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

Question

Christopher Agresta · Accepted Answer

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

Charles Arocha · Answer

undefined To determine if the improper integral $ \int_{1}^{\infty} 5x^2e^{-x^3} \, dx $ converges or diverges:

1. First, check for convergence at $ x = 1 $. Evaluate $ 5x^2e^{-x^3} $ at $ x = 1 $ to see if it's finite.
2. Next, investigate the behavior of the integrand as $ x $ approaches infinity. Analyze whether $ 5x^2e^{-x^3} $ approaches zero as $ x $ goes to infinity, as this is a necessary condition for convergence.
3. If the integrand approaches zero as $ x $ approaches infinity, apply the Limit Comparison Test or the Comparison Test with a known convergent or divergent integral to determine convergence or divergence of the original integral.
4. If the integral passes both convergence tests, then it converges. If it fails one or both tests, it diverges.

By following these steps, you can determine if the improper integral converges or diverges.