Man berechne: ∫12(x(xx34))5dx{\displaystyle \int _{1}^{2}({\sqrt[{4}]{x({\sqrt[{3}]{x{\sqrt {x}}}}}}))^{5}dx}
Man berechne:
(x(xx3)4)5=(x(xx3))54=(x(x323))54=(x(x32)13)54=(x⋅x36)54=(x96)54=x4524=x158{\displaystyle {\begin{aligned}{\left({\sqrt[{4}]{x\left({\sqrt[{3}]{x{\sqrt {x}}}}\right)}}\right)}^{5}={\left({x\left({\sqrt[{3}]{x{\sqrt {x}}}}\right)}\right)}^{\frac {5}{4}}\\={\left({x\left({\sqrt[{3}]{x^{\frac {3}{2}}}}\right)}\right)}^{\frac {5}{4}}\\={\left({x\left({x^{\frac {3}{2}}}\right)^{\frac {1}{3}}}\right)}^{\frac {5}{4}}\\=\left({x\cdot x^{\frac {3}{6}}}\right)^{\frac {5}{4}}\\=\left({x^{\frac {9}{6}}}\right)^{\frac {5}{4}}\\=x^{\frac {45}{24}}\\=x^{\frac {15}{8}}\end{aligned}}}
∫x158dx=x158+1238=8x23823+C{\displaystyle {\begin{aligned}\int {x^{\frac {15}{8}}}dx={\frac {x^{{\frac {15}{8}}+1}}{\frac {23}{8}}}={\frac {8x^{\frac {23}{8}}}{23}}+C\end{aligned}}}
F(x)|12=F(b)−F(a)8x23823|12=8⋅223823−8⋅123823≈2.2 (2.2038...){\displaystyle {\begin{aligned}F(x)|_{1}^{2}=F(b)-F(a)\\{\frac {8x^{\frac {23}{8}}}{23}}|_{1}^{2}={\frac {8\cdot 2^{\frac {23}{8}}}{23}}-{\frac {8\cdot 1^{\frac {23}{8}}}{23}}\approx 2.2{\text{ (2.2038...)}}\end{aligned}}}
f.thread:52984 / SS07 Beispiel 24
![{\displaystyle \int _{1}^{2}({\sqrt[{4}]{x({\sqrt[{3}]{x{\sqrt {x}}}}}}))^{5}dx}](/index.php?title=Spezial:MathShowImage&hash=888461ce5ef472a6d4c683cd8eafbd91&mode=mathml)
![{\displaystyle {\begin{aligned}{\left({\sqrt[{4}]{x\left({\sqrt[{3}]{x{\sqrt {x}}}}\right)}}\right)}^{5}={\left({x\left({\sqrt[{3}]{x{\sqrt {x}}}}\right)}\right)}^{\frac {5}{4}}\\={\left({x\left({\sqrt[{3}]{x^{\frac {3}{2}}}}\right)}\right)}^{\frac {5}{4}}\\={\left({x\left({x^{\frac {3}{2}}}\right)^{\frac {1}{3}}}\right)}^{\frac {5}{4}}\\=\left({x\cdot x^{\frac {3}{6}}}\right)^{\frac {5}{4}}\\=\left({x^{\frac {9}{6}}}\right)^{\frac {5}{4}}\\=x^{\frac {45}{24}}\\=x^{\frac {15}{8}}\end{aligned}}}](/index.php?title=Spezial:MathShowImage&hash=0ace9caf41e43aea9197e6c0026d0f60&mode=mathml)
