integral from-3 to 0 of x(x+1)(x+3)

解

\int_{- 3}^{0} x (x + 1) (x + 3) d x

\frac{9}{4}

十進法表記

2.25

解答ステップ

\int_{- 3}^{0} x (x + 1) (x + 3) d x

拡張

x (x + 1) (x + 3) : x^{3} + 4 x^{2} + 3 x

= \int_{- 3}^{0} x^{3} + 4 x^{2} + 3 x 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_{- 3}^{0} x^{3} d x + \int_{- 3}^{0} 4 x^{2} d x + \int_{- 3}^{0} 3 x d x

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

\int_{- 3}^{0} 4 x^{2} d x = 36

\int_{- 3}^{0} 3 x d x = - \frac{27}{2}

= - \frac{81}{4} + 36 - \frac{27}{2}

簡素化

- \frac{81}{4} + 36 - \frac{27}{2} : \frac{9}{4}

= \frac{9}{4}