integral from 0 to 1 of 6xy(2-x-y)

解

\int_{0}^{1} 6 x y (2 - x - y) d x

6 y (1 - \frac{y}{2} - \frac{1}{3})

解答ステップ

\int_{0}^{1} 6 x y (2 - x - y) d x

定数を除く:

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

= 6 y \cdot \int_{0}^{1} (2 - y - x) x d x

拡張

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

= 6 y \cdot \int_{0}^{1} 2 x - y 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

= 6 y (\int_{0}^{1} 2 x d x - \int_{0}^{1} y x d x - \int_{0}^{1} x^{2} d x)

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

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

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

= 6 y (1 - \frac{y}{2} - \frac{1}{3})