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

解

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

\frac{2}{3}

十進法表記

0.66666 \dots

解答ステップ

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

拡張

(x - 2) (x + 1) : x^{2} - x - 2

= \int_{- 2}^{0} x^{2} - x - 2 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_{- 2}^{0} x^{2} d x - \int_{- 2}^{0} x d x - \int_{- 2}^{0} 2 d x

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

\int_{- 2}^{0} x d x = - 2

\int_{- 2}^{0} 2 d x = 4

= \frac{8}{3} - (- 2) - 4

簡素化

\frac{8}{3} - (- 2) - 4 : \frac{2}{3}

= \frac{2}{3}