integral from-1 to 1 of 6/(\sqrt[3]{x)}

解答

\int_{- 1}^{1} \frac{6}{3 x} d x

0

求解步骤

\int_{- 1}^{1} \frac{6}{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_{- 1}^{0} \frac{6}{3 x} d x + \int_{0}^{1} \frac{6}{3 x} d x

\int_{- 1}^{0} \frac{6}{3 x} d x = - 9 (- 1)^{\frac{2}{3}}

\int_{0}^{1} \frac{6}{3 x} d x = 9

= - 9 (- 1)^{\frac{2}{3}} + 9

化简

- 9 (- 1)^{\frac{2}{3}} + 9 : 0

= 0

\int_{y}^{2 y} 8 x + e^{y} d x

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

\int_{2}^{4} - x^{3} + 6 x^{2} - 8 x d x

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\int_{0}^{10} - 6 x^{2} - 12 x + 720 d x