The area enclosed by y 2 = 8 x and y = 2 x that lies outside the triangle formed by y = 2 x , x = 1 , y = 2 2 , is equal to

Intersection point of y 2 = 8 x & y = 2 x is 0 , 0 & 2 , 4 2 Area of ∆ A B C = 1 2 2 · 1 = 1 2 So required area = area between the curve y 2 = 8 x and the line y = 2 x - ar △ A B C = ∫ 0 4 8 x - 2 x d x - 1 2 = 32 2 3 - 8 2 - 1 2 = 13 2 6

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MathematicsArea Under CurvesArea bounded by two curvesMedium2 minPYQ_2022

MathematicsMediumsingle choice

The area enclosed by $y^{2} = 8 x$ and $y = \sqrt{2} x$ that lies outside the triangle formed by $y = \sqrt{2} x, x = 1, y = 2 \sqrt{2}$ , is equal to

Question diagram: The area enclosed by y 2 = 8 x and y = 2 x that lies outside

Options:

Answer:

Solution:

Intersection point of $y^{2} = 8 x & y = \sqrt{2} x$ is $(0, 0) & (2, 4 \sqrt{2})$

Area of $∆ A B C = \frac{1}{2} (\sqrt{2}) \cdot 1 = \frac{1}{\sqrt{2}}$

So required area $=$ area between the curve $y^{2} = 8 x$ and the line $y = \sqrt{2} x$ $- ar △ A B C$

$= [\int_{0}^{4} (\sqrt{8 x} - \sqrt{2} x) d x] - \frac{1}{\sqrt{2}}$

$= \frac{32 \sqrt{2}}{3} - 8 \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{13 \sqrt{2}}{6}$

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

⏱ 2mℹ️ Source: PYQ_2022

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