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

解

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

\frac{16}{3}

十進法表記

5.33333 \dots

解答ステップ

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

拡張

(x - 2) (x - 4) : x^{2} - 6 x + 8

= \int_{0}^{4} x^{2} - 6 x + 8 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} x^{2} d x - \int_{0}^{4} 6 x d x + \int_{0}^{4} 8 d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 6 x d x = 48

\int_{0}^{4} 8 d x = 32

= \frac{64}{3} - 48 + 32

簡素化

\frac{64}{3} - 48 + 32 : \frac{16}{3}

= \frac{16}{3}