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

解

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

- \frac{1}{4}

十進法表記

- 0.25

解答ステップ

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

拡張

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

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

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

\int_{1}^{2} 3 x d x = \frac{9}{2}

\int_{1}^{2} 3 x^{2} d x = 7

\int_{1}^{2} x^{3} d x = \frac{15}{4}

= 1 - \frac{9}{2} + 7 - \frac{15}{4}

簡素化

1 - \frac{9}{2} + 7 - \frac{15}{4} : - \frac{1}{4}

= - \frac{1}{4}