Area lying between the curves y2-2 x and y-x isA- s- sq- unitsB- 1- s sq- unitsC- 1-A sq- unitsD- -4 sq- units

The correct option is A 23 sq. units nbsp; Point of intersection of the curves : x^2=2x ⇒ x=0,2 Therefore, the curves intersect at (0,0) and (2,2). Required ...

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Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to Be Perpendicular

Area lying be...

Question

Area lying between the curves $y^{2} = 2 x$ and $y = x$ is

\frac{2}{3}

sq. units

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\frac{1}{3}

sq. units

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\frac{1}{4}

sq. units

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\frac{3}{4}

sq. units

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Solution

The correct option is A $\frac{2}{3}$ sq. units

Point of intersection of the curves :
$x^{2} = 2 x$
$\Rightarrow x = 0, 2$
Therefore, the curves intersect at $(0, 0)$ and $(2, 2) .$
$Required area = 2 \int 0 (\sqrt{2 x} - x) d x$
$= \frac{2}{3}$ sq. units

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Similar questions

The area between the curves y = x² and y = x³ is:

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and

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Q. The area bounded by the curves y = sin x between the ordinates x = 0, x = π and the x-axis is
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Area lying between the curves y2=2x and y=x is

The correct option is A 23 sq. units Point of intersection of the curves : x2=2x ⇒x=0,2 Therefore, the curves intersect at (0,0) and (2,2). Required area=2∫0(√2x−x)dx =23 sq. units

Area lying between the curves $y^{2} = 2 x$ and $y = x$ is

The correct option is A $\frac{2}{3}$ sq. units

Point of intersection of the curves :
$x^{2} = 2 x$
$\Rightarrow x = 0, 2$
Therefore, the curves intersect at $(0, 0)$ and $(2, 2) .$
$Required area = 2 \int 0 (\sqrt{2 x} - x) d x$
$= \frac{2}{3}$ sq. units