«K12/K13» 2. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 2. Hausaufgabe

0.0.1.1 ↑ Analysis-Buch Seite 14, Aufgabe 4

Berechne

a)

\int \mathrm{f}(x) \,\mathrm{d}x = \int \left|x\right| \mathrm{d}x = \mathrm{F}_C(x) = \begin{cases} {} \frac{1}{2} x^2 + C & \text{f"ur } x > 0; \\ {} C & \text{f"ur } x = 0; \\ {} -\frac{1}{2} x^2 + C & \text{f"ur } x < 0; \end{cases}$\int \; <!--nolimits-->\text{f}\left(x\right)\text{d}x=\int \; <!--nolimits-->\left|x\right|\text{d}x={\text{F}}_C\left(x\right)=\left\{\begin{array}{cc}\frac{1}{2}{x}^2+C\hfill & \text{f”ur }x>0;\hfill \\ C\hfill & \text{f”ur }x=0;\hfill \\ -\frac{1}{2}{x}^2+C\hfill & \text{f”ur }x<0;\hfill \end{array}\right.$

Diffbarkeit für x > 0$x>0$ und x < 0$x<0$ gesichert, Nachweis des Falles für x = 0$x=0$:

\left.\begin{array}{l} {} \lim\limits_{x \to 0+} \mathrm{F}'(x) = \lim\limits_{x \to 0-} \mathrm{F}'(x) = 0 = \mathrm{f}(0); \\ {} \mathrm{f} \text{ stetig bei } 0; \end{array}\right\} \Rightarrow \text{Ermitteltes } \mathrm{F}_C \text{ ok}$\left.\begin{array}{c}\underset{x\to 0+}{lim}{\text{F}}^{\prime }\left(x\right)=\underset{x\to 0-}{lim}{\text{F}}^{\prime }\left(x\right)=0=\text{f}\left(0\right);\hfill \\ \text{f}\text{ stetig bei }0;\hfill \end{array}\right\}\Rightarrow \text{Ermitteltes }{\text{F}}_C\text{ ok}$

b)

\int \mathrm{sgn}\, x \,\mathrm{d}x = \begin{cases} {} x + C_1 & \text{f"ur } x > 0; \\ {} -x + C_2 & \text{f"ur } x < 0; \end{cases} \quad x \neq 0;$\int \; <!--nolimits-->\text{sgn}x\text{d}x=\left\{\begin{array}{cc}x+C_1\hfill & \text{f”ur }x>0;\hfill \\ -x+C_2\hfill & \text{f”ur }x<0;\hfill \end{array}\right.x\ne 0;$

0.0.1.2 ↑ Analysis-Buch Seite 14, Aufgabe 5

\mathrm{f}\colon x \mapsto \begin{cases} {} x & \text{f"ur } x \leq 0; \\ {} x^2 & \text{f"ur } x > 0; \end{cases}$\text{f}:x\mapsto \left\{\begin{array}{cc}x\hfill & \text{f”ur }x\le 0;\hfill \\ x^2\hfill & \text{f”ur }x>0;\hfill \end{array}\right.$

Berechne \int \mathrm{f}(x) \,\mathrm{d}x$\int \; <!--nolimits-->\text{f}\left(x\right)\text{d}x$.

\int \mathrm{f}(x) \,\mathrm{d}x = \mathrm{F}_C(x) = \begin{cases} {} \frac{1}{2}x^2 + C & \text{f"ur } x \leq 0; \\ {} \frac{1}{3}x^3 + C & \text{f"ur } x > 0; \end{cases}$\int \; <!--nolimits-->\text{f}\left(x\right)\text{d}x={\text{F}}_C\left(x\right)=\left\{\begin{array}{cc}\frac{1}{2}{x}^2+C\hfill & \text{f”ur }x\le 0;\hfill \\ \frac{1}{3}{x}^3+C\hfill & \text{f”ur }x>0;\hfill \end{array}\right.$

Überprüfung des Falles für x = 0$x=0$:

