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

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

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integral from-8 to 27 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

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

Simplify

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

= \frac{15}{2}

2 y^{^{''}} + 35 y = 0, y (0) = 9, y^{^{'}} (0) = 6

\frac{\partial}{\partial x} (x e^{x^{2} - y^{2}})

y^{^{''}} + y = 5 e^{- 7 t}

e^{- 5 t} sin (2 t)

\int - x^{2} + 4 x d x

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