Berechnen Sie die Stammfunktion der folgenden Funktionen. Berechnen Sie auch den ungefähren Wert der Funktion an der Stelle 2,4 und ihrer Stammfunktion zwischen den Stellen −2,3 und 1 A) b ( v ) = 7 sin v − 3 v + 0 , 4 v 3 + 3 {\displaystyle \ b(v)=7\ \sin v-{\frac {3}{v}}+{\frac {0{,}4}{v^{3}}}+3} B) v ( t ) = 5 t 3 4 − cos t + t − t − 3 {\displaystyle \ v(t)={\frac {5}{\sqrt[{4}]{t^{3}}}}-\cos t+t-t^{-3}} C) t ( b ) = 5 e b − 27 b 7 3 + k {\displaystyle \ t(b)=5\ e^{b}-{\sqrt[{3}]{\frac {27}{b^{7}}}}+k\quad }
0 , 4 v 3 = 2 5 v − 3 ⇒ {\displaystyle \ \ {\frac {0{,}4}{v^{3}}}={\frac {2}{5}}\ v^{-3}\ \Rightarrow \ } b ( v ) = 7 sin v − 3 v + 2 5 v − 3 + 3 ⇒ {\displaystyle \ b(v)=7\ \sin v-{\frac {3}{v}}+{\frac {2}{5}}\ v^{-3}+3\ \Rightarrow \ } ∫ b ( v ) d v = 7 ⋅ ( − cos v ) − 3 ln v + 2 v − 3 + 1 5 ⋅ ( − 3 + 1 ) + c {\displaystyle \ \int b(v)dv={7}\cdot (-\cos v)-3\ \ln v+{\frac {2\ v^{-3+1}}{5\cdot (-3+1)}}+c\ \quad } ∫ b ( v ) d v = − 7 cos v − 3 ln v − 1 5 v 2 + c {\displaystyle \ \int b(v)dv=-{7}\ \cos v-3\ \ln v-{\frac {1}{5v^{2}}}+c\ \quad } Hoch zum Anfang
5 t 3 4 = 5 t − 3 4 ⇒ {\displaystyle \ \ {\frac {5}{\sqrt[{4}]{t^{3}}}}={5}\ t^{-{\frac {3}{4}}}\ \Rightarrow \ } v ( t ) = 5 t − 3 4 − cos t + t − t − 3 ⇒ {\displaystyle \ v(t)={5}\ t^{-{\frac {3}{4}}}-\cos t+t-t^{-3}\ \Rightarrow \ } ∫ v ( t ) d t = 5 t − 3 4 + 1 − 3 4 + 1 − sin t + t 1 + 1 1 + 1 − t − 3 + 1 − 3 + 1 + c {\displaystyle \ \int v(t)dt={\frac {5\ t^{-{\frac {3}{4}}+1}}{-{\frac {3}{4}}+1}}-\sin t+{\frac {t^{1+1}}{1+1}}-{\frac {t^{-3+1}}{-3+1}}+c} ∫ v ( t ) d t = 20 t 1 4 − sin t + t 2 2 + t − 2 2 + c {\displaystyle \ \int v(t)dt=20\ t^{\frac {1}{4}}-\sin t+{\frac {t^{2}}{2}}+{\frac {t^{-2}}{2}}+c} ∫ v ( t ) d t = 20 t 4 − sin t + t 2 2 + 1 2 t 2 + c {\displaystyle \ \int v(t)dt=20\ {\sqrt[{4}]{t}}-\sin t+{\frac {t^{2}}{2}}+{\frac {1}{2\ t^{2}}}+c\ } Hoch zum Anfang
27 b 7 3 = 3 b − 7 3 ⇒ {\displaystyle \ \ {\sqrt[{3}]{\frac {27}{b^{7}}}}={3}\ b^{-{\frac {7}{3}}}\ \Rightarrow \ } t ( b ) = 5 e b − 3 b − 7 3 + k ⇒ {\displaystyle \ t(b)=5\ e^{b}-{3}\ b^{-{\frac {7}{3}}}+k\quad \ \Rightarrow \ } ∫ t ( b ) d b = 5 e b − 3 b − 7 3 + 1 − 7 3 + 1 + k ⋅ b + c {\displaystyle \ \int t(b)db=5\ e^{b}-{\frac {3\ b^{-{\frac {7}{3}}+1}}{-{\frac {7}{3}}+1}}+k\cdot b+c\ \quad } ∫ t ( b ) d b = 5 e b − 9 b − 4 3 4 + k ⋅ b + c {\displaystyle \ \int t(b)db=5\ e^{b}-{\frac {9\ b^{-{\frac {4}{3}}}}{4}}+k\cdot b+c\ \quad } Hoch zum Anfang
![{\displaystyle \ v(t)={\frac {5}{\sqrt[{4}]{t^{3}}}}-\cos t+t-t^{-3}}](../_assets_/eb734a37dd21ce173a46342d1cc64c92/200551b46994e1983804d38546981a8f3e987a65.svg)
![{\displaystyle \ t(b)=5\ e^{b}-{\sqrt[{3}]{\frac {27}{b^{7}}}}+k\quad }](../_assets_/eb734a37dd21ce173a46342d1cc64c92/030777a0812c106c679259931b463188172bb68f.svg)
![{\displaystyle \ \ {\frac {5}{\sqrt[{4}]{t^{3}}}}={5}\ t^{-{\frac {3}{4}}}\ \Rightarrow \ }](../_assets_/eb734a37dd21ce173a46342d1cc64c92/bac277e6aa4a3a6ff3417e491b3dd9a5dd36245c.svg)
![{\displaystyle \ \int v(t)dt=20\ {\sqrt[{4}]{t}}-\sin t+{\frac {t^{2}}{2}}+{\frac {1}{2\ t^{2}}}+c\ }](../_assets_/eb734a37dd21ce173a46342d1cc64c92/221e708b92e137d56efea0c48ce7c42cb2c45f43.svg)
![{\displaystyle \ \ {\sqrt[{3}]{\frac {27}{b^{7}}}}={3}\ b^{-{\frac {7}{3}}}\ \Rightarrow \ }](../_assets_/eb734a37dd21ce173a46342d1cc64c92/d2cddbf905109c894e02136b9eaa994c98362616.svg)
