integral from 0 to 1 of 4/3 (1-xy)

解

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

\frac{4}{3} (1 - \frac{x}{2})

解答ステップ

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

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}^{1} 1 - x y 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

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

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

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

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