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

解

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

\frac{8 x ^{\frac{1}{3}} ( - x ^{\frac{4}{3}} + 1 )}{3}

解答ステップ

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

定数を除く:

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

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

定数を除く:

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

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

べき乗則を適用する

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

限度を計算する:

\frac{3}{4} - \frac{3}{4} x^{\frac{4}{3}}

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

簡素化

= \frac{8 x ^{\frac{1}{3}} ( - x ^{\frac{4}{3}} + 1 )}{3}