integral from 1 to 2 of (3x-7)(x-3)

解

\int_{1}^{2} (3 x - 7) (x - 3) d x

4

解答ステップ

\int_{1}^{2} (3 x - 7) (x - 3) d x

拡張

(3 x - 7) (x - 3) : 3 x^{2} - 16 x + 21

= \int_{1}^{2} 3 x^{2} - 16 x + 21 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

= \int_{1}^{2} 3 x^{2} d x - \int_{1}^{2} 16 x d x + \int_{1}^{2} 21 d x

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

\int_{1}^{2} 16 x d x = 24

\int_{1}^{2} 21 d x = 21

= 7 - 24 + 21

簡素化

= 4