What is the integral from-1 to 64 of 1/(x^{1/3)} ?

The integral from-1 to 64 of 1/(x^{1/3)} is 45/2

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integral from-1 to 64 of 1/(x^{1/3)}

\int_{- 1}^{64} \frac{1}{x ^{\frac{1}{3}}} d x

\frac{45}{2}

Decimal

22.5

Solution steps

\int_{- 1}^{64} \frac{1}{x ^{\frac{1}{3}}} d x

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

= \int_{- 1}^{0} \frac{1}{x ^{\frac{1}{3}}} d x + \int_{0}^{64} \frac{1}{x ^{\frac{1}{3}}} d x

\int_{- 1}^{0} \frac{1}{x ^{\frac{1}{3}}} d x = - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2}

\int_{0}^{64} \frac{1}{x ^{\frac{1}{3}}} d x = 24

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

Simplify

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

= \frac{45}{2}

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