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

解

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

\frac{1}{30}

十進法表記

0.03333 \dots

解答ステップ

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

拡張

x (x - 1)^{4} : x^{5} - 4 x^{4} + 6 x^{3} - 4 x^{2} + x

= \int_{0}^{1} x^{5} - 4 x^{4} + 6 x^{3} - 4 x^{2} + x 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

= \int_{0}^{1} x^{5} d x - \int_{0}^{1} 4 x^{4} d x + \int_{0}^{1} 6 x^{3} d x - \int_{0}^{1} 4 x^{2} d x + \int_{0}^{1} x d x

\int_{0}^{1} x^{5} d x = \frac{1}{6}

\int_{0}^{1} 4 x^{4} d x = \frac{4}{5}

\int_{0}^{1} 6 x^{3} d x = \frac{3}{2}

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

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

= \frac{1}{6} - \frac{4}{5} + \frac{3}{2} - \frac{4}{3} + \frac{1}{2}

簡素化

\frac{1}{6} - \frac{4}{5} + \frac{3}{2} - \frac{4}{3} + \frac{1}{2} : \frac{1}{30}

= \frac{1}{30}