integral from 0 to y/3 of ((y^2-9x^2)/2)

解

\int_{0}^{\frac{y}{3}} (\frac{y ^{2} - 9 x ^{2}}{2}) d x

\frac{y ^{3}}{9}

解答ステップ

\int_{0}^{\frac{y}{3}} \frac{y ^{2} - 9 x ^{2}}{2} d x

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{y}{3}} y^{2} - 9 x^{2} d x

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

= \frac{1}{2} (\int_{0}^{\frac{y}{3}} y^{2} d x - \int_{0}^{\frac{y}{3}} 9 x^{2} d x)

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

\int_{0}^{\frac{y}{3}} 9 x^{2} d x = \frac{y ^{3}}{9}

= \frac{1}{2} (\frac{y ^{3}}{3} - \frac{y ^{3}}{9})

簡素化

\frac{1}{2} (\frac{y ^{3}}{3} - \frac{y ^{3}}{9}) : \frac{y ^{3}}{9}

= \frac{y ^{3}}{9}