What is the integral from-1 to 27 of 1/(\sqrt[3]{x)} ?

The integral from-1 to 27 of 1/(\sqrt[3]{x)} is 12

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integral from-1 to 27 of 1/(\sqrt[3]{x)}

12

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

= - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2} + \frac{27}{2}

Simplify

- \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2} + \frac{27}{2} : 12

= 12

\int \frac{e ^{\frac{9}{t}}}{t ^{2}} d t

\frac{d}{d x} (\frac{x ^{3}}{3 !})

\int (6 x + 5)^{2} d x

b

2 y (x^{2} - y + x) d x + (x^{2} - 2 y) d y = 0

What is the integral from-1 to 27 of 1/(\sqrt[3]{x)} ?
The integral from-1 to 27 of 1/(\sqrt[3]{x)} is 12