«K12/K13» 7. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 7. Hausaufgabe

0.0.1.1 ↑ Analysis-Buch Seite 35, Aufgabe 6

\mathrm{f}(x) := -x^2 + 4x - 3; \quad D_{\mathrm{f}} = \left[1, 3\right];$\text{f}\left(x\right):=-x^2+4x-3;D_{\text{f}}=\left[1,3\right];$

Gib die Flächenfunktion \mathrm{A}_{\frac{3}{3}}${\text{A}}_{\frac{3}{3}}$ an und berechne damit die Flächen

a)

\left\{(x,y) | \frac{3}{2} \leq x \leq 2 \wedge 0 \leq y \leq \mathrm{f}(x) \right\}$\left\{\left(x,y\right)\mid \frac{3}{2}\le x\le 2\wedge 0\le y\le \text{f}\left(x\right)\right\}$

A = \int\limits_{\frac{3}{2}}^2 \mathrm{f}(x) \,\mathrm{d}x = \frac{11}{24};$A=\underset{\frac{3}{2}}{\overset{2}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{11}{24};$

b)

\left\{(x,y) | \frac{3}{2} \leq x \leq \frac{5}{2} \wedge 0 \leq y \leq \mathrm{f}(x) \right\}$\left\{\left(x,y\right)\mid \frac{3}{2}\le x\le \frac{5}{2}\wedge 0\le y\le \text{f}\left(x\right)\right\}$

A = \int\limits_{\frac{3}{2}}^{\frac{5}{2}} \mathrm{f}(x) \,\mathrm{d}x = \frac{11}{12};$A=\underset{\frac{3}{2}}{\overset{\frac{5}{2}}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{11}{12};$

c)

\left\{(x,y) | \frac{3}{2} \leq x \leq 3 \wedge 0 \leq y \leq \mathrm{f}(x) \right\}$\left\{\left(x,y\right)\mid \frac{3}{2}\le x\le 3\wedge 0\le y\le \text{f}\left(x\right)\right\}$

A = \int\limits_{\frac{3}{2}}^3 \mathrm{f}(x) \,\mathrm{d}x = \frac{9}{8};$A=\underset{\frac{3}{2}}{\overset{3}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{9}{8};$

0.0.1.2 ↑ Analysis-Buch Seite 35, Aufgabe 7

\mathrm{f}(x) := -x^2 + 4x - 3; \quad D_{\mathrm{f}} = \left[1, 3\right];$\text{f}\left(x\right):=-x^2+4x-3;D_{\text{f}}=\left[1,3\right];$

Berechne die Flächenfunktionen

a)

\mathrm{A}_1(b) = \int\limits_1^b \mathrm{f}(x) \,\mathrm{d}x = \frac{4}{3} - \frac{b^3 - 6b^2 + 9b}{3};${\text{A}}_1\left(b\right)=\underset{1}{\overset{b}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{4}{3}-\frac{b^3-6b^2+9b}{3};$

b)

\mathrm{A}_2(b) = \int\limits_2^b \mathrm{f}(x) \,\mathrm{d}x = \frac{2}{3} - \frac{b^3 - 6b^2 + 9b}{3};${\text{A}}_2\left(b\right)=\underset{2}{\overset{b}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{2}{3}-\frac{b^3-6b^2+9b}{3};$

c)

\mathrm{A}_{\frac{3}{2}}(b) = \int\limits_{\frac{3}{2}}^b \mathrm{f}(x) \,\mathrm{d}x = \frac{5}{24} - \frac{b^3 - 6b^2 + 9b}{3};${\text{A}}_{\frac{3}{2}}\left(b\right)=\underset{\frac{3}{2}}{\overset{b}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{5}{24}-\frac{b^3-6b^2+9b}{3};$

0.0.1.3 ↑ Analysis-Buch Seite 36, Aufgabe 8

\mathrm{f}(x) := -x^2 + 4x - 3; \quad D_{\mathrm{f}} = \left[1, 3\right];$\text{f}\left(x\right):=-x^2+4x-3;D_{\text{f}}=\left[1,3\right];$

Berechne folgende Flächen (vgl. Aufgabe 7!)

a)

\left\{(x,y) | 1 \leq x \leq 2 \wedge 0 \leq y \leq \mathrm{f}(x) \right\}$\left\{\left(x,y\right)\mid 1\le x\le 2\wedge 0\le y\le \text{f}\left(x\right)\right\}$

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

b)

\left\{(x,y) | 2 \leq x \leq 3 \wedge 0 \leq y \leq \mathrm{f}(x) \right\}$\left\{\left(x,y\right)\mid 2\le x\le 3\wedge 0\le y\le \text{f}\left(x\right)\right\}$

A = \int\limits_2^3 \mathrm{f}(x) \,\mathrm{d}x = \frac{2}{3};$A=\underset{2}{\overset{3}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{2}{3};$

c)

\left\{(x,y) | \frac{5}{2} \leq x \leq 3 \wedge 0 \leq y \leq \mathrm{f}(x) \right\}$\left\{\left(x,y\right)\mid \frac{5}{2}\le x\le 3\wedge 0\le y\le \text{f}\left(x\right)\right\}$

A = \int\limits_{\frac{5}{2}}^3 \mathrm{f}(x) \,\mathrm{d}x = \frac{5}{24};$A=\underset{\frac{5}{2}}{\overset{3}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=\frac{5}{24};$

d)

\left\{(x,y) | 2{,}9 \leq x \leq 2{,}9 \wedge 0 \leq y \leq \mathrm{f}(x) \right\}$\left\{\left(x,y\right)\mid 2,9\le x\le 2,9\wedge 0\le y\le \text{f}\left(x\right)\right\}$

A = \int\limits_{2{,}9}^{2{,}9} \mathrm{f}(x) \,\mathrm{d}x = 0;$A=\underset{2,9}{\overset{2,9}{\int \; <!--nolimits-->}}\text{f}\left(x\right)\text{d}x=0;$