integral from 1 to 2 of 6(x-1)(2-x)

解

\int_{1}^{2} 6 (x - 1) (2 - x) d x

1

解答ステップ

\int_{1}^{2} 6 (x - 1) (2 - x) d x

定数を除く:

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

= 6 \cdot \int_{1}^{2} (x - 1) (2 - x) d x

拡張

(x - 1) (2 - x) : 3 x - x^{2} - 2

= 6 \cdot \int_{1}^{2} 3 x - x^{2} - 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

= 6 (\int_{1}^{2} 3 x d x - \int_{1}^{2} x^{2} d x - \int_{1}^{2} 2 d x)

\int_{1}^{2} 3 x d x = \frac{9}{2}

\int_{1}^{2} x^{2} d x = \frac{7}{3}

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

= 6 (\frac{9}{2} - \frac{7}{3} - 2)

簡素化

6 (\frac{9}{2} - \frac{7}{3} - 2) : 1

= 1