The area enclosed by y - x - and y -1- x - is in sq- units A- -2B- 1 C- 1-2D- 2

The correct option is D 12The point of intersection of the two curves, |x| = 1 |x| ⇒ 2|x| =1 ⇒ x = ± 1 2 nbsp; So, the required area under the curve is, A = ...

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Standard XII

Mathematics

Area between x=g(y) and y Axis

The area encl...

Question

The area enclosed by $y = | x |$ and $y = 1 - | x |$ is (in sq. units)

\frac{1}{2}

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\sqrt{2}

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1

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\frac{1}{\sqrt{2}}

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Solution

The correct option is A $\frac{1}{2}$
The point of intersection of the two curves,
$| x | = 1 - | x | \Rightarrow 2 | x | = 1 \Rightarrow x = \pm \frac{1}{2}$
So, the required area under the curve is,
$A = 2 0.5 \int 0 (1 - x - x) d x = 2 {[x - x^{2}]}_{0}^{2} = 2 [\frac{1}{2} - \frac{1}{4}] = \frac{1}{2}$

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The area enclosed by y=|x| and y=1−|x| is (in sq. units)

The correct option is A 12The point of intersection of the two curves, |x|=1−|x|⇒2|x|=1⇒x=±12 So, the required area under the curve is, A=20.5∫0(1−x−x) dx =2[x−x2]0.50 =2[12−14]=12

The area enclosed by $y = | x |$ and $y = 1 - | x |$ is (in sq. units)