\left.\begin{array}{l} {} \lim\limits_{x \to 0-} \mathrm{F}'(x) = \lim\limits_{x \to 0+} \mathrm{F}'(x) = \mathrm{f}(0); \\ {} \mathrm{F}_C \text{ stetig bei } 0; \end{array}\right\} \Rightarrow \text{Ermitteltes } \mathrm{F}_C \text{ ok}$\left.\begin{array}{c}\underset{x\to 0-}{lim}{\text{F}}^{\prime }\left(x\right)=\underset{x\to 0+}{lim}{\text{F}}^{\prime }\left(x\right)=\text{f}\left(0\right);\hfill \\ {\text{F}}_C\text{ stetig bei }0;\hfill \end{array}\right\}\Rightarrow \text{Ermitteltes }{\text{F}}_C\text{ ok}$

0.0.1.3 ↑ Analysis-Buch Seite 14, Aufgabe 6

Gegeben sind die Funktionen \mathrm{a}$\text{a}$ big \mathrm{g}$\text{g}$, bestimme die Scharen der zugehörigen Stammfunktionen \mathrm{A}_c${\text{A}}_c$ bis \mathrm{G}_c${\text{G}}_c$.

\renewcommand{\arraystretch}{1.5}\begin{array}{ll} {} \mathrm{a}(x) = 6 - \frac{1}{24}x^2; & \Rightarrow {} \mathrm{A}_c(x) = 6x - \frac{1}{72}x^3 + C; \\ {} \mathrm{b}(x) = x^3 - 3x - 2; & \Rightarrow {} \mathrm{B}_c(x) = \frac{1}{4}x^4 - \frac{3}{2}x^2 - 2x + C; \\ {} \mathrm{c}(x) = -x^3 + 3x^2 - 2; & \Rightarrow {} \mathrm{C}_c(x) = -\frac{1}{4}x^4 + x^3 - 2x + C; \\ {} \mathrm{d}(x) = x^4 - 6x^2 + 5; & \Rightarrow {} \mathrm{D}_c(x) = \frac{1}{5}x^5 - 2x^3 + 5x + C; \\ {} \mathrm{e}(x) = \frac{1}{9}x^4 - \frac{8}{9}x^3 + 2x^2; & \Rightarrow {} \mathrm{E}_c(x) = \frac{1}{45}x^5 - \frac{2}{9}x^4 + \frac{2}{3}x^3 + C; \\ {} \mathrm{f}(x) = \frac{1}{40}x^5 - \frac{1}{8}x^4; & \Rightarrow {} \mathrm{F}_c(x) = \frac{1}{240}x^6 - \frac{1}{40}x^5 + C; \\ {} \mathrm{g}(x) = -\frac{1}{16}x^6 + \frac{3}{8}x^4; & \Rightarrow {} \mathrm{G}_c(x) = -\frac{1}{112}x^7 + \frac{3}{40}x^5 + C; \end{array}$\begin{array}{cc}\text{a}\left(x\right)=6-\frac{1}{24}{x}^2;\hfill & \Rightarrow {\text{A}}_c\left(x\right)=6x-\frac{1}{72}{x}^3+C;\hfill \\ \text{b}\left(x\right)=x^3-3x-2;\hfill & \Rightarrow {\text{B}}_c\left(x\right)=\frac{1}{4}{x}^4-\frac{3}{2}{x}^2-2x+C;\hfill \\ \text{c}\left(x\right)=-x^3+3x^2-2;\hfill & \Rightarrow {\text{C}}_c\left(x\right)=-\frac{1}{4}{x}^4+x^3-2x+C;\hfill \\ \text{d}\left(x\right)=x^4-6x^2+5;\hfill & \Rightarrow {\text{D}}_c\left(x\right)=\frac{1}{5}{x}^5-2x^3+5x+C;\hfill \\ \text{e}\left(x\right)=\frac{1}{9}{x}^4-\frac{8}{9}{x}^3+2x^2;\hfill & \Rightarrow {\text{E}}_c\left(x\right)=\frac{1}{45}{x}^5-\frac{2}{9}{x}^4+\frac{2}{3}{x}^3+C;\hfill \\ \text{f}\left(x\right)=\frac{1}{40}{x}^5-\frac{1}{8}{x}^4;\hfill & \Rightarrow {\text{F}}_c\left(x\right)=\frac{1}{240}{x}^6-\frac{1}{40}{x}^5+C;\hfill \\ \text{g}\left(x\right)=-\frac{1}{16}{x}^6+\frac{3}{8}{x}^4;\hfill & \Rightarrow {\text{G}}_c\left(x\right)=-\frac{1}{112}{x}^7+\frac{3}{40}{x}^5+C;\hfill \end{array}$