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人気のある微分積分 >

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

解

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

解

\frac{4}{15}

+1

十進法表記

0.26666 \dots

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

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

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

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

定数を除く:

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

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

拡張

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

= 2 (- (- \frac{1}{2} \cdot \int_{0}^{1} - 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

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

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

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

= 2 (- (- \frac{1}{2} (- \frac{2}{5} + \frac{2}{3})))

簡素化

2 (- (- \frac{1}{2} (- \frac{2}{5} + \frac{2}{3}))) : \frac{4}{15}

= \frac{4}{15}

グラフ

人気の例

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