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:

$2 (1 + 2 {ln}^{2} 2)$

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$2 (2 ln 2 - 1)$

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$(2 ln 2 - 1)$

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$1 + 2 {ln}^{2} 2$

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

The correct option is B 2(2ln2−1) Area =∫21lnxdx+∫21lnydy=2∫21lnx=2[xlnx−x]21=2[2ln2−2+1]=4ln2−2=2(2ln2−1)

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 $= \int_{1}^{2} ln x d x + \int_{1}^{2} ln y d y = 2 \int_{1}^{2} ln x = 2 {[x ln x - x]}_{1}^{2} = 2 [2 ln 2 - 2 + 1] = 4 ln 2 - 2 = 2 (2 ln 2 - 1)$