integral from x to x+2 of 1/24 (x+y)

解

\int_{x}^{x + 2} \frac{1}{24} (x + y) d y

\frac{2 x + 1}{12}

解答ステップ

\int_{x}^{x + 2} \frac{1}{24} (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}{24} \cdot \int_{x}^{2 + x} 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}{24} (\int_{x}^{2 + x} x d y + \int_{x}^{2 + x} y d y)

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

\int_{x}^{2 + x} y d y = 2 (x + 1)

= \frac{1}{24} (2 x + 2 (x + 1))

簡素化

\frac{1}{24} (2 x + 2 (x + 1)) : \frac{2 x + 1}{12}

= \frac{2 x + 1}{12}