integral from-1 to 3 of (3x-5)(x+2)

解

\int_{- 1}^{3} (3 x - 5) (x + 2) d x

- 8

解答ステップ

\int_{- 1}^{3} (3 x - 5) (x + 2) d x

拡張

(3 x - 5) (x + 2) : 3 x^{2} + x - 10

= \int_{- 1}^{3} 3 x^{2} + x - 10 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}^{3} 3 x^{2} d x + \int_{- 1}^{3} x d x - \int_{- 1}^{3} 10 d x

\int_{- 1}^{3} 3 x^{2} d x = 28

\int_{- 1}^{3} x d x = 4

\int_{- 1}^{3} 10 d x = 40

= 28 + 4 - 40

簡素化

= - 8