What is the integral from-64 to 8 of 1/(x^{1/3)} ?

The integral from-64 to 8 of 1/(x^{1/3)} is -18

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integral from-64 to 8 of 1/(x^{1/3)}

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

- 18

Solution steps

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

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

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

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

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

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

Simplify

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

= - 18

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