integral from 1 to 5 of (t^2-t)^2

解

\int_{1}^{5} (t^{2} - t)^{2} d t

\frac{5312}{15}

十進法表記

354.13333 \dots

解答ステップ

\int_{1}^{5} (t^{2} - t)^{2} d t

拡張

(t^{2} - t)^{2} : t^{4} - 2 t^{3} + t^{2}

= \int_{1}^{5} t^{4} - 2 t^{3} + 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_{1}^{5} t^{4} d t - \int_{1}^{5} 2 t^{3} d t + \int_{1}^{5} t^{2} d t

\int_{1}^{5} t^{4} d t = \frac{3124}{5}

\int_{1}^{5} 2 t^{3} d t = 312

\int_{1}^{5} t^{2} d t = \frac{124}{3}

= \frac{3124}{5} - 312 + \frac{124}{3}

簡素化

\frac{3124}{5} - 312 + \frac{124}{3} : \frac{5312}{15}

= \frac{5312}{15}