integral from 1 to 8 of-1/8 x+1

解

\int_{1}^{8} - \frac{1}{8} x + 1 d x

\frac{49}{16}

十進法表記

3.0625

解答ステップ

\int_{1}^{8} - \frac{1}{8} x + 1 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_{1}^{8} \frac{1}{8} x d x + \int_{1}^{8} 1 d x

\int_{1}^{8} \frac{1}{8} x d x = \frac{63}{16}

\int_{1}^{8} 1 d x = 7

= - \frac{63}{16} + 7

簡素化

- \frac{63}{16} + 7 : \frac{49}{16}

= \frac{49}{16}