The area (in sq. , units) of the region bound | vedclass

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(B) The curves are $x^2 = 1 - 2y$,$x = \frac{4-y^2}{4}$,and $x = \frac{y^2-4}{4}$.For the upper half plane $(y \ge 0)$,the region is bounded by $x = \

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$2$

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(B) The curves are $x^2 = 1 - 2y$,$x = \frac{4-y^2}{4}$,and $x = \frac{y^2-4}{4}$.For the upper half plane $(y \ge 0)$,the region is bounded by $x = \frac{4-y^2}{4}$ on the right,$x = \frac{y^2-4}{4}$ on the left,and $y = \frac{1-x^2}{2}$ from below.Due to symmetry about the $y$-axis,the area is $2 \int_{0}^{2} \left( \frac{4-y^2}{4} \right) dy - 2 \int_{0}^{1} \left( \frac{1-x^2}{2} \right) dx$.$= 2 \left[ y - \frac{y^3}{12} \right]_{0}^{2} - 2 \left[ \frac{x}{2} - \frac{x^3}{6} \right]_{0}^{1}$$= 2 \left( 2 - \frac{8}{12} \right) - 2 \left( \frac{1}{2} - \frac{1}{6} \right)$$= 2 \left( \frac{4}{3} \right) - 2 \left( \frac{1}{3} \right) = \frac{8}{3} - \frac{2}{3} = \frac{6}{3} = 2 \, sq. \, units$.

The area (in $sq. \, units$) of the region bounded by the curves $x^{2}+2y-1=0$,$y^{2}+4x-4=0$,and $y^{2}-4x-4=0$ in the upper half plane is $....$

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