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

解

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

\frac{128}{15}

十進法表記

8.53333 \dots

解答ステップ

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

拡張

(4 - t) t : 4 t - t^{\frac{3}{2}}

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

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

\int_{0}^{4} t^{\frac{3}{2}} d t = \frac{64}{5}

= \frac{64}{3} - \frac{64}{5}

簡素化

\frac{64}{3} - \frac{64}{5} : \frac{128}{15}

= \frac{128}{15}