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

解

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

\frac{5}{6}

十進法表記

0.83333 \dots

解答ステップ

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

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

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

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

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

簡素化

\frac{1}{3} - \frac{1}{2} + 1 : \frac{5}{6}

= \frac{5}{6}