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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?

- Thread starter semc
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- #1

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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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Mark44

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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.

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?

[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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