Integral Evaluate if the integral of y1x13 is convergent or

Integral

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

Solution

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.