Integration of (e[SUP]-√x[/SUP])/√x

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In summary, when integrating (e-√x)/√x with upper bound infinity and lower bound 1, the correct limits are 1 and +∞, not -1 and -∞. The final integrated answer is 2/e^2.
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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.
 
  • #3
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. [itex]\sqrt{x}[/itex] is the positive root. Thus [itex]\sqrt{1} = 1[/itex] and [itex]\sqrt{+\infty} = +\infty[/itex].
 
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  • #4
pasmith said:
You have the wrong limits. [itex]\sqrt{x}[/itex] is the positive root. Thus [itex]\sqrt{1} = 1[/itex] and [itex]\sqrt{+\infty} = +\infty[/itex].
Thank you, I originally let my substitution = -√x
But √x is better
Final answer now is 2/e
 
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Related to Integration of (e[SUP]-√x[/SUP])/√x

1. What is the purpose of integrating (e-√x)/√x?

The purpose of integrating (e-√x)/√x is to find the area under the curve of the function. This can be useful in solving real-world problems and understanding the behavior of the function.

2. What is the general formula for integrating (e-√x)/√x?

The general formula for integrating (e-√x)/√x is ∫(e-√x)/√x dx = -2e-√x + C.

3. Can (e-√x)/√x be integrated using substitution?

Yes, (e-√x)/√x can be integrated using substitution. One possible substitution is u = √x, which will result in du = (1/2√x)dx. The integral then becomes ∫2e-udu = -2e-u + C = -2e-√x + C.

4. Is there a specific range of x values for which (e-√x)/√x can be integrated?

No, there is no specific range of x values for which (e-√x)/√x can be integrated. It can be integrated for all values of x greater than or equal to 0.

5. How is the integration of (e-√x)/√x related to the inverse function?

The integration of (e-√x)/√x is related to the inverse function in that the inverse function of e-√x is √(-lnx), which is the same as the integrand. This means that the integral is essentially finding the inverse function of e-√x.

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