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人気のある微分積分 >

integral from 0 to 1 of x^2\sqrt[3]{1-x}

解

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

解

\frac{27}{140}

+1

十進法表記

0.19285 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{0} - (- u + 1)^{2} 3 u d u

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

= - \int_{0}^{1} - (- u + 1)^{2} 3 u d u

定数を除く:

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

= - (- \int_{0}^{1} (- u + 1)^{2} 3 u d u)

拡張

(- u + 1)^{2} 3 u : u^{\frac{7}{3}} - 2 u^{\frac{4}{3}} + 3 u

= - (- \int_{0}^{1} u^{\frac{7}{3}} - 2 u^{\frac{4}{3}} + 3 u 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_{0}^{1} u^{\frac{7}{3}} d u - \int_{0}^{1} 2 u^{\frac{4}{3}} d u + \int_{0}^{1} 3 u d u))

\int_{0}^{1} u^{\frac{7}{3}} d u = \frac{3}{10}

\int_{0}^{1} 2 u^{\frac{4}{3}} d u = \frac{6}{7}

\int_{0}^{1} 3 u d u = \frac{3}{4}

= - (- (\frac{3}{10} - \frac{6}{7} + \frac{3}{4}))

簡素化

- (- (\frac{3}{10} - \frac{6}{7} + \frac{3}{4})) : \frac{27}{140}

= \frac{27}{140}

グラフ

人気の例

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