integral from 0 to 2 of pi(4y^2-y^4)

解

\int_{0}^{2} π (4 y^{2} - y^{4}) d y

π \frac{64}{15}

十進法表記

13.40412 \dots

解答ステップ

\int_{0}^{2} π (4 y^{2} - y^{4}) d y

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{0}^{2} 4 y^{2} - y^{4} 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

= π (\int_{0}^{2} 4 y^{2} d y - \int_{0}^{2} y^{4} d y)

\int_{0}^{2} 4 y^{2} d y = \frac{32}{3}

\int_{0}^{2} y^{4} d y = \frac{32}{5}

= π (\frac{32}{3} - \frac{32}{5})

結合

\frac{32}{3} - \frac{32}{5} : \frac{64}{15}

= π \frac{64}{15}