integral from 0 to 1 of 6/7 (x^2+(xy)/2)

解

\int_{0}^{1} \frac{6}{7} (x^{2} + \frac{x y}{2}) d x

\frac{3 y}{14} + \frac{2}{7}

解答ステップ

\int_{0}^{1} \frac{6}{7} (x^{2} + \frac{x y}{2}) d x

定数を除く:

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

= \frac{6}{7} \cdot \int_{0}^{1} \frac{y x}{2} + 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

= \frac{6}{7} (\int_{0}^{1} \frac{y x}{2} d x + \int_{0}^{1} x^{2} d x)

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

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

= \frac{6}{7} (\frac{y}{4} + \frac{1}{3})

簡素化

\frac{6}{7} (\frac{y}{4} + \frac{1}{3}) : \frac{3 y}{14} + \frac{2}{7}

= \frac{3 y}{14} + \frac{2}{7}