integral from 0 to 2 of (2-t)^2t^3

解

\int_{0}^{2} (2 - t)^{2} t^{3} d t

\frac{16}{15}

十進法表記

1.06666 \dots

解答ステップ

\int_{0}^{2} (2 - t)^{2} t^{3} d t

部分積分を適用する

= [\frac{1}{4} t^{4} (2 - t)^{2} - \int - \frac{1}{2} t^{4} (- t + 2) d t]_{0}^{2}

\int - \frac{1}{2} t^{4} (- t + 2) d t = \frac{t ^{6}}{12} - \frac{t ^{5}}{5}

= [\frac{1}{4} t^{4} (2 - t)^{2} - (\frac{t ^{6}}{12} - \frac{t ^{5}}{5})]_{0}^{2}

簡素化

[\frac{1}{4} t^{4} (2 - t)^{2} - (\frac{t ^{6}}{12} - \frac{t ^{5}}{5})]_{0}^{2} : [\frac{1}{30} t^{4} (5 t^{2} - 24 t + 30)]_{0}^{2}

= [\frac{1}{30} t^{4} (5 t^{2} - 24 t + 30)]_{0}^{2}

限度を計算する:

\frac{16}{15}

= \frac{16}{15}