Integration of (e[SUP]-√x[/SUP])/√x

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Homework Help Overview

The discussion revolves around the integration of the function (e-√x)/√x, specifically evaluating the definite integral with bounds from 1 to infinity. Participants are examining the implications of their substitution choices and the behavior of the function at the limits of integration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the integration process, including substitution methods and the resulting limits. There is a focus on understanding the behavior of the integrated function at the specified bounds and questioning the correctness of the limits used.

Discussion Status

The conversation includes attempts to clarify the limits of integration and the nature of the square root function. Some participants have provided guidance on the correct interpretation of the limits, while others are exploring the implications of their initial assumptions.

Contextual Notes

There is mention of confusion regarding the limits of integration, particularly the treatment of the square root function and its positive root. Participants are navigating through these assumptions without reaching a definitive conclusion.

Anne5632
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Homework Statement
Compute the integral with upper bound infinity and lower bound 1
Relevant Equations
Integral of (e[SUP]-√x[/SUP])/√x
(e-√x)/√x (integral from title)

I integrated by substituting and the bounds changed with inf changing to -inf and 1 changing to -1

My final integrated answer is -2lim[e-√x]. What happens to this equation at -inf and -1? As I can't put them into the roots
 
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I'm not sure I understand the difficulty.
 
Anne5632 said:
Homework Statement:: Compute the integral with upper bound infinity and lower bound 1
Relevant Equations:: Integral of (e-√x)/√x

(e-√x)/√x (integral from title)

I integrated by substituting and the bounds changed with inf changing to -inf and 1 changing to -1

My final integrated answer is -2lim[e-√x]. What happens to this equation at -inf and -1? As I can't put them into the roots
You have the wrong limits. \sqrt{x} is the positive root. Thus \sqrt{1} = 1 and \sqrt{+\infty} = +\infty.
 
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pasmith said:
You have the wrong limits. \sqrt{x} is the positive root. Thus \sqrt{1} = 1 and \sqrt{+\infty} = +\infty.
Thank you, I originally let my substitution = -√x
But √x is better
Final answer now is 2/e
 
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