integral from-64 to 8 of 1/(\sqrt[3]{x)}

解答

\int_{- 64}^{8} \frac{1}{3 x} d x

- 18

求解步骤

\int_{- 64}^{8} \frac{1}{3 x} d x

找到无定义的点（奇点）:

x = 0

如果存在

b, a < b < c, f (b) =

无定义

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{- 64}^{0} \frac{1}{3 x} d x + \int_{0}^{8} \frac{1}{3 x} d x

\int_{- 64}^{0} \frac{1}{3 x} d x = - 24 (- 1)^{\frac{2}{3}}

\int_{0}^{8} \frac{1}{3 x} d x = 6

= - 24 (- 1)^{\frac{2}{3}} + 6

化简

- 24 (- 1)^{\frac{2}{3}} + 6 : - 18

= - 18

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