What is the integral from-1 to 32 of 1/(x^{1/5)} ?

The integral from-1 to 32 of 1/(x^{1/5)} is 75/4

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integral from-1 to 32 of 1/(x^{1/5)}

\int_{- 1}^{32} \frac{1}{x ^{\frac{1}{5}}} d x

\frac{75}{4}

Decimal Notation

18.75

Solution steps

\int_{- 1}^{32} \frac{1}{x ^{\frac{1}{5}}} 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}{5}}} d x + \int_{0}^{32} \frac{1}{x ^{\frac{1}{5}}} d x

\int_{- 1}^{0} \frac{1}{x ^{\frac{1}{5}}} d x = - \frac{5 ( - 1 ) ^{\frac{4}{5}}}{4}

\int_{0}^{32} \frac{1}{x ^{\frac{1}{5}}} d x = 20

= - \frac{5 ( - 1 ) ^{\frac{4}{5}}}{4} + 20

Simplify

- \frac{5 ( - 1 ) ^{\frac{4}{5}}}{4} + 20 : \frac{75}{4}

= \frac{75}{4}

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What is the integral from-1 to 32 of 1/(x^{1/5)} ?
The integral from-1 to 32 of 1/(x^{1/5)} is 75/4