1. Problem from Seite 51 Nummer 8: Given the function $f(x) = -x^3 + 3x^2$, we are to split the area enclosed by the graph of $f$ and the x-axis by a vertical line so that the two resulting areas have equal size. 2. First, find the points where $f(x)$ intersects the x-axis by solving $f(x) = 0$: $$-x^3 + 3x^2 = 0 \Rightarrow x^2(-x + 3) = 0$$ This gives $x=0$ and $x=3$ as roots. 3. The enclosed area is between $x=0$ and $x=3$. Calculate the total area: $$A = \int_0^3 |f(x)| \, dx$$ Since $f(x) \geq 0$ on $[0,3]$ (check sign), we have: $$A = \int_0^3 (-x^3 + 3x^2) \, dx$$ 4. Compute the integral: $$\int_0^3 (-x^3 + 3x^2) \, dx = \left[-\frac{1}{4}x^4 + x^3\right]_0^3 = \left(-\frac{1}{4} \cdot 81 + 27\right) - 0 = -20.25 + 27 = 6.75$$ 5. We want to find $c$ in $[0,3]$ such that: $$\int_0^c (-x^3 + 3x^2) \, dx = \frac{6.75}{2} = 3.375$$ 6. Calculate the integral from 0 to $c$: $$\int_0^c (-x^3 + 3x^2) \, dx = -\frac{1}{4}c^4 + c^3 = 3.375$$ 7. Solve for $c$: $$-\frac{1}{4}c^4 + c^3 - 3.375 = 0$$ Multiply both sides by 4 to clear denominator: $$-c^4 + 4c^3 - 13.5 = 0$$ Or $$c^4 - 4c^3 + 13.5 = 0$$ 8. This quartic can be solved numerically. Approximate $c \approx 2.18$. --- Next, from Seite 54 Nummer 1 a and b: Problem a) Find the area enclosed between $f(x) = x^3 - 4x + 3$ and $g(x) = 3x - 3$ over $I = [-3, 2]$. 1. Find intersection points by solving $f(x) = g(x)$: $$x^3 - 4x + 3 = 3x - 3 \Rightarrow x^3 - 7x + 6 = 0$$ 2. Factor or find roots: Try $x=1$: $1 -7 +6=0$, root found. Divide polynomial by $(x-1)$: $$x^3 -7x +6 = (x-1)(x^2 + x -6)$$ 3. Solve quadratic: $$x^2 + x -6 = 0 \Rightarrow x = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$$ Roots: $x=2$ and $x=-3$. 4. So intersection points are at $x=-3, 1, 2$. 5. Determine which function is on top in each subinterval: - For $x$ in $[-3,1]$, test $x=0$: $f(0)=3$, $g(0)=-3$, so $f(x) > g(x)$. - For $x$ in $[1,2]$, test $x=1.5$: $f(1.5) = 1.5^3 - 4(1.5) + 3 = 3.375 - 6 + 3 = 0.375$ $g(1.5) = 3(1.5) - 3 = 4.5 - 3 = 1.5$ So $g(x) > f(x)$ on $[1,2]$. 6. Calculate area: $$A = \int_{-3}^1 (f(x) - g(x)) \, dx + \int_1^2 (g(x) - f(x)) \, dx$$ 7. Compute $f(x) - g(x)$: $$x^3 - 4x + 3 - (3x - 3) = x^3 - 7x + 6$$ 8. Compute $g(x) - f(x)$: $$3x - 3 - (x^3 - 4x + 3) = -x^3 + 7x - 6$$ 9. Integrate: $$\int (x^3 - 7x + 6) dx = \frac{1}{4}x^4 - \frac{7}{2}x^2 + 6x + C$$ $$\int (-x^3 + 7x - 6) dx = -\frac{1}{4}x^4 + \frac{7}{2}x^2 - 6x + C$$ 10. Calculate first integral: $$\left[\frac{1}{4}x^4 - \frac{7}{2}x^2 + 6x\right]_{-3}^1 = \left(\frac{1}{4} - \frac{7}{2} + 6\right) - \left(\frac{81}{4} - \frac{7}{2} \cdot 9 - 18\right) = (0.25 - 3.5 + 6) - (20.25 - 31.5 - 18) = 2.75 - (-29.25) = 32$$ 11. Calculate second integral: $$\left[-\frac{1}{4}x^4 + \frac{7}{2}x^2 - 6x\right]_1^2 = \left(-4 + 14 - 12\right) - \left(-\frac{1}{4} + \frac{7}{2} - 6\right) = (-2) - (-3.25) = 1.25$$ 12. Total area: $$A = 32 + 1.25 = 33.25$$ --- Problem b) Find the area enclosed between $f(x) = x^2 - 8x + 14$ and $g(x) = -x^2 + 6x - 6$ over $I = [2,5]$. 1. Find intersection points by solving $f(x) = g(x)$: $$x^2 - 8x + 14 = -x^2 + 6x - 6 \Rightarrow 2x^2 - 14x + 20 = 0$$ 2. Simplify: $$x^2 - 7x + 10 = 0$$ 3. Solve quadratic: $$x = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm 3}{2}$$ Roots: $x=2$ and $x=5$. 4. Determine which function is on top in $[2,5]$: Test $x=3$: $f(3) = 9 - 24 + 14 = -1$ $g(3) = -9 + 18 - 6 = 3$ So $g(x) > f(x)$ on $[2,5]$. 5. Calculate area: $$A = \int_2^5 (g(x) - f(x)) \, dx = \int_2^5 (-x^2 + 6x - 6 - (x^2 - 8x + 14)) \, dx = \int_2^5 (-2x^2 + 14x - 20) \, dx$$ 6. Integrate: $$\int (-2x^2 + 14x - 20) dx = -\frac{2}{3}x^3 + 7x^2 - 20x + C$$ 7. Evaluate: $$\left(-\frac{2}{3}x^3 + 7x^2 - 20x\right)_2^5 = \left(-\frac{2}{3} \cdot 125 + 7 \cdot 25 - 100\right) - \left(-\frac{2}{3} \cdot 8 + 7 \cdot 4 - 40\right) = (-\frac{250}{3} + 175 - 100) - (-\frac{16}{3} + 28 - 40) = (-83.33 + 75) - (-5.33 - 12) = (-8.33) - (-17.33) = 9$$ --- Finally, from Seite 55 Nummer 6: Problem: Express the areas $A_1, A_2, ..., A_6$ of the colored regions as integrals without determining the function terms of $f$ and $g$. 1. The areas $A_i$ are defined as integrals of the absolute difference between $f(x)$ and $g(x)$ over their respective intervals. 2. For each area $A_i$ over interval $[a_i, b_i]$ where $f(x) \geq g(x)$ or vice versa, write: $$A_i = \int_{a_i}^{b_i} |f(x) - g(x)| \, dx = \int_{a_i}^{b_i} (\text{upper function} - \text{lower function}) \, dx$$ 3. Since the exact functions are not given, the answer is the set of integrals: $$A_1 = \int_{x_0}^{x_1} |f(x) - g(x)| \, dx$$ $$A_2 = \int_{x_1}^{x_2} |f(x) - g(x)| \, dx$$ $$\ldots$$ $$A_6 = \int_{x_5}^{x_6} |f(x) - g(x)| \, dx$$ where $x_0, x_1, ..., x_6$ are the x-coordinates of the boundaries of the colored areas. --- "slug": "area split and integrals", "subject": "calculus", "desmos": {"latex": "f(x)=-x^3+3x^2","features": {"intercepts": true,"extrema": true}}, "q_count": 3