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人気のある微分積分 >

integral from 0 to 1 of x^2sqrt(9-x^3)

解

\int_{0}^{1} x^{2} 9 - x^{3} d x

解

\frac{54 - 32 2}{9}

+1

十進法表記

0.97168 \dots

解答ステップ

\int_{0}^{1} x^{2} 9 - x^{3} d x

u置換積分法を適用する

= \int_{9}^{8} - \frac{u}{3} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{8}^{9} - \frac{u}{3} d u

定数を除く:

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

= - (- \frac{1}{3} \cdot \int_{8}^{9} u d u)

べき乗則を適用する

= - (- \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{9})

簡素化

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{9}

限度を計算する:

18 - \frac{32 2}{3}

= \frac{1}{3} (18 - \frac{32 2}{3})

簡素化

= \frac{54 - 32 2}{9}

グラフ

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