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

The integral from-1 to 1 of 6/(\sqrt[3]{x)} is 0

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

0

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

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

Simplify

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

= 0

What is the integral from-1 to 1 of 6/(\sqrt[3]{x)} ?
The integral from-1 to 1 of 6/(\sqrt[3]{x)} is 0