Area Parabola 533D4B

1. **State the problem:** We need to find the total area between the curve $$f(x) = -x^2 + 4$$ and the x-axis from $$x = -1$$ to $$x = 3$$. 2. **Understand the curve:** The function is a downward-opening parabola with vertex at $$(0,4)$$ and roots at $$x = -2$$ and $$x = 2$$ because $$-x^2 + 4 = 0 \implies x^2 = 4 \implies x = \pm 2$$. 3. **Identify intervals where the curve is above or below the x-axis:** - From $$x = -1$$ to $$x = 2$$, the curve is above the x-axis (since $$f(x) > 0$$). - From $$x = 2$$ to $$x = 3$$, the curve is below the x-axis (since $$f(x) < 0$$). 4. **Formula for area between curve and x-axis:** The total area is the sum of the absolute values of the definite integrals over these intervals: $$\text{Area} = \int_{-1}^{2} f(x) \, dx - \int_{2}^{3} f(x) \, dx$$ Note the minus sign in the second integral because the curve is below the x-axis there. 5. **Calculate the integrals:** $$\int f(x) \, dx = \int (-x^2 + 4) \, dx = -\frac{x^3}{3} + 4x + C$$ 6. **Evaluate the first integral:** $$\int_{-1}^{2} (-x^2 + 4) \, dx = \left[-\frac{x^3}{3} + 4x\right]_{-1}^{2} = \left(-\frac{8}{3} + 8\right) - \left(\frac{1}{3} - 4\right) = \left(\frac{16}{3}\right) - \left(-\frac{11}{3}\right) = \frac{16}{3} + \frac{11}{3} = \frac{27}{3} = 9$$ 7. **Evaluate the second integral:** $$\int_{2}^{3} (-x^2 + 4) \, dx = \left[-\frac{x^3}{3} + 4x\right]_{2}^{3} = \left(-\frac{27}{3} + 12\right) - \left(-\frac{8}{3} + 8\right) = (-9 + 12) - (-\frac{8}{3} + 8) = 3 - \left(\frac{16}{3}\right) = 3 - \frac{16}{3} = \frac{9}{3} - \frac{16}{3} = -\frac{7}{3}$$ 8. **Take absolute value for area below x-axis:** $$\left| -\frac{7}{3} \right| = \frac{7}{3}$$ 9. **Total area:** $$9 + \frac{7}{3} = \frac{27}{3} + \frac{7}{3} = \frac{34}{3}$$ **Final answer:** The total area between the curve and the x-axis from $$x = -1$$ to $$x = 3$$ is $$\boxed{\frac{34}{3}}$$ units squared.