How do you determine if the improper integral converges or diverges #int 5x^(2)e^(-x^(3))# from 1 to infinity?

1 Answer
Mar 2, 2017

Hence, the integral converges,
and the value of the integral is #5/(3e) #

Explanation:

As this is an improper integral (we have an infinite upper bound), we must examine the behaviour of the integral at the upper limit:

First note that #d/dx e^(-x^3) = -3x^2e^(-x^3) #

And so:

# int \ 5x^2 e^(-x^3) \ dx = -5/3 e^(-x^3) + c#

So for our improper integral we have:

# int_1^oo \ 5x^2 e^(-x^3) \ dx =lim_(a rarr oo) int_1^a \ 5x^2 e^(-x^3) \ dx #
# " " = lim_(a rarr oo) [-5/3 e^(-x^3)]_1^a #

# " " = -5/3 \ lim_(a rarr oo) [ 1/e^(x^3) ]_1^a #

# " " = -5/3 \ lim_(a rarr oo) (1/e^(a^2) - 1/e) #
# " " = -5/3 (0 - 1/e) #
# " " = 5/(3e) #