Is [0,∞] e^(-x^3) dx an Improper Integral?

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Homework Help Overview

The discussion revolves around determining whether the integral from 0 to infinity of e^(-x^3) is classified as an improper integral. Participants are exploring the definitions and reasoning behind this classification.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants assert that the integral is improper due to the upper limit of infinity, while others question this reasoning and seek clarification on the definitions being used. There is also a mention of differing interpretations of what constitutes an improper integral.

Discussion Status

The discussion is active, with participants expressing differing views on the classification of the integral. Some have provided their reasoning, while others are asking for clarification on definitions and assumptions related to improper integrals.

Contextual Notes

Participants reference specific definitions from calculus texts, indicating that the understanding of "improper integral" may vary based on the source material. There is an ongoing exploration of these definitions in the context of the integral in question.

Jbreezy
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Homework Statement


It just wants me to tell whether this is improper or not.
[0,infinity] e^(-x^3) dx

Homework Equations




I say Yes

The Attempt at a Solution

 
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I don't see why it would be improper. What's your reasoning?
 
I'm not 100 percent sure. I said yes because I think it keeps going it's domain is all real.
 
Jbreezy said:

Homework Statement


It just wants me to tell whether this is improper or not.
[0,infinity] e^(-x^3) dx

Homework Equations




I say Yes

The Attempt at a Solution


What definition of "improper integral" are you using?
 
Jbreezy said:

Homework Statement


It just wants me to tell whether this is improper or not.
[0,infinity] e^(-x^3) dx

Homework Equations




I say Yes

The Attempt at a Solution

I would call this integral --
$$\int_0^{\infty}e^{-x^3}dx$$
-- improper, because of the upper limit of integration.
 
I'm just using the def in stewart calc. I think it is improper thanks.
 

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