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

The integral from-8 to 64 of 1/(\sqrt[3]{x)} is 18

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

18

Solution steps

Find undefined (singularity) points:

x = 0

If exist

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

undefined

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

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

Simplify

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

= 18

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

What is the integral from-8 to 64 of 1/(\sqrt[3]{x)} ?
The integral from-8 to 64 of 1/(\sqrt[3]{x)} is 18