55 Multiple Choice 1 point Find the area between the curves: \( y=x^{2}-2 \) and \( y=-(x-1)^{2}+3 \) \( 1 / 4 \) \( 1 / 2 \) None of these \( 1 / 8 \)

To find the area between the curves \( y = x^2 - 2 \) and \( y = -(x-1)^2 + 3 \), follow these steps: 1. **Find the Points of Intersection:** Set \( x^2 - 2 = -(x-1)^2 + 3 \). \[ x^2 - 2 = - (x^2 - 2x + 1) + 3 \\ x^2 - 2 = -x^2 + 2x - 1 + 3 \\ x^2 - 2 = -x^2 + 2x + 2 \\ 2x^2 - 2x - 4 = 0 \\ x^2 - x - 2 = 0 \\ x = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \\ x = 2 \text{ or } x = -1 \] 2. **Determine Which Function is on Top:** Choose a test point between \( x = -1 \) and \( x = 2 \), say \( x = 0 \). \[ y_1 = 0^2 - 2 = -2 \\ y_2 = -(0-1)^2 + 3 = 2 \] Since \( y_2 > y_1 \), \( y_2 \) is the upper function. 3. **Set Up the Integral for the Area:** \[ \text{Area} = \int_{-1}^{2} [y_2 - y_1] \, dx = \int_{-1}^{2} [- (x-1)^2 + 3 - (x^2 - 2)] \, dx \] Simplify the integrand: \[ - (x^2 - 2x + 1) + 3 - x^2 + 2 = -2x^2 + 2x + 4 \] 4. **Evaluate the Integral:** \[ \int_{-1}^{2} (-2x^2 + 2x + 4) \, dx = \left[ -\frac{2}{3}x^3 + x^2 + 4x \right]_{-1}^{2} \] Calculate the definite integral: \[ \left(-\frac{16}{3} + 4 + 8 \right) - \left(\frac{2}{3} + 1 - 4\right) = \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{27}{3} = 9 \] The calculated area is **9**, which does not match any of the provided options. **Answer:** None of these