The area (in square units) of the region boun | vedclass

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(D) Given curves are $x=y^2$ and $x=3-2y^2$.To find the points of intersection, set $y^2 = 3-2y^2$, which gives $3y^2 = 3$, so $y^2 = 1$, meaning $y =

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$4$

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(D) Given curves are $x=y^2$ and $x=3-2y^2$.To find the points of intersection, set $y^2 = 3-2y^2$, which gives $3y^2 = 3$, so $y^2 = 1$, meaning $y = \pm 1$.When $y = 1$, $x = 1$. When $y = -1$, $x = 1$.Thus, the points of intersection are $(1, 1)$ and $(1, -1)$.The area is symmetric about the $x$-axis.$	ext{Required Area} = 2 \int_0^1 (x_2 - x_1) dy$$= 2 \int_0^1 ((3-2y^2) - y^2) dy$$= 2 \int_0^1 (3-3y^2) dy$$= 2 [3y - y^3]_0^1$$= 2 [3(1) - (1)^3 - (0)]$$= 2 [3 - 1] = 2 	imes 2 = 4$ square units.

The area (in square units) of the region bounded by the curves $x=y^2$ and $x=3-2y^2$ is

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