integral from 0 to 3 of (x-1)(x-2)(x-3)

解

\int_{0}^{3} (x - 1) (x - 2) (x - 3) d x

- \frac{9}{4}

十進法表記

- 2.25

解答ステップ

\int_{0}^{3} (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_{0}^{3} 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_{0}^{3} x^{3} d x - \int_{0}^{3} 6 x^{2} d x + \int_{0}^{3} 11 x d x - \int_{0}^{3} 6 d x

\int_{0}^{3} x^{3} d x = \frac{81}{4}

\int_{0}^{3} 6 x^{2} d x = 54

\int_{0}^{3} 11 x d x = \frac{99}{2}

\int_{0}^{3} 6 d x = 18

= \frac{81}{4} - 54 + \frac{99}{2} - 18

簡素化

\frac{81}{4} - 54 + \frac{99}{2} - 18 : - \frac{9}{4}

= - \frac{9}{4}