integral from 0 to 3 of 4x(x-4)

解

\int_{0}^{3} 4 x (x - 4) d x

- 36

解答ステップ

\int_{0}^{3} 4 x (x - 4) d x

定数を除く:

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

= 4 \cdot \int_{0}^{3} x (x - 4) d x

拡張

x (x - 4) : x^{2} - 4 x

= 4 \cdot \int_{0}^{3} x^{2} - 4 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

= 4 (\int_{0}^{3} x^{2} d x - \int_{0}^{3} 4 x d x)

\int_{0}^{3} x^{2} d x = 9

\int_{0}^{3} 4 x d x = 18

= 4 (9 - 18)

簡素化

= - 36