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 identify which curve is on top and which is on the bottom in the given interval. 1. **Identify the curves**: The curve \( y = 4 - x^2 \) is a downward-opening parabola, while \( y = x^2 \) is an upward-opening parabola. 2. **Finding the points of intersection**: To find where the curves intersect, we set them equal to each other: \[ x^2 = 4 - x^2 \] This simplifies to: \[ 2x^2 = 4 \quad \Rightarrow \quad x^2 = 2 \quad \Rightarrow \quad x = \pm \sqrt{2} \] 3. **Determine the area**: To find the area between the curves from \( x = -2 \) to \( x = 2 \), we can integrate the top curve minus the bottom curve. Since \( 4 - x^2 \) is above \( x^2 \) in this interval, the area \( A \) can be written as: \[ A = \int_{-2}^{2} \left[(4 - x^2) - (x^2)\right] \, dx = \int_{-2}^{2} (4 - 2x^2) \, dx \] 4. **Calculating the integral**: \[ A = \int_{-2}^{2} (4 - 2x^2) \, dx \] This can be computed in two separate parts: \[ A = \left[ 4x - \frac{2}{3}x^3 \right]_{-2}^{2} \] Now, evaluate at the bounds: \[ = \left( 4(2) - \frac{2}{3}(2^3) \right) - \left( 4(-2) - \frac{2}{3}(-2)^3 \right) \] \[ = \left( 8 - \frac{16}{3} \right) - \left( -8 + \frac{16}{3} \right) \] \[ = (8 - \frac{16}{3}) + (8 - \frac{16}{3}) = 16 - \frac{32}{3} = \frac{48}{3} - \frac{32}{3} = \frac{16}{3} \] Thus, the total area between the curves \( y = x^2 \) and \( y = 4 - x^2 \) from \( x = -2 \) to \( x = 2 \) is \( \frac{16}{3} \) square units.