The area of the region S = x , y : y 2 ≤ 8 x , y ≥ 2 x , x ≥ 1 is

Plotting the region for the given inequalities y 2 ≤ 8 x , y ≥ 2 x , x ≤ 1 , we get Required area will be ∫ 1 4 2 2 x - 2 x d x = 2 2 x 3 2 3 / 2 1 4 - 2 x 2 2 1 4 = 4 2 3 8 - 1 - 2 3 16 - 1 = 28 2 3 - 15 2 2 = 11 2 6

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MathematicsArea Under CurvesArea bounded by two curvesMedium2 minPYQ_2022

MathematicsMediumsingle choice

The area of the region $S = \{(x, y) : y^{2} \leq 8 x, y \geq \sqrt{2} x, x \geq 1\}$ is

Question diagram: The area of the region S = x , y : y 2 ≤ 8 x , y ≥ 2 x , x ≥

Options:

Answer:

Solution:

Plotting the region for the given inequalities $y^{2} \leq 8 x, y \geq \sqrt{2} x, x \leq 1$ , we get

Required area will be $\int_{1}^{4} (2 \sqrt{2} \sqrt{x} - \sqrt{2} x) d x$

$= 2 \sqrt{2} {(\frac{x^{\frac{3}{2}}}{3 / 2})}_{1}^{4} - \sqrt{2} {(\frac{x^{2}}{2})}_{1}^{4}$

$= \frac{4 \sqrt{2}}{3} (8 - 1) - \frac{\sqrt{2}}{3} (16 - 1)$

$= \frac{28 \sqrt{2}}{3} - \frac{15 \sqrt{2}}{2} = \frac{11 \sqrt{2}}{6}$

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

⏱ 2mℹ️ Source: PYQ_2022

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