Consider the region formed by the lines x = 0, y = 0, x = 2, y = 2. Area enclosed by the curves $y = e^{x}$ and $y = ln x$ , within this region is removed, then the area of the remaining region is

$2 (1 + 2 ln)$

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$2 (2 ln 2 - 1)$

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$(2 ln 2 - 1)$

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$1 + 2 ln$

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Solution

The correct option is B
$2 (2 ln 2 - 1)$

Area $= \int_{1}^{2} ln x d x + \int_{1}^{2} ln y d y$

$= 2 \int_{1}^{2} ln x = 2 [x ln x - x]_{1}^{2} = 2 [2 ln 2 - 2 + 1]$

$= 4 ln 2 - 2 = 2 (2 ln 2 - 1)$

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Similar questions

Q. Consider the origin formed by the lines

x = 0, y = 0, x = 2, y = 2

. If the area enclosed by the curves

y = e^{x}

and

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, within this region, is being removed, then find the area of the remaining region.

Consider the region formed by the lines $x = 0, y =$ $0, x = 2, y = 2$ . Area enclosed by the curves $y = e^{x}$ and $y = ln x$ , within this region is removed, then the area of the remaining region is:

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.if the area enclosed by the curve and

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Consider the region formed by the lines x = 0, y = 0, x = 2, y = 2. Area enclosed by the curves y=ex and y=ln x, within this region is removed, then the area of the remaining region is

The correct option is B 2(2 ln 2−1) Area =∫21 ln x dx+∫21 ln y dy =2∫21 ln x=2[x ln x−x]21=2[2 ln 2−2+1] =4 ln 2−2=2(2 ln 2−1)

Consider the region formed by the lines x = 0, y = 0, x = 2, y = 2. Area enclosed by the curves $y = e^{x}$ and $y = ln x$ , within this region is removed, then the area of the remaining region is

The correct option is B
$2 (2 ln 2 - 1)$

Area $= \int_{1}^{2} ln x d x + \int_{1}^{2} ln y d y$

$= 2 \int_{1}^{2} ln x = 2 [x ln x - x]_{1}^{2} = 2 [2 ln 2 - 2 + 1]$

$= 4 ln 2 - 2 = 2 (2 ln 2 - 1)$