«K12/K13» 3. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 3. Hausaufgabe

0.0.1.1 ↑ Analysis-Buch Seite 14, Aufgabe 7

Gegeben sind die Funktionen \mathrm{h}_a${\text{h}}_a$ bis \mathrm{n}_a${\text{n}}_a$; Bestimme die Scharen der zugehörigen Stammfunktionen \mathrm{H}_a${\text{H}}_a$ bis \mathrm{N}_a${\text{N}}_a$.

\renewcommand{\arraystretch}{1.5}\begin{array}{ll} {} \mathrm{h}_a(x) = \frac{1}{2} x \left(x - a\right)^2; & \Rightarrow {} \mathrm{H}_a(x) = \frac{1}{36}x^3 - \frac{1}{18}ax^3 + \frac{1}{24}ax^2 + C; \\ {} \mathrm{i}_a(x) = \frac{2}{3a^2} x^3 - \frac{2}{a} x^2; & \Rightarrow {} \mathrm{I}_a(x) = \frac{1}{6a^2} x^4 - \frac{2}{3a} x^3 + C; \\ {} \mathrm{j}_s(x) = \frac{1}{6s} x^3 - x^2 + \frac{3}{2}sx; & \Rightarrow {} \mathrm{J}_s(x) = \frac{1}{24s} x^4 - \frac{1}{3}x^3 + \frac{3}{4}sx^2 + C; \\ {} \mathrm{k}_t(x) = \frac{1}{81} x^2 \left(3x^2 - tx + 3t\right); & \Rightarrow {} \mathrm{K}_t(x) = \frac{1}{135} x^5 - \frac{1}{324} tx^4 + \frac{1}{81} tx^3 + C; \\ {} \mathrm{l}_k(x) = \frac{1}{2}x \left[x^2 - 2kx + k^2 - 4\right]; & \Rightarrow {} \mathrm{L}_k(x) = \frac{1}{8}x^4 - \frac{1}{9}kx^3 + \frac{1}{4}k^2x^2 - x^2 + C; \\ {} \mathrm{m}_k(x) = \frac{x}{729} \left(8k^3x^2 - 216k^2x + 1215k + 729\right); & \Rightarrow {} \mathrm{M}_k(x) = \frac{2}{729}k^3x^4 - \frac{8}{81}k^2x^3 + \frac{5}{6}kx^2 + \frac{1}{2}x^2 + C; \\ {} \mathrm{n}_a(x) = \frac{1}{8}x \left(x^2 - 3ax + 3a^2x - 12\right); & \Rightarrow {} \mathrm{N}_a(x) = \frac{1}{32}x^4 - \frac{1}{8}ax^3 + \frac{1}{8}a^2x^3 - \frac{3}{4}x^2 + C; \end{array}$\begin{array}{cc}{\text{h}}_a\left(x\right)=\frac{1}{2}x{\left(x-a\right)}^2;\hfill & \Rightarrow {\text{H}}_a\left(x\right)=\frac{1}{36}{x}^3-\frac{1}{18}a{x}^3+\frac{1}{24}a{x}^2+C;\hfill \\ {\text{i}}_a\left(x\right)=\frac{2}{3a^2}{x}^3-\frac{2}{a}{x}^2;\hfill & \Rightarrow {\text{I}}_a\left(x\right)=\frac{1}{6a^2}{x}^4-\frac{2}{3a}{x}^3+C;\hfill \\ {\text{j}}_s\left(x\right)=\frac{1}{6s}{x}^3-x^2+\frac{3}{2}sx;\hfill & \Rightarrow {\text{J}}_s\left(x\right)=\frac{1}{24s}{x}^4-\frac{1}{3}{x}^3+\frac{3}{4}s{x}^2+C;\hfill \\ {\text{k}}_t\left(x\right)=\frac{1}{81}{x}^2\left(3x^2-tx+3t\right);\hfill & \Rightarrow {\text{K}}_t\left(x\right)=\frac{1}{135}{x}^5-\frac{1}{324}t{x}^4+\frac{1}{81}t{x}^3+C;\hfill \\ {\text{l}}_k\left(x\right)=\frac{1}{2}x\left[x^2-2kx+k^2-4\right];\hfill & \Rightarrow {\text{L}}_k\left(x\right)=\frac{1}{8}{x}^4-\frac{1}{9}k{x}^3+\frac{1}{4}{k}^2{x}^2-x^2+C;\hfill \\ {\text{m}}_k\left(x\right)=\frac{x}{729}\left(8k^3{x}^2-216k^{2}x+1215k+729\right);\hfill & \Rightarrow {\text{M}}_k\left(x\right)=\frac{2}{729}{k}^3{x}^4-\frac{8}{81}{k}^2{x}^3+\frac{5}{6}k{x}^2+\frac{1}{2}{x}^2+C;\hfill \\ {\text{n}}_a\left(x\right)=\frac{1}{8}x\left(x^2-3ax+3a^{2}x-12\right);\hfill & \Rightarrow {\text{N}}_a\left(x\right)=\frac{1}{32}{x}^4-\frac{1}{8}a{x}^3+\frac{1}{8}{a}^2{x}^3-\frac{3}{4}{x}^2+C;\hfill \end{array}$

0.0.1.2 ↑ Analysis-Buch Seite 15, Aufgabe 8

Berechne

a)

\int \left(ax + a\right) \mathrm{d}x = \frac{1}{2}ax^2 + ax + C;$\int \; <!--nolimits-->\left(ax+a\right)\text{d}x=\frac{1}{2}a{x}^2+ax+C;$

b)

\int \left(ax + a\right) \mathrm{d}a = \frac{1}{2}xa^2 + \frac{1}{2}a^2 + C;$\int \; <!--nolimits-->\left(ax+a\right)\text{d}a=\frac{1}{2}x{a}^2+\frac{1}{2}{a}^2+C;$

c)

\int \left(ax + a\right) \mathrm{d}t = \left(ax + a\right)t + C;$\int \; <!--nolimits-->\left(ax+a\right)\text{d}t=\left(ax+a\right)t+C;$