integral from 0 to 4 of t(4-t)

解

\int_{0}^{4} t (4 - t) d t

\frac{32}{3}

十進法表記

10.66666 \dots

解答ステップ

\int_{0}^{4} t (4 - t) d t

拡張

t (4 - t) : 4 t - t^{2}

= \int_{0}^{4} 4 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}^{4} 4 t d t - \int_{0}^{4} t^{2} d t

\int_{0}^{4} 4 t d t = 32

\int_{0}^{4} t^{2} d t = \frac{64}{3}

= 32 - \frac{64}{3}

簡素化

32 - \frac{64}{3} : \frac{32}{3}

= \frac{32}{3}