Area Bounded 353240

1. **State the problem:** We need to find the area of the region bounded by the line $$y = x$$ and the parabola $$y = 6 - x^2$$. 2. **Find the points of intersection:** Set the two functions equal to find the limits of integration. $$x = 6 - x^2$$ Rearranging: $$x^2 + x - 6 = 0$$ 3. **Solve the quadratic equation:** $$x^2 + x - 6 = 0$$ Using the quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$$ So the roots are: $$x = \frac{-1 + 5}{2} = 2$$ $$x = \frac{-1 - 5}{2} = -3$$ 4. **Set up the integral for the area:** The parabola is above the line between $$x = -3$$ and $$x = 2$$, so the area is: $$\text{Area} = \int_{-3}^{2} \big[(6 - x^2) - x\big] \, dx = \int_{-3}^{2} (6 - x^2 - x) \, dx$$ 5. **Integrate:** $$\int (6 - x^2 - x) \, dx = 6x - \frac{x^3}{3} - \frac{x^2}{2} + C$$ 6. **Evaluate the definite integral:** $$\text{Area} = \left[6x - \frac{x^3}{3} - \frac{x^2}{2}\right]_{-3}^{2}$$ Calculate at $$x=2$$: $$6(2) - \frac{2^3}{3} - \frac{2^2}{2} = 12 - \frac{8}{3} - 2 = 12 - 2.6667 - 2 = 7.3333$$ Calculate at $$x=-3$$: $$6(-3) - \frac{(-3)^3}{3} - \frac{(-3)^2}{2} = -18 - \frac{-27}{3} - \frac{9}{2} = -18 + 9 - 4.5 = -13.5$$ 7. **Subtract:** $$7.3333 - (-13.5) = 7.3333 + 13.5 = 20.8333$$ 8. **Final answer:** $$\boxed{\frac{125}{6} \approx 20.8333}$$ This is the area of the region bounded by the line and the parabola.