 3 e - 3 e + 3 = 3 - e$

Suggest Corrections

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