integral of (4-x^{2/3})^3

解

\int (4 - x^{\frac{2}{3}})^{3} d x

64 x - \frac{144}{5} x^{\frac{5}{3}} + \frac{36}{7} x^{\frac{7}{3}} - \frac{x ^{3}}{3} + C

解答ステップ

\int (4 - x^{\frac{2}{3}})^{3} d x

拡張

(4 - x^{\frac{2}{3}})^{3} : 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

= \int 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 64 d x - \int 48 x^{\frac{2}{3}} d x + \int 12 x^{\frac{4}{3}} d x - \int x^{2} d x

\int 64 d x = 64 x

\int 48 x^{\frac{2}{3}} d x = \frac{144}{5} x^{\frac{5}{3}}

\int 12 x^{\frac{4}{3}} d x = \frac{36}{7} x^{\frac{7}{3}}

\int x^{2} d x = \frac{x ^{3}}{3}

= 64 x - \frac{144}{5} x^{\frac{5}{3}} + \frac{36}{7} x^{\frac{7}{3}} - \frac{x ^{3}}{3}

定数を解答に追加する

= 64 x - \frac{144}{5} x^{\frac{5}{3}} + \frac{36}{7} x^{\frac{7}{3}} - \frac{x ^{3}}{3} + C