The area bounded by curve $y = ln x, x = 1, x = 2$ and x-axis is

ln 4 e

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ln \frac{4}{e}

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ln \frac{e}{4}

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ln \frac{3}{e}

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Solution

The correct option is B $ln \frac{4}{e}$
Area bounded $\int_{01}^{2} l n x d x$
$\Rightarrow \int l n x d x = x l n x - x$
$[x l n x - x]_{1}^{2} \Rightarrow (2 l n 2 - 2) - (1 l n 1 - 1)$
$\Rightarrow 2 l n 2 - 2 - 1 l n 1 + 1$
$\Rightarrow 2 l n 2 - 1$
$\Rightarrow l n 4 - l n e$
$\Rightarrow l n (\frac{4}{e})$

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Similar questions

y = x (1 - l n x)

x = e^{- 1}

x = e

y = ln x, x = 1, x = 2

y = log x

x = 1

x = e

y = {log}_{e} x

x = e

Find the area bounded by the curve y = $e^{- x}$ , the X -axis and the Y-axis.

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