The area (in square units) bounded by the cur | vedclass

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Question

(C) Given curves are $x = -2y^2$ and $x = 1 - 3y^2$.To find the intersection points,set the equations equal to each other:$-2y^2 = 1 - 3y^2$$y^2 = 1$$

Vedclass · Accepted Answer

$4/3$

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(C) Given curves are $x = -2y^2$ and $x = 1 - 3y^2$.To find the intersection points,set the equations equal to each other:$-2y^2 = 1 - 3y^2$$y^2 = 1$$y = \pm 1$When $y = 1$,$x = -2(1)^2 = -2$. When $y = -1$,$x = -2(-1)^2 = -2$.The intersection points are $(-2, 1)$ and $(-2, -1)$.The area $A$ is given by the integral with respect to $y$ from $-1$ to $1$ of the right curve minus the left curve:$A = \int_{-1}^{1} ((1 - 3y^2) - (-2y^2)) \, dy$$A = \int_{-1}^{1} (1 - y^2) \, dy$Since the function is even,$A = 2 \int_{0}^{1} (1 - y^2) \, dy$$A = 2 [y - \frac{y^3}{3}]_{0}^{1}$$A = 2 (1 - \frac{1}{3}) = 2 (\frac{2}{3}) = \frac{4}{3}$ square units.

The area (in square units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ is

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