integral from 0 to 2 of (2x-3)(4x^2+5)

解

\int_{0}^{2} (2 x - 3) (4 x^{2} + 5) d x

- 10

解答ステップ

\int_{0}^{2} (2 x - 3) (4 x^{2} + 5) d x

拡張

(2 x - 3) (4 x^{2} + 5) : 8 x^{3} + 10 x - 12 x^{2} - 15

= \int_{0}^{2} 8 x^{3} + 10 x - 12 x^{2} - 15 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_{0}^{2} 8 x^{3} d x + \int_{0}^{2} 10 x d x - \int_{0}^{2} 12 x^{2} d x - \int_{0}^{2} 15 d x

\int_{0}^{2} 8 x^{3} d x = 32

\int_{0}^{2} 10 x d x = 20

\int_{0}^{2} 12 x^{2} d x = 32

\int_{0}^{2} 15 d x = 30

= 32 + 20 - 32 - 30

簡素化

= - 10