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

解

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

\frac{19}{6}

十進法表記

3.16666 \dots

解答ステップ

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

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

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

= - (\frac{15}{2} - \frac{14}{3} - 6)

簡素化

- (\frac{15}{2} - \frac{14}{3} - 6) : \frac{19}{6}

= \frac{19}{6}