Simple integration but i cant get the answer

  • Thread starter semc
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  • #1
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[tex]\int xe^-^x^/^3dx[/tex] Integrating from 0 to infinity.

So i did this by parts and end up with -3( [tex]\frac{x}{e^x^/^3}[/tex] + [tex]\frac{3}{e^x^/^3}[/tex] ) x is from 0 to infinity. So for [tex]\frac{x}{e^x^/^3}[/tex] if we sub in infinity it would be infinity/infinity? How do we do that?
 

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  • #2
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I am not really sure if it applies here but remember with horizontal asymptotes, if we were to have for example 3x/x, as x goes to infinity, it is a horizontal asymptote at 3 right?
 
  • #3
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[tex]\int xe^-^x^/^3dx[/tex] Integrating from 0 to infinity.

So i did this by parts and end up with -3( [tex]\frac{x}{e^x^/^3}[/tex] + [tex]\frac{3}{e^x^/^3}[/tex] ) x is from 0 to infinity. So for [tex]\frac{x}{e^x^/^3}[/tex] if we sub in infinity it would be infinity/infinity? How do we do that?
No, you don't substitute infinity - ever. Because infinity is one of the limits of integration, this is what is called an improper integral, so you need to use limits to evaluate it.
[tex]\int_0^{\infty} xe^{-x/3}dx~=~\lim_{b \to \infty}\int_0^b xe^{-x/3}dx[/tex]

Since you already have found the antiderivative, evaluate it at b and at 0, and take the limit as x approaches infinity. You will probably need to use L'Hopital's Rule for the x/e^(x/3) term.
 

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