integral from 0 to 1 of t+t^2

解

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

\frac{5}{6}

十進法表記

0.83333 \dots

解答ステップ

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

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

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

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

簡素化

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

= \frac{5}{6}