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

解

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

\frac{1}{3}

十進法表記

0.33333 \dots

解答ステップ

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

拡張

(1 - t)^{2} : 1 - 2 t + t^{2}

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

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

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

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

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

簡素化

= \frac{1}{3}