The area bounded by the parabolas y=x^2 and y | vedclass

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(B) Given the parabolas $y=x^2$ and $y=1-x^2$.To find the intersection points,set $x^2 = 1-x^2$,which gives $2x^2 = 1$,so $x^2 = \frac{1}{2}$,implying

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$\frac{2 \sqrt{2}}{3}$

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(B) Given the parabolas $y=x^2$ and $y=1-x^2$.To find the intersection points,set $x^2 = 1-x^2$,which gives $2x^2 = 1$,so $x^2 = \frac{1}{2}$,implying $x = \pm \frac{1}{\sqrt{2}}$.The intersection points are $A\left(\frac{1}{\sqrt{2}}, \frac{1}{2}ight)$ and $C\left(-\frac{1}{\sqrt{2}}, \frac{1}{2}ight)$.The area of the shaded region is given by the integral of the upper curve minus the lower curve from $x = -\frac{1}{\sqrt{2}}$ to $x = \frac{1}{\sqrt{2}}$.Area $= \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} [(1-x^2) - x^2] dx = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (1-2x^2) dx$.Since the function is even,Area $= 2 \int_{0}^{\frac{1}{\sqrt{2}}} (1-2x^2) dx$.$= 2 \left[ x - \frac{2x^3}{3} ight]_{0}^{\frac{1}{\sqrt{2}}} = 2 \left[ \frac{1}{\sqrt{2}} - \frac{2}{3} \left( \frac{1}{\sqrt{2}} ight)^3 ight] = 2 \left[ \frac{1}{\sqrt{2}} - \frac{2}{3 \cdot 2 \sqrt{2}} ight] = 2 \left[ \frac{1}{\sqrt{2}} - \frac{1}{3 \sqrt{2}} ight]$.$= 2 \left[ \frac{3-1}{3 \sqrt{2}} ight] = 2 \left[ \frac{2}{3 \sqrt{2}} ight] = \frac{4}{3 \sqrt{2}} = \frac{2 \sqrt{2}}{3} 	ext{ sq units}$.

The area bounded by the parabolas $y=x^2$ and $y=1-x^2$ is equal to

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