integral from 1 to 2 of f(x+1)

解

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

\frac{5}{2} f

解答ステップ

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

定数を除く:

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

= f \cdot \int_{1}^{2} 1 + x 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

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

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

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

= f (1 + \frac{3}{2})

結合

1 + \frac{3}{2} : \frac{5}{2}

= \frac{5}{2} f