integral from 0 to 1 of (t^2-t+1/6)^2

解

\int_{0}^{1} (t^{2} - t + \frac{1}{6})^{2} d t

\frac{1}{180}

十進法表記

0.00555 \dots

解答ステップ

\int_{0}^{1} (t^{2} - t + \frac{1}{6})^{2} d t

拡張

(t^{2} - t + \frac{1}{6})^{2} : \frac{1}{36} + t^{4} - 2 t^{3} + \frac{4 t ^{2}}{3} - \frac{t}{3}

= \int_{0}^{1} \frac{1}{36} + t^{4} - 2 t^{3} + \frac{4 t ^{2}}{3} - \frac{t}{3} 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}^{1} \frac{1}{36} d t + \int_{0}^{1} t^{4} d t - \int_{0}^{1} 2 t^{3} d t + \int_{0}^{1} \frac{4 t ^{2}}{3} d t - \int_{0}^{1} \frac{t}{3} d t

\int_{0}^{1} \frac{1}{36} d t = \frac{1}{36}

\int_{0}^{1} t^{4} d t = \frac{1}{5}

\int_{0}^{1} 2 t^{3} d t = \frac{1}{2}

\int_{0}^{1} \frac{4 t ^{2}}{3} d t = \frac{4}{9}

\int_{0}^{1} \frac{t}{3} d t = \frac{1}{6}

= \frac{1}{36} + \frac{1}{5} - \frac{1}{2} + \frac{4}{9} - \frac{1}{6}

簡素化

\frac{1}{36} + \frac{1}{5} - \frac{1}{2} + \frac{4}{9} - \frac{1}{6} : \frac{1}{180}

= \frac{1}{180}