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

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

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$\frac{4}{3}$

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(C) Given curves are $y = -2x^2$ and $y = 1 - 3x^2$.To find the points of intersection,set the equations equal to each other:$-2x^2 = 1 - 3x^2$$x^2 = 1$$x = \pm 1$.Since the curves are symmetric about the $y$-axis,the area $A$ is given by:$A = 2 \int_{0}^{1} (y_{upper} - y_{lower}) dx$Here,$y_{upper} = 1 - 3x^2$ and $y_{lower} = -2x^2$.$A = 2 \int_{0}^{1} ((1 - 3x^2) - (-2x^2)) dx$$A = 2 \int_{0}^{1} (1 - x^2) dx$$A = 2 [x - \frac{x^3}{3}]_{0}^{1}$$A = 2 (1 - \frac{1}{3}) = 2 (\frac{2}{3}) = \frac{4}{3}$ square units.

The area (in square units) of the region bounded by the curves $y + 2x^2 = 0$ and $y + 3x^2 = 1$ is equal to

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