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

解

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

\frac{8 y ^{\frac{5}{3}}}{3}

解答ステップ

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

定数を除く:

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

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

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

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

定数を除く:

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

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

べき乗則を適用する

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

限度を計算する:

\frac{3}{4} y^{\frac{4}{3}}

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

簡素化

= \frac{8 y ^{\frac{5}{3}}}{3}