integral from-1 to 1 of (7x^3-2x^2+5x-4)

解

\int_{- 1}^{1} (7 x^{3} - 2 x^{2} + 5 x - 4) d x

- \frac{28}{3}

十進法表記

- 9.33333 \dots

解答ステップ

\int_{- 1}^{1} 7 x^{3} - 2 x^{2} + 5 x - 4 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}^{1} 7 x^{3} d x - \int_{- 1}^{1} 2 x^{2} d x + \int_{- 1}^{1} 5 x d x - \int_{- 1}^{1} 4 d x

\int_{- 1}^{1} 7 x^{3} d x = 0

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

\int_{- 1}^{1} 5 x d x = 0

\int_{- 1}^{1} 4 d x = 8

= 0 - \frac{4}{3} + 0 - 8

簡素化

0 - \frac{4}{3} + 0 - 8 : - \frac{28}{3}

= - \frac{28}{3}