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 $y = ln x$ , within this region, is being removed, then find the area of the remaining region.

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Solution

$\Rightarrow \int_{0}^{2} ln x - e^{x}$
$= {[x ln x - x - e^{x}]}_{0}^{x}$
$= 2 ln 2 - 2 - e^{2} + 1$
$= 2 ln 2 - e^{2} - 1$
$\Rightarrow$ total area $= 4 = 2 ln 2 - e^{2} - 1$
remaining area $= 4 - 2 ln 2 - e^{2} - 1$
$= 3 - 2 ln 2 - e^{2}$

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Similar questions

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

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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y = 0, x = 2, y = 2

.if the area enclosed by the curve and

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

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x = 0, y = 0, x = 2, y = 2, y \leq e^{x}

and

y \geq ln x

, is

Q. Area of region bounded by

x = 0

y = 0

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y \leq e^{x}

y \geq l n x

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

⇒∫20lnx−ex=[xlnx−x−ex]20=2ln2−2−e2+1=2ln2−e2−1⇒ total area =4=2ln2−e2−1remaining area =4−2ln2−e2−1 =3−2ln2−e2

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 $y = ln x$ , within this region, is being removed, then find the area of the remaining region.

$\Rightarrow \int_{0}^{2} ln x - e^{x}$
$= {[x ln x - x - e^{x}]}_{0}^{x}$
$= 2 ln 2 - 2 - e^{2} + 1$
$= 2 ln 2 - e^{2} - 1$
$\Rightarrow$ total area $= 4 = 2 ln 2 - e^{2} - 1$
remaining area $= 4 - 2 ln 2 - e^{2} - 1$
$= 3 - 2 ln 2 - e^{2}$