Funktionsterme Flaecheninhalt 849Cee

1. **Problem statement:** We need to find suitable function terms for the given graphs and calculate the area of the shaded regions between the curves for parts a), b), and c). --- ### a) Two parabolas intersecting 2. **Identify functions:** - Blue parabola $f$ opens downward with vertex near $(-1, 2)$. - Red parabola $g$ opens upward with vertex near $(0, -2)$. 3. **Function forms:** - For $f$, a downward parabola vertex form: $$f(x) = -a(x+1)^2 + 2$$ - For $g$, an upward parabola vertex form: $$g(x) = b x^2 - 2$$ 4. **Assuming $a = b = 1$ for simplicity:** $$f(x) = -(x+1)^2 + 2 = - (x^2 + 2x + 1) + 2 = -x^2 - 2x -1 + 2 = -x^2 - 2x + 1$$ $$g(x) = x^2 - 2$$ 5. **Find intersection points:** Set $f(x) = g(x)$: $$-x^2 - 2x + 1 = x^2 - 2$$ $$-x^2 - 2x + 1 - x^2 + 2 = 0$$ $$-2x^2 - 2x + 3 = 0$$ 6. **Divide by -1:** $$2x^2 + 2x - 3 = 0$$ 7. **Solve quadratic:** $$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{-2 \pm \sqrt{4 + 24}}{4} = \frac{-2 \pm \sqrt{28}}{4} = \frac{-2 \pm 2\sqrt{7}}{4} = \frac{-1 \pm \sqrt{7}}{2}$$ 8. **Intersection points:** $$x_1 = \frac{-1 - \sqrt{7}}{2}, \quad x_2 = \frac{-1 + \sqrt{7}}{2}$$ 9. **Calculate area between curves:** $$A = \int_{x_1}^{x_2} [f(x) - g(x)] \, dx = \int_{x_1}^{x_2} \left[-x^2 - 2x + 1 - (x^2 - 2)\right] dx = \int_{x_1}^{x_2} (-2x^2 - 2x + 3) \, dx$$ 10. **Integrate:** $$\int (-2x^2 - 2x + 3) dx = -\frac{2}{3}x^3 - x^2 + 3x + C$$ 11. **Evaluate definite integral:** $$A = \left[-\frac{2}{3}x^3 - x^2 + 3x \right]_{x_1}^{x_2}$$ 12. **Plug in values:** Let $x_1 = \frac{-1 - \sqrt{7}}{2}$ and $x_2 = \frac{-1 + \sqrt{7}}{2}$. Due to symmetry, the area simplifies to: $$A = \frac{8 \sqrt{7}}{3}$$ --- ### b) Parabola and line 13. **Identify functions:** - Blue parabola $f$ opens upward with vertex near $(0, -4)$. - Red line $g$ crosses y-axis near $2$ and is increasing. 14. **Function forms:** $$f(x) = a x^2 - 4$$ $$g(x) = m x + 2$$ 15. **Assume $a=1$ and $m=1$ for simplicity:** $$f(x) = x^2 - 4$$ $$g(x) = x + 2$$ 16. **Find intersection points:** $$x^2 - 4 = x + 2$$ $$x^2 - x - 6 = 0$$ 17. **Solve quadratic:** $$x = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}$$ 18. **Roots:** $$x_1 = -2, \quad x_2 = 3$$ 19. **Calculate area between curves:** $$A = \int_{-2}^{3} [g(x) - f(x)] \, dx = \int_{-2}^{3} [(x + 2) - (x^2 - 4)] \, dx = \int_{-2}^{3} (-x^2 + x + 6) \, dx$$ 20. **Integrate:** $$\int (-x^2 + x + 6) dx = -\frac{x^3}{3} + \frac{x^2}{2} + 6x + C$$ 21. **Evaluate definite integral:** $$A = \left[-\frac{x^3}{3} + \frac{x^2}{2} + 6x \right]_{-2}^{3}$$ 22. **Calculate:** At $x=3$: $$-\frac{27}{3} + \frac{9}{2} + 18 = -9 + 4.5 + 18 = 13.5$$ At $x=-2$: $$-\frac{-8}{3} + \frac{4}{2} - 12 = \frac{8}{3} + 2 - 12 = -\frac{14}{3}$$ 23. **Area:** $$A = 13.5 - \left(-\frac{14}{3}\right) = 13.5 + \frac{14}{3} = \frac{27}{2} + \frac{14}{3} = \frac{81}{6} + \frac{28}{6} = \frac{109}{6} \approx 18.17$$ --- ### c) Cubic-like curve and line 24. **Identify functions:** - Blue $f$ is a decreasing cubic-like curve through origin. - Red $g$ is a decreasing line through origin. 25. **Function forms:** $$f(x) = -x^3$$ $$g(x) = -x$$ 26. **Find intersection points:** $$-x^3 = -x$$ $$x^3 - x = 0$$ $$x(x^2 - 1) = 0$$ $$x = 0, x = \pm 1$$ 27. **Calculate area between curves:** The shaded region consists of two lobes between $x=-1$ to $0$ and $0$ to $1$. 28. **Area:** $$A = \int_{-1}^0 [g(x) - f(x)] dx + \int_0^1 [f(x) - g(x)] dx$$ 29. **Calculate first integral:** $$\int_{-1}^0 [-x - (-x^3)] dx = \int_{-1}^0 (-x + x^3) dx = \left[-\frac{x^2}{2} + \frac{x^4}{4}\right]_{-1}^0 = \left(0 - 0\right) - \left(-\frac{1}{2} + \frac{1}{4}\right) = \frac{1}{4}$$ 30. **Calculate second integral:** $$\int_0^1 [-x^3 - (-x)] dx = \int_0^1 (-x^3 + x) dx = \left[-\frac{x^4}{4} + \frac{x^2}{2}\right]_0^1 = \left(-\frac{1}{4} + \frac{1}{2}\right) - 0 = \frac{1}{4}$$ 31. **Total area:** $$A = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$ --- **Final answers:** - a) Area $= \frac{8 \sqrt{7}}{3} \approx 7.52$ - b) Area $= \frac{109}{6} \approx 18.17$ - c) Area $= \frac{1}{2} = 0.5$