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人気のある微分積分 >

integral of x^3sqrt(9-x^2)

解

\int x^{3} 9 - x^{2} d x

解

\frac{1}{5} (9 - x^{2})^{\frac{5}{2}} - 3 (9 - x^{2})^{\frac{3}{2}} + C

解答ステップ

\int x^{3} 9 - x^{2} d x

u置換積分法を適用する

= \int - (- u^{2} + 9) u^{2} d u

定数を除く:

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

= - \int (- u^{2} + 9) u^{2} d u

拡張

(- u^{2} + 9) u^{2} : - u^{4} + 9 u^{2}

= - \int - u^{4} + 9 u^{2} d u

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 u^{4} d u + \int 9 u^{2} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int 9 u^{2} d u = 3 u^{3}

= - (- \frac{u ^{5}}{5} + 3 u^{3})

代用を戻す

u = 9 - x^{2}

= - (- \frac{( 9 - x ^{2} ) ^{5}}{5} + 3 (9 - x^{2})^{3})

簡素化

- (- \frac{( 9 - x ^{2} ) ^{5}}{5} + 3 (9 - x^{2})^{3}) : \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} - 3 (9 - x^{2})^{\frac{3}{2}}

= \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} - 3 (9 - x^{2})^{\frac{3}{2}}

定数を解答に追加する

= \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} - 3 (9 - x^{2})^{\frac{3}{2}} + C

グラフ

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