integral from 0 to 1 of cy^2(1-y)^4

解

\int_{0}^{1} c y^{2} (1 - y)^{4} d y

\frac{1}{105} c

解答ステップ

\int_{0}^{1} c y^{2} (1 - y)^{4} d y

定数を除く:

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

= c \cdot \int_{0}^{1} y^{2} (1 - y)^{4} d y

拡張

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

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

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

= c (\int_{0}^{1} y^{2} d y - \int_{0}^{1} 4 y^{3} d y + \int_{0}^{1} 6 y^{4} d y - \int_{0}^{1} 4 y^{5} d y + \int_{0}^{1} y^{6} d y)

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

\int_{0}^{1} 4 y^{3} d y = 1

\int_{0}^{1} 6 y^{4} d y = \frac{6}{5}

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

\int_{0}^{1} y^{6} d y = \frac{1}{7}

= c (\frac{1}{3} - 1 + \frac{6}{5} - \frac{2}{3} + \frac{1}{7})

簡素化

c (\frac{1}{3} - 1 + \frac{6}{5} - \frac{2}{3} + \frac{1}{7}) : \frac{1}{105} c

= \frac{1}{105} c