integral from 0 to 3 of (x^3-6x^2-7x)/x

解

\int_{0}^{3} \frac{x ^{3} - 6 x ^{2} - 7 x}{x} d x

- 39

解答ステップ

\int_{0}^{3} \frac{x ^{3} - 6 x ^{2} - 7 x}{x} d x

簡素化

\frac{x ^{3} - 6 x ^{2} - 7 x}{x} : x^{2} - 6 x - 7

= \int_{0}^{3} x^{2} - 6 x - 7 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_{0}^{3} x^{2} d x - \int_{0}^{3} 6 x d x - \int_{0}^{3} 7 d x

\int_{0}^{3} x^{2} d x = 9

\int_{0}^{3} 6 x d x = 27

\int_{0}^{3} 7 d x = 21

= 9 - 27 - 21

簡素化

= - 39