integral from 1 to 2 of (2x^3-2x^2-4x-1)

解

\int_{1}^{2} (2 x^{3} - 2 x^{2} - 4 x - 1) d x

- \frac{25}{6}

十進法表記

- 4.16666 \dots

解答ステップ

\int_{1}^{2} 2 x^{3} - 2 x^{2} - 4 x - 1 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_{1}^{2} 2 x^{3} d x - \int_{1}^{2} 2 x^{2} d x - \int_{1}^{2} 4 x d x - \int_{1}^{2} 1 d x

\int_{1}^{2} 2 x^{3} d x = \frac{15}{2}

\int_{1}^{2} 2 x^{2} d x = \frac{14}{3}

\int_{1}^{2} 4 x d x = 6

\int_{1}^{2} 1 d x = 1

= \frac{15}{2} - \frac{14}{3} - 6 - 1

簡素化

\frac{15}{2} - \frac{14}{3} - 6 - 1 : - \frac{25}{6}

= - \frac{25}{6}