integral from 0 to t^6 of sqrt(u^5)

Lösung

\int_{0}^{t^{6}} u^{5} d u

\frac{2 t ^{21}}{7}

Schritte zur Lösung

\int_{0}^{t^{6}} u^{5} d u

u^{5} = u^{\frac{5}{2}},

angenommen

u \geq 0

= \int_{0}^{t^{6}} u^{\frac{5}{2}} d u

Wende die Potenzregel an

= [\frac{2}{7} u^{\frac{7}{2}}]_{0}^{t^{6}}

Berechne die Grenzen:

\frac{2}{7} t^{21}

= \frac{2}{7} t^{21}

Vereinfache

= \frac{2 t ^{21}}{7}

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