# Mathematrix: Aufgabenbeispiele/ Integral von Potenzfunktionen

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 Berechnen Sie die Stammfunktionen der folgenden Funktionen. A)${\displaystyle \ b(v)=7\ v^{5}\quad }$ B)${\displaystyle \ a(c)=b\ c^{d}}$ C)${\displaystyle \ f(x)=65\ x^{12}}$ D)${\displaystyle \ x(y)=y^{2}-2x+1\quad }$ E)${\displaystyle \ v(t)={\sqrt[{4}]{t^{3}}}\quad }$ F)${\displaystyle \ f(x)={\frac {4}{x^{5}}}\quad }$ G)${\displaystyle \ V(h)={\sqrt[{4}]{\frac {1}{h^{5}}}}\quad }$ H)${\displaystyle \ H(x)={\sqrt[{3}]{\frac {27}{x^{7}}}}\quad }$ I)${\displaystyle \ z(w)={\frac {5}{w}}\ }$ J)${\displaystyle \ m(n)={\frac {5}{n}}+{\sqrt[{5}]{\frac {3125}{n^{30}}}}-{\frac {5}{13}}n^{12}\ }$.

Regel:${\displaystyle \ \int ax^{n}dx={\frac {a\ x^{n+1}}{n+1}}\ +c\ \quad }$

## A

${\displaystyle \ b(v)=7\ v^{5}\ \Rightarrow \ \int b(v)dv={\frac {7}{6}}\ v^{6}+c\ \quad }$
Hoch zum Anfang

## B

${\displaystyle \ a(c)=b\ c^{d}\ \Rightarrow \ \int a(c)dc={\frac {b}{d+1}}\ c^{d+1}+k}$
Hoch zum Anfang

## C

${\displaystyle \ f(x)={65}\ x^{12}\ \Rightarrow \ \int f(x)dx=65\cdot {\frac {x^{12+1}}{12+1}}=5\ x^{13}+c\ \quad }$
Hoch zum Anfang

## D

${\displaystyle \ x(y)=y^{2}-2y+1\ \Rightarrow \ \int x(y)dy={\frac {y^{3}}{3}}\ -2\cdot {\frac {y^{2}}{2}}\ +y+c\ \quad }$
Hoch zum Anfang

## E

${\displaystyle \ v(t)={\sqrt[{4}]{t^{3}}}\ =t^{\frac {3}{4}}\ \ \Rightarrow \ \int v(t)dt=\left({\dfrac {t^{{\frac {3}{4}}+1}}{{\frac {3}{4}}+1}}\right)+c\ }$

${\displaystyle \qquad ={\frac {4}{7}}t^{\frac {7}{4}}+c\left(={\frac {4\ {\sqrt[{4}]{t^{7}}}}{7}}+c\right)\quad }$
Hoch zum Anfang

## F

${\displaystyle \ f(x)={\frac {4}{x^{5}}}=4\ x^{-5}\ \ \Rightarrow \ \int f(x)dx=}$

${\displaystyle \qquad {\frac {4\cdot x^{-5+1}}{-5+1}})\ =-x^{-4}+c\quad }$
Hoch zum Anfang

## G

${\displaystyle \ V(h)={\sqrt[{4}]{\frac {1}{h^{5}}}}\ ={\sqrt[{4}]{h^{-5}}}\ =h^{-{5 \over 4}}\ \Rightarrow \ }$

${\displaystyle \qquad \int V(h)dh={\dfrac {h^{-{\frac {5}{4}}+1}}{-{\frac {5}{4}}+1}}+c\ =-{4}\ h^{-{\frac {1}{4}}}+c\left(=-{\frac {4}{\sqrt[{4}]{h}}}+c\right)}$
Hoch zum Anfang

## H

${\displaystyle \ H(x)={\sqrt[{3}]{\frac {27}{x^{7}}}}\ ={\sqrt[{3}]{27}}{\sqrt[{3}]{x^{-7}}}=3\ x^{-{\frac {7}{3}}}\ \Rightarrow \ }$

${\displaystyle \qquad \int H(x)dx={\dfrac {3\cdot x^{-{\frac {7}{3}}+1}}{-{\frac {7}{3}}+1}}+c=-{\frac {9}{4}}\ x^{-{\frac {4}{3}}}+c\left(=-{\frac {9}{4\ {\sqrt[{3}]{x^{4}\ }}}}+c\right)}$
Hoch zum Anfang

## I

${\displaystyle \ z(w)={\frac {5}{w}}=5\cdot w^{-1}\ \Rightarrow \ \int z(w)dw=5\ln w+c\ \quad }$
Hoch zum Anfang

## J

${\displaystyle \ m(n)={\frac {5}{n}}-{\sqrt[{5}]{\frac {3125}{n^{30}}}}-{13}\ n^{12}\ }$

${\displaystyle \qquad {\sqrt[{5}]{\frac {1}{3125\ n^{30}}}}\ ={\sqrt[{5}]{{\frac {1}{3125}}\ }}\ {\sqrt[{5}]{n^{-30}\ \ }}={\frac {1}{5}}\ n^{-{\frac {30}{5}}}={\frac {1}{5}}\ x^{-6}\ }$

${\displaystyle \qquad m(n)={\frac {5}{n}}-5\ x^{-6}-{13}\ n^{12}\ }$

${\displaystyle \qquad \int m(n)dn=5\ \ln n-5\cdot {\frac {n^{-6+1}}{-6+1}}\ -13\cdot {\frac {n^{12+1}}{12+1}}\ +c}$

${\displaystyle \qquad \int m(n)dn=5\ln n+n^{5}-n^{13}+c}$
Hoch zum Anfang