integral from 0 to t^6 of sqrt(u^3)

Lösung

\int_{0}^{t^{6}} u^{3} d u

\frac{2 t ^{15}}{5}

Schritte zur Lösung

\int_{0}^{t^{6}} u^{3} d u

u^{3} = u^{\frac{3}{2}},

angenommen

u \geq 0

= \int_{0}^{t^{6}} u^{\frac{3}{2}} d u

Wende die Potenzregel an

= [\frac{2}{5} u^{\frac{5}{2}}]_{0}^{t^{6}}

Berechne die Grenzen:

\frac{2}{5} t^{15}

= \frac{2}{5} t^{15}

Vereinfache

= \frac{2 t ^{15}}{5}

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