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

The integral from-1 to 1 of 1/(x^{1/3)} is 0

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

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

0

Solution steps

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

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

Simplify

- \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2} + \frac{3}{2} : 0

= 0

\int_{0}^{\infty} e^{- b t} d t

\int \frac{x ^{2}}{4} - 1 d x

\frac{d}{d x} (2 e^{- x} x - e^{- x} x^{2})

\frac{d}{d x} ((4 x - x^{2})^{3})

y^{^{'}} + 8 y = 4, y (0) = 3

What is the integral from-1 to 1 of 1/(x^{1/3)} ?
The integral from-1 to 1 of 1/(x^{1/3)} is 0