integral from 0 to 1 of 1/4 (2x+y)

解

\int_{0}^{1} \frac{1}{4} (2 x + y) d y

\frac{1}{4} (\frac{1}{2} + 2 x)

解答ステップ

\int_{0}^{1} \frac{1}{4} (2 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}{4} \cdot \int_{0}^{1} y + 2 x 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}{4} (\int_{0}^{1} y d y + \int_{0}^{1} 2 x d y)

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

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

= \frac{1}{4} (\frac{1}{2} + 2 x)