integral from 0 to 1/2 of 4/3 (1-x^3)

解

\int_{0}^{\frac{1}{2}} \frac{4}{3} (1 - x^{3}) d x

\frac{31}{48}

十進法表記

0.64583 \dots

解答ステップ

\int_{0}^{\frac{1}{2}} \frac{4}{3} (1 - x^{3}) d x

定数を除く:

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

= \frac{4}{3} \cdot \int_{0}^{\frac{1}{2}} 1 - x^{3} 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

= \frac{4}{3} (\int_{0}^{\frac{1}{2}} 1 d x - \int_{0}^{\frac{1}{2}} x^{3} d x)

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

\int_{0}^{\frac{1}{2}} x^{3} d x = \frac{1}{64}

= \frac{4}{3} (\frac{1}{2} - \frac{1}{64})

簡素化

\frac{4}{3} (\frac{1}{2} - \frac{1}{64}) : \frac{31}{48}

= \frac{31}{48}