Determine the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \) for the interval where they intersect.

To find the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \), we first need to determine the points of intersection. We can do this by setting the equations equal to each other: \[ x^2 = 4 - x^2 \] Adding \( x^2 \) to both sides gives: \[ 2x^2 = 4 \] Dividing by 2: \[ x^2 = 2 \] Taking the square root of both sides, we find: \[ x = \pm \sqrt{2} \] So, the curves intersect at the points \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). Next, we will find the area between the curves from \( x = -\sqrt{2} \) to \( x = \sqrt{2} \). The area \( A \) can be calculated using the integral: \[ A = \int_{-\sqrt{2}}^{\sqrt{2}} ((4 - x^2) - (x^2)) \, dx \] This simplifies to: \[ A = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) \, dx \] We can now compute the integral: \[ A = \left[ 4x - \frac{2}{3}x^3 \right]_{-\sqrt{2}}^{\sqrt{2}} \] Evaluating the antiderivative at the limits: 1. At \( x = \sqrt{2} \): \[ 4(\sqrt{2}) - \frac{2}{3}(\sqrt{2})^3 = 4\sqrt{2} - \frac{2}{3}(2\sqrt{2}) = 4\sqrt{2} - \frac{4\sqrt{2}}{3} = \frac{12\sqrt{2}}{3} - \frac{4\sqrt{2}}{3} = \frac{8\sqrt{2}}{3} \] 2. At \( x = -\sqrt{2} \): \[ 4(-\sqrt{2}) - \frac{2}{3}(-\sqrt{2})^3 = -4\sqrt{2} + \frac{4\sqrt{2}}{3} = \frac{-12\sqrt{2}}{3} + \frac{4\sqrt{2}}{3} = \frac{-8\sqrt{2}}{3} \] Now, combining these results: \[ A = \left(\frac{8\sqrt{2}}{3}\right) - \left(\frac{-8\sqrt{2}}{3}\right) = \frac{8\sqrt{2}}{3} + \frac{8\sqrt{2}}{3} = \frac{16\sqrt{2}}{3} \] Thus, the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \) from \( x = -\sqrt{2} \) to \( x = \sqrt{2} \) is: \[ \boxed{\frac{16\sqrt{2}}{3}} \]