integral from 0 to 1 of (u+5)(u-6)

解

\int_{0}^{1} (u + 5) (u - 6) d u

- \frac{181}{6}

十進法表記

- 30.16666 \dots

解答ステップ

\int_{0}^{1} (u + 5) (u - 6) d u

拡張

(u + 5) (u - 6) : u^{2} - u - 30

= \int_{0}^{1} u^{2} - u - 30 d u

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}^{1} u^{2} d u - \int_{0}^{1} u d u - \int_{0}^{1} 30 d u

\int_{0}^{1} u^{2} d u = \frac{1}{3}

\int_{0}^{1} u d u = \frac{1}{2}

\int_{0}^{1} 30 d u = 30

= \frac{1}{3} - \frac{1}{2} - 30

簡素化

\frac{1}{3} - \frac{1}{2} - 30 : - \frac{181}{6}

= - \frac{181}{6}