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 \( y = 4 - x^2 \), we first need to determine their intersection points. Setting the two equations equal to each other: \[ x^2 = 4 - x^2 \] This simplifies to: \[ 2x^2 = 4 \implies x^2 = 2 \implies x = \pm \sqrt{2} \] Now, we will integrate to find the area between the curves. The area \( A \) between the curves from \( x = -2 \) to \( x = 2 \) can be calculated using the formula: \[ A = \int_{-2}^{2} ((4 - x^2) - x^2) \, dx \] This simplifies to: \[ A = \int_{-2}^{2} (4 - 2x^2) \, dx \] We can evaluate the integral: 1. Calculate the integral of \( 4 \): \[ \int 4 \, dx = 4x \] 2. Calculate the integral of \( -2x^2 \): \[ \int -2x^2 \, dx = -\frac{2}{3} x^3 \] Now we combine these results and evaluate from \( -2 \) to \( 2 \): \[ A = \left[ 4x - \frac{2}{3} x^3 \right]_{-2}^{2} \] Calculating at \( x = 2 \): \[ 4(2) - \frac{2}{3}(2^3) = 8 - \frac{2}{3}(8) = 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} \] Calculating at \( x = -2 \): \[ 4(-2) - \frac{2}{3}(-2^3) = -8 + \frac{2}{3}(8) = -8 + \frac{16}{3} = -\frac{24}{3} + \frac{16}{3} = -\frac{8}{3} \] Subtract these results: \[ A = \left( \frac{8}{3} - (-\frac{8}{3}) \right) = \frac{8}{3} + \frac{8}{3} = \frac{16}{3} \] Thus, the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \) from \( x = -2 \) to \( x = 2 \) is \[ \boxed{\frac{16}{3}} \]