Find the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \) from \( x = -2 \) to \( x = 2 \).

To find the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \) from \( x = -2 \) to \( x = 2 \), we first need to determine where these two curves intersect. We set the equations equal to each other: \[ x^2 = 4 - x^2 \] Combining like terms, we get: \[ 2x^2 = 4 \] Dividing both sides by 2: \[ x^2 = 2 \] Taking the square root of both sides gives us the x-coordinates of the points of intersection: \[ x = \pm \sqrt{2} \] Next, we can set up the integral for the area between the curves. The upper curve is \( y = 4 - x^2 \) and the lower curve is \( y = x^2 \). We can write the area \( A \) as: \[ A = \int_{-2}^{2} \left( (4 - x^2) - (x^2) \right) \, dx \] This simplifies to: \[ A = \int_{-2}^{2} (4 - 2x^2) \, dx \] Now, we can compute the integral: \[ A = \int_{-2}^{2} 4 \, dx - \int_{-2}^{2} 2x^2 \, dx \] Calculating each integral: 1. For \( \int_{-2}^{2} 4 \, dx \): \[ = 4[x]_{-2}^{2} = 4(2 - (-2)) = 4(4) = 16 \] 2. For \( \int_{-2}^{2} 2x^2 \, dx \): \[ = 2 \left[ \frac{x^3}{3} \right]_{-2}^{2} = 2 \left( \frac{(2)^3}{3} - \frac{(-2)^3}{3} \right) = 2 \left( \frac{8}{3} - \frac{-8}{3} \right) = 2 \left( \frac{16}{3} \right) = \frac{32}{3} \] Now, substituting back into our area calculation: \[ A = 16 - \frac{32}{3} \] To combine these, we convert 16 to a fraction with a denominator of 3: \[ = \frac{48}{3} - \frac{32}{3} = \frac{16}{3} \] Thus, the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \) from \( x = -2 \) to \( x = 2 \) is: \[ \boxed{\frac{16}{3}} \]