What is the integral from-64 to 1 of 1/(\sqrt[3]{x)} ?

The integral from-64 to 1 of 1/(\sqrt[3]{x)} is -45/2

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integral from-64 to 1 of 1/(\sqrt[3]{x)}

- \frac{45}{2}

Decimal Notation

- 22.5

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

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

Simplify

- 24 (- 1)^{\frac{2}{3}} + \frac{3}{2} : - \frac{45}{2}

= - \frac{45}{2}

(3 - x) \frac{d y}{d x} = y

\int (\frac{t ^{2} - 8 t + 1}{( 2 t ) ^{4}}) d t

f (x) = - 3 x^{2} + 2, at x = 1

f (x) = \frac{4 x ^{3}}{5 - 2 x}

x \to 6 + lim (\frac{4}{x - 6})

What is the integral from-64 to 1 of 1/(\sqrt[3]{x)} ?
The integral from-64 to 1 of 1/(\sqrt[3]{x)} is -45/2