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人気のある微分積分 >

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

解

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

解

\frac{16}{105}

+1

十進法表記

0.15238 \dots

解答ステップ

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

u置換積分法を適用する

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

拡張

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

= - (- \int_{0}^{1} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + 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{5}{2}} d u - \int_{0}^{1} 2 u^{\frac{3}{2}} d u + \int_{0}^{1} u d u))

\int_{0}^{1} u^{\frac{5}{2}} d u = \frac{2}{7}

\int_{0}^{1} 2 u^{\frac{3}{2}} d u = \frac{4}{5}

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

= - (- (\frac{2}{7} - \frac{4}{5} + \frac{2}{3}))

簡素化

- (- (\frac{2}{7} - \frac{4}{5} + \frac{2}{3})) : \frac{16}{105}

= \frac{16}{105}

グラフ

人気の例

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\int_{1}^{2} \frac{2 x + 1}{12} d x

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\int_{0}^{\frac{π}{2}} θ^{2} d θ

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