integral from 0 to 3 of (x^3-4x^2+3x)

解

\int_{0}^{3} (x^{3} - 4 x^{2} + 3 x) d x

- \frac{9}{4}

十進法表記

- 2.25

解答ステップ

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

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

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

\int_{0}^{3} 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}