integral from 0 to 4 of (x-2)(2x+1)

解

\int_{0}^{4} (x - 2) (2 x + 1) d x

\frac{32}{3}

十進法表記

10.66666 \dots

解答ステップ

\int_{0}^{4} (x - 2) (2 x + 1) d x

拡張

(x - 2) (2 x + 1) : 2 x^{2} - 3 x - 2

= \int_{0}^{4} 2 x^{2} - 3 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_{0}^{4} 2 x^{2} d x - \int_{0}^{4} 3 x d x - \int_{0}^{4} 2 d x

\int_{0}^{4} 2 x^{2} d x = \frac{128}{3}

\int_{0}^{4} 3 x d x = 24

\int_{0}^{4} 2 d x = 8

= \frac{128}{3} - 24 - 8

簡素化

\frac{128}{3} - 24 - 8 : \frac{32}{3}

= \frac{32}{3}