integral from-1 to 2 of 6x(x^2-1)^2

解

\int_{- 1}^{2} 6 x (x^{2} - 1)^{2} d x

27

解答ステップ

\int_{- 1}^{2} 6 x (x^{2} - 1)^{2} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{- 1}^{2} x (x^{2} - 1)^{2} d x

拡張

x (x^{2} - 1)^{2} : x^{5} - 2 x^{3} + x

= 6 \cdot \int_{- 1}^{2} x^{5} - 2 x^{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

= 6 (\int_{- 1}^{2} x^{5} d x - \int_{- 1}^{2} 2 x^{3} d x + \int_{- 1}^{2} x d x)

\int_{- 1}^{2} x^{5} d x = \frac{21}{2}

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

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

= 6 (\frac{21}{2} - \frac{15}{2} + \frac{3}{2})

簡素化

6 (\frac{21}{2} - \frac{15}{2} + \frac{3}{2}) : 27

= 27