The area of the region R = {(x, y) : 5x^2 leq | vedclass

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(B) To find the area of the region $R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$, we first find the intersection points of the curves $y = 5x^2$ and $y

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) To find the area of the region $R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$, we first find the intersection points of the curves $y = 5x^2$ and $y = 2x^2 + 9$.Setting $5x^2 = 2x^2 + 9$, we get $3x^2 = 9$, which implies $x^2 = 3$, so $x = \pm \sqrt{3}$.The region is symmetric about the $y$-axis.The required area is given by the integral:Area $= \int_{-\sqrt{3}}^{\sqrt{3}} ((2x^2 + 9) - 5x^2) dx$$= 2 \int_{0}^{\sqrt{3}} (9 - 3x^2) dx$$= 2 [9x - x^3]_{0}^{\sqrt{3}}$$= 2 [9\sqrt{3} - (\sqrt{3})^3]$$= 2 [9\sqrt{3} - 3\sqrt{3}]$$= 2 [6\sqrt{3}] = 12\sqrt{3} 	ext{ square units.}$

The area of the region $R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$ is ........ $\text{square units}$. (in $\sqrt{3}$)

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