What is the integral from-27 to 8 of 1/(\sqrt[3]{x)} ?

The integral from-27 to 8 of 1/(\sqrt[3]{x)} is -15/2

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integral from-27 to 8 of 1/(\sqrt[3]{x)}

- \frac{15}{2}

Decimal Notation

- 7.5

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{27 ( - 1 ) ^{\frac{2}{3}}}{2} + 6

Simplify

- \frac{27 ( - 1 ) ^{\frac{2}{3}}}{2} + 6 : - \frac{15}{2}

= - \frac{15}{2}

2 \frac{d y}{d x} - 4 x y = 8 x

x \to 2 lim ((x + 5) (x - 3))

\frac{d}{d x} (x - x^{2} y - y + x y^{2})

\int 1 + x^{3} d x

What is the integral from-27 to 8 of 1/(\sqrt[3]{x)} ?
The integral from-27 to 8 of 1/(\sqrt[3]{x)} is -15/2