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

解

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

\frac{7}{3}

十進法表記

2.33333 \dots

解答ステップ

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

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

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

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

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

簡素化

3 - 1 + \frac{1}{3} : \frac{7}{3}

= \frac{7}{3}