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

解

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

16

解答ステップ

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

拡張

(x + 1) (x + 2) (x + 3) : x^{3} + 6 x^{2} + 11 x + 6

= \int_{- 1}^{1} x^{3} + 6 x^{2} + 11 x + 6 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}^{1} x^{3} d x + \int_{- 1}^{1} 6 x^{2} d x + \int_{- 1}^{1} 11 x d x + \int_{- 1}^{1} 6 d x

\int_{- 1}^{1} x^{3} d x = 0

\int_{- 1}^{1} 6 x^{2} d x = 4

\int_{- 1}^{1} 11 x d x = 0

\int_{- 1}^{1} 6 d x = 12

= 0 + 4 + 0 + 12

簡素化

= 16