integral from 3 to 0 of (4x-x^3)

解

\int_{3}^{0} (4 x - x^{3}) d x

\frac{9}{4}

十進法表記

2.25

解答ステップ

\int_{3}^{0} 4 x - x^{3} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{3} 4 x - x^{3} 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

= - (\int_{0}^{3} 4 x d x - \int_{0}^{3} x^{3} d x)

\int_{0}^{3} 4 x d x = 18

\int_{0}^{3} x^{3} d x = \frac{81}{4}

= - (18 - \frac{81}{4})

簡素化

- (18 - \frac{81}{4}) : \frac{9}{4}

= \frac{9}{4}