integral from 2 to 4 of (2x-3)(3x+1)

解

\int_{2}^{4} (2 x - 3) (3 x + 1) d x

64

解答ステップ

\int_{2}^{4} (2 x - 3) (3 x + 1) d x

拡張

(2 x - 3) (3 x + 1) : 6 x^{2} - 7 x - 3

= \int_{2}^{4} 6 x^{2} - 7 x - 3 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_{2}^{4} 6 x^{2} d x - \int_{2}^{4} 7 x d x - \int_{2}^{4} 3 d x

\int_{2}^{4} 6 x^{2} d x = 112

\int_{2}^{4} 7 x d x = 42

\int_{2}^{4} 3 d x = 6

= 112 - 42 - 6

簡素化

= 64