integral from 0 to 1 of 32/9 (xy)^{1/3}

Lösung

\int_{0}^{1} \frac{32}{9} (x y)^{\frac{1}{3}} d y

\frac{8 x ^{\frac{1}{3}}}{3}

Schritte zur Lösung

\int_{0}^{1} \frac{32}{9} (x y)^{\frac{1}{3}} d y

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{32}{9} \cdot \int_{0}^{1} (x y)^{\frac{1}{3}} d y

(x y)^{\frac{1}{3}} = x^{\frac{1}{3}} y^{\frac{1}{3}}

= \frac{32}{9} \cdot \int_{0}^{1} x^{\frac{1}{3}} y^{\frac{1}{3}} d y

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{32}{9} x^{\frac{1}{3}} \cdot \int_{0}^{1} y^{\frac{1}{3}} d y

Wende die Potenzregel an

= \frac{32}{9} x^{\frac{1}{3}} [\frac{3}{4} y^{\frac{4}{3}}]_{0}^{1}

Berechne die Grenzen:

\frac{3}{4}

= \frac{32}{9} x^{\frac{1}{3}} \frac{3}{4}

Vereinfache

= \frac{8 x ^{\frac{1}{3}}}{3}

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