«K12/K13» 14. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 14. Hausaufgabe

0.0.1.1 ↑ Analysis-Buch Seite 37, Aufgabe 32

\mathrm{f}(x) := \begin{cases} {} \frac{1}{2}x + \frac{1}{2} & \text{f"ur } x \in \left[0, 3\right]; \\ {} -x + 5 & \text{f"ur } x \in \left]3, 4\right]; \end{cases}$\text{f}\left(x\right):=\left\{\begin{array}{cc}\frac{1}{2}x+\frac{1}{2}\hfill & \text{f”ur }x\in \left[0,3\right];\hfill \\ -x+5\hfill & \text{f”ur }x\in \left]3,4\right];\hfill \end{array}\right.$

⇒ \mathrm{F}(x) = \begin{cases} {} \frac{1}{4}x^2 + \frac{1}{2}x & \text{f"ur } x \in \left[0, 3\right]; \\ {} \frac{15}{4} & \text{f"ur } x = 3; \\ {} -\frac{1}{2}x^2 + 5x - \frac{27}{4} & \text{f"ur } x \in \left]3, 4\right]; \end{cases}$\text{F}\left(x\right)=\left\{\begin{array}{cc}\frac{1}{4}{x}^2+\frac{1}{2}x\hfill & \text{f”ur }x\in \left[0,3\right];\hfill \\ \frac{15}{4}\hfill & \text{f”ur }x=3;\hfill \\ -\frac{1}{2}{x}^2+5x-\frac{27}{4}\hfill & \text{f”ur }x\in \left]3,4\right];\hfill \end{array}\right.$

\renewcommand{\arraystretch}{1.5} \left.\begin{array}{l} {} \mathrm{F} \text{ stetig in } D_{\mathrm{f}}; \\ {} \left.\begin{array}{l} {} \lim\limits_{x \to 3-} \mathrm{F}'(x) = 2 \\ {} \lim\limits_{x \to 3+} \mathrm{F}'(x) = 2 {} \end{array}\right\} = {} \mathrm{F}'(3) = \mathrm{f}(x); \end{array}\right\} \Rightarrow \mathrm{F}' = \mathrm{f}';$\left.\begin{array}{c}\text{F}\text{ stetig in }{D}_{\text{f}};\hfill \\ \left.\begin{array}{c}\underset{x\to 3-}{lim}{\text{F}}^{\prime }\left(x\right)=2\hfill \\ \underset{x\to 3+}{lim}{\text{F}}^{\prime }\left(x\right)=2\hfill \end{array}\right\}={\text{F}}^{\prime }\left(3\right)=\text{f}\left(x\right);\hfill \end{array}\right\}\Rightarrow {\text{F}}^{\prime }={\text{f}}^{\prime };$

Berechne

a)

\int\limits_0^3 \mathrm{f}(x) \,\mathrm{d}x = \mathrm{F}(3) - \mathrm{F}(0) = \frac{15}{4};$\underset{0}{\overset{3}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\text{F}\left(3\right)-\text{F}\left(0\right)=\frac{15}{4};$

b)

\int\limits_3^4 \mathrm{f}(x) \,\mathrm{d}x = \mathrm{F}(4) - \mathrm{F}(3) = \frac{3}{2};$\underset{3}{\overset{4}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\text{F}\left(4\right)-\text{F}\left(3\right)=\frac{3}{2};$

c)

\int\limits_0^4 \mathrm{f}(x) \,\mathrm{d}x = \mathrm{F}(4) - \mathrm{F}(0) = \int\limits_0^3 \mathrm{f}(x) \,\mathrm{d}x + \int\limits_3^4 \mathrm{f}(x) \,\mathrm{d}x = \mathrm{F}(3) - \mathrm{F}(0) + \mathrm{F}(4) - \mathrm{F}(3) = \mathrm{F}(4) - \mathrm{F}(0) = \frac{21}{4};$\underset{0}{\overset{4}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\text{F}\left(4\right)-\text{F}\left(0\right)=\underset{0}{\overset{3}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x+\underset{3}{\overset{4}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\text{F}\left(3\right)-\text{F}\left(0\right)+\text{F}\left(4\right)-\text{F}\left(3\right)=\text{F}\left(4\right)-\text{F}\left(0\right)=\frac{21}{4};$

0.0.1.2 ↑ Analysis-Buch Seite 37, Aufgabe 34

\mathrm{f}(x) := \sqrt{4 - x^2}; \quad D_{\mathrm{f}} = \left[-2, 2\right];$\text{f}\left(x\right):=\sqrt{4-x^2};D_{\text{f}}=\left[-2,2\right];$

a)

