The area of the region: R=x,y:5x2≤y≤2x2+9 is

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The area of the region:

$R = \{(x, y) : 5 x^{2} \leq y \leq 2 x^{2} + 9\}$ is

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$6 \sqrt{3}$ square units

$9 \sqrt{3}$ square units

$11 \sqrt{3}$ square units

$12 \sqrt{3}$ square units

answer is D.

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Detailed Solution

$5 x^{2} = 2 x^{2} + 9 \Rightarrow 3 x^{2} = 9$

$x^{2} = 3 \Rightarrow x = \pm \sqrt{3}$

Area= $2 \int_{0}^{\sqrt{3}} (2 x^{2} + 9 - 5 x^{2}) d x = 2 \int_{0}^{\sqrt{3}} (9 - 3 x^{2}) d x$

$A r e a = 2 {(9 x - x^{3})}_{0}^{\sqrt{3}} = 2 (9 \sqrt{3} - 3 \sqrt{3}) = 12 \sqrt{3}$

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