Area Parabolas Cd33D2

1. **State the problem:** We need to find the area bounded by the two parabolas given by $$y = 2 - x^2$$ and $$y = x^2$$. 2. **Find the points of intersection:** Set the two equations equal to find the limits of integration: $$2 - x^2 = x^2$$ $$2 = 2x^2$$ $$x^2 = 1$$ $$x = \pm 1$$ So, the parabolas intersect at $$x = -1$$ and $$x = 1$$. 3. **Set up the integral for the area:** The area bounded between the curves is the integral of the difference of the top function and the bottom function from $$-1$$ to $$1$$. Here, $$y = 2 - x^2$$ is above $$y = x^2$$ in this interval. So, the area $$A$$ is: $$A = \int_{-1}^{1} \left[(2 - x^2) - x^2\right] dx = \int_{-1}^{1} (2 - 2x^2) dx$$ 4. **Integrate:** $$\int (2 - 2x^2) dx = 2x - \frac{2x^3}{3} + C$$ 5. **Evaluate the definite integral:** $$A = \left[2x - \frac{2x^3}{3}\right]_{-1}^{1} = \left(2(1) - \frac{2(1)^3}{3}\right) - \left(2(-1) - \frac{2(-1)^3}{3}\right)$$ $$= \left(2 - \frac{2}{3}\right) - \left(-2 + \frac{2}{3}\right) = \left(\frac{6}{3} - \frac{2}{3}\right) - \left(-\frac{6}{3} + \frac{2}{3}\right) = \frac{4}{3} - \left(-\frac{4}{3}\right) = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$$ 6. **Final answer:** The area bounded by the parabolas is $$\boxed{\frac{8}{3}}$$.