integral from 0 to 1 of (x^2-3x+8)

解

\int_{0}^{1} (x^{2} - 3 x + 8) d x

\frac{41}{6}

十進法表記

6.83333 \dots

解答ステップ

\int_{0}^{1} x^{2} - 3 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}^{1} x^{2} d x - \int_{0}^{1} 3 x d x + \int_{0}^{1} 8 d x

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

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

\int_{0}^{1} 8 d x = 8

= \frac{1}{3} - \frac{3}{2} + 8

簡素化

\frac{1}{3} - \frac{3}{2} + 8 : \frac{41}{6}

= \frac{41}{6}