integral from 0 to 12 of 1/36 (t(12-t))

解

\int_{0}^{12} \frac{1}{36} (t (12 - t)) d t

8

解答ステップ

\int_{0}^{12} \frac{1}{36} t (12 - t) d t

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{36} \cdot \int_{0}^{12} t (12 - t) d t

拡張

t (12 - t) : 12 t - t^{2}

= \frac{1}{36} \cdot \int_{0}^{12} 12 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

= \frac{1}{36} (\int_{0}^{12} 12 t d t - \int_{0}^{12} t^{2} d t)

\int_{0}^{12} 12 t d t = 864

\int_{0}^{12} t^{2} d t = 576

= \frac{1}{36} (864 - 576)

簡素化

\frac{1}{36} (864 - 576) : 8

= 8