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人気のある微分積分 >

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

解

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

解

\frac{1}{3}

+1

十進法表記

0.33333 \dots

解答ステップ

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

拡張

x (1 - x) : x - x^{2}

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

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

\int_{0}^{1} x d x = \frac{1}{2}

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

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

簡素化

2 (\frac{1}{2} - \frac{1}{3}) : \frac{1}{3}

= \frac{1}{3}

グラフ

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