Zeige, dass für (x_0, y_0) \in G_{\mathrm{f}}$\left(x_0,y_0\right)\in G_{\text{f}}$ gilt: x_0^2 + y_0^2 = 4;$x_0^2+y_0^2=4;$

x_0^2 + y_0^2 = x_0^2 + \mathrm{f}(x_0) = x_0^2 + 4 - x_0^2 = 4;$x_0^2+y_0^2=x_0^2+\text{f}\left(x_0\right)=x_0^2+4-x_0^2=4;$

Was folgt daraus für die Form von G_{\mathrm{f}}$G_{\text{f}}$?

\mathrm{f}$\text{f}$ beschreibt einen Halbkreis mit Radius r = \sqrt{4} = 2$r=\sqrt{4}=2$.

b)

Berechne mit Holfe geometrischer Überlegungen

• \int\limits_{-2}^{+2} \mathrm{f}(x) \,\mathrm{d}x = \frac{\pi r^2}{2} = 2\pi;$\underset{-2}{\overset{+2}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{\pi r^2}{2}=2\pi ;$

• \int\limits_0^{+2} \mathrm{f}(x) \,\mathrm{d}x = \frac{\pi r^2}{4} = \pi;$\underset{0}{\overset{+2}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{\pi r^2}{4}=\pi ;$

• \int\limits_0^1 \mathrm{f}(x) \,\mathrm{d}x = r^2 \frac{\alpha}{2} + \frac{1}{2} \cdot 1 \cdot \mathrm{f}(1) = \frac{1}{2}\left(\frac{\pi}{2} - \operatorname{arctan} \frac{\mathrm{f}(1)}{1}\right) \, r^2 + \frac{\sqrt{3}}{2} = \frac{\frac{\pi}{2} - \frac{\pi}{3}}{2} r^2 + \frac{\sqrt{3}}{2} = \frac{\pi}{3} + \frac{\sqrt{3}}{2};$\underset{0}{\overset{1}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=r^2\frac{\alpha }{2}+\frac{1}{2}\cdot 1\cdot \text{f}\left(1\right)=\frac{1}{2}\left(\frac{\pi }{2}-arctan\frac{\text{f}\left(1\right)}{1}\right)r^2+\frac{\sqrt{3}}{2}=\frac{\frac{\pi }{2}-\frac{\pi }{3}}{2}{r}^2+\frac{\sqrt{3}}{2}=\frac{\pi }{3}+\frac{\sqrt{3}}{2};$

• \int\limits_1^2 \mathrm{f}(x) \,\mathrm{d}x = \int\limits_0^{+2} \mathrm{f}(x) \,\mathrm{d}x - \int\limits_0^1 \mathrm{f}(x) \,\mathrm{d}x = \pi - \frac{\pi}{3} - \frac{\sqrt{3}}{2} = \frac{2}{3}\pi - \frac{\sqrt{3}}{2};$\underset{1}{\overset{2}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\underset{0}{\overset{+2}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x-\underset{0}{\overset{1}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\pi -\frac{\pi }{3}-\frac{\sqrt{3}}{2}=\frac{2}{3}\pi -\frac{\sqrt{3}}{2};$

0.0.1.3 ↑ Analysis-Buch Seite 37, Aufgabe 35

Berechne \int\limits_0^2 \sqrt{9 - x^2} \,\mathrm{d}x$\underset{0}{\overset{2}{\int \; <!--nolimits-->}}\sqrt{9-x^2}\text{d}x$, indem du den Kreis x^2 + y^2 = 9$x^2+y^2=9$ betrachtest.

r = \sqrt{9} = 3;$r=\sqrt{9}=3;$

\int\limits_0^2 \sqrt{9 - x^2} \,\mathrm{d}x = r^2 \frac{\alpha}{2} + \frac{1}{2} \cdot 2 \cdot \mathrm{f}(2) = \frac{1}{2}\left(\frac{\pi}{2} - \operatorname{arctan} \frac{\mathrm{f}(2)}{2}\right) \, r^2 + \sqrt{5} \approx 5{,}52;$\underset{0}{\overset{2}{\int \; <!--nolimits-->}}\sqrt{9-x^2}\text{d}x=r^2\frac{\alpha }{2}+\frac{1}{2}\cdot 2\cdot \text{f}\left(2\right)=\frac{1}{2}\left(\frac{\pi }{2}-arctan\frac{\text{f}\left(2\right)}{2}\right)r^2+\sqrt{5}\approx 5,52;$