integral from 2 to 4 of 1/8 (6-x-y)

解

\int_{2}^{4} \frac{1}{8} (6 - x - y) d y

\frac{- x + 3}{4}

解答ステップ

\int_{2}^{4} \frac{1}{8} (6 - x - y) d y

定数を除く:

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

= \frac{1}{8} \cdot \int_{2}^{4} 6 - 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{1}{8} (\int_{2}^{4} 6 d y - \int_{2}^{4} x d y - \int_{2}^{4} y d y)

\int_{2}^{4} 6 d y = 12

\int_{2}^{4} x d y = 2 x

\int_{2}^{4} y d y = 6

= \frac{1}{8} (12 - 2 x - 6)

簡素化

\frac{1}{8} (12 - 2 x - 6) : \frac{- x + 3}{4}

= \frac{- x + 3}{4}