integral from-3 to-1 of (x^3+6x^2+9x+9)

解

\int_{- 3}^{- 1} (x^{3} + 6 x^{2} + 9 x + 9) d x

14

解答ステップ

\int_{- 3}^{- 1} x^{3} + 6 x^{2} + 9 x + 9 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_{- 3}^{- 1} x^{3} d x + \int_{- 3}^{- 1} 6 x^{2} d x + \int_{- 3}^{- 1} 9 x d x + \int_{- 3}^{- 1} 9 d x

\int_{- 3}^{- 1} x^{3} d x = - 20

\int_{- 3}^{- 1} 6 x^{2} d x = 52

\int_{- 3}^{- 1} 9 x d x = - 36

\int_{- 3}^{- 1} 9 d x = 18

= - 20 + 52 - 36 + 18

簡素化

= 14