integral from 0 to 1 of t^2-t

解

\int_{0}^{1} t^{2} - t d t

- \frac{1}{6}

十進法表記

- 0.16666 \dots

解答ステップ

\int_{0}^{1} t^{2} - t d t

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} t^{2} d t - \int_{0}^{1} t d t

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

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

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

簡素化

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

= - \frac{1}{6}