# IntegralEvaluate if the integral of y=1/(x^1/3) is convergent or divergent when  x = 1 to x = infinite.

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have the function y = 1/(x^1/3) = x^(-1/3)

The integral of the function for x = 1 to x = inf. is convergent if the value is finite.

Int [ x^(-1/3) dx]

=> x^(2/3) / (2/3)

For the limits x = 1 to x = infinity the area is

inf - 3/2 = infinity

As the area is not a finite value the definite integral is divergent

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll apply the definition of improper integrals:

Int f(x)dx (x = a to x = infinite) = lim (N -> infinite) Int f(x)dx( x = a to x = N)

By definition, we'll get:

Int dx/(x^1/3) (x = 1 to x = infinite) = lim (N -> infinite) Int dx/(x^1/3) ( x = a to x = N)

We'll determine  Int dx/(x^1/3):

Int dx/(x^1/3) = Int x^(-1/3) dx

Int x^(-1/3) dx = x^(-1/3 + 1)/(-1/3 + 1) + C

Int x^(-1/3) dx = 3x^(2/3)/2, for x = 1 to x = N

Now, we'll determine the limit:

lim [3N^(2/3)/2 - 3*1^(2/3)/2] = lim [3N^(2/3)/2 - 3/2]

lim [3N^(2/3)/2 - 3/2] = infinite

Since the improper integral is infinite, therefore the integral is divergent.