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

解答

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

18

求解步骤

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

\int_{- 8}^{0} \frac{1}{3 x} d x = - 6 (- 1)^{\frac{2}{3}}

\int_{0}^{64} \frac{1}{3 x} d x = 24

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

化简

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

= 18

d er i v a t i v e 24 x - 0.01 x^{2}

y^{^{''}} + 121 y = sec (11 x)

x \to 0 - lim (\frac{1 - e ^{- 3 x}}{x})

\frac{d}{d x} (\frac{e ^{x}}{7 x ^{2} + 8})

\int_{0}^{5} ∣ x - 3 ∣ d x