The area (in sq. units) bounded by the curves | vedclass

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(A) The 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 $y^2 = 1$, implying

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The 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 $y^2 = 1$, implying $y = \pm 1$.The area is symmetric about the $x$-axis.$	ext{Required Area} = 2 \int_{-1}^{1} (x_{	ext{right}} - x_{	ext{left}}) dy = 2 \int_{-1}^{1} ((3-2y^2) - y^2) dy$$= 2 \int_{-1}^{1} (3-3y^2) dy = 6 \int_{-1}^{1} (1-y^2) dy$$= 6 \left[ y - \frac{y^3}{3} ight]_{-1}^{1} = 6 \left( (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) ight)$$= 6 \left( \frac{2}{3} - (-\frac{2}{3}) ight) = 6 \left( \frac{4}{3} ight) = 8 	ext{ sq. units.}$

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

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