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

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### 1 Answer

Since it is an improper integral, we'll evaluate it using 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)

According to the rule, 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, then the integral is divergent.**