integral from 0 to 1 of (2-2y^2)^3

解

\int_{0}^{1} (2 - 2 y^{2})^{3} d y

\frac{128}{35}

十進法表記

3.65714 \dots

解答ステップ

\int_{0}^{1} (2 - 2 y^{2})^{3} d y

拡張

(2 - 2 y^{2})^{3} : 8 - 24 y^{2} + 24 y^{4} - 8 y^{6}

= \int_{0}^{1} 8 - 24 y^{2} + 24 y^{4} - 8 y^{6} d y

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} 8 d y - \int_{0}^{1} 24 y^{2} d y + \int_{0}^{1} 24 y^{4} d y - \int_{0}^{1} 8 y^{6} d y

\int_{0}^{1} 8 d y = 8

\int_{0}^{1} 24 y^{2} d y = 8

\int_{0}^{1} 24 y^{4} d y = \frac{24}{5}

\int_{0}^{1} 8 y^{6} d y = \frac{8}{7}

= 8 - 8 + \frac{24}{5} - \frac{8}{7}

簡素化

8 - 8 + \frac{24}{5} - \frac{8}{7} : \frac{128}{35}

= \frac{128}{35}