Simple integration but i cant get the answer

[semc]

$$\int xe^-^x^/^3dx$$ Integrating from 0 to infinity.

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

 

[lax1113]

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?

 

[Mark44]

$$\int xe^-^x^/^3dx$$ Integrating from 0 to infinity.

So i did this by parts and end up with -3( $$\frac{x}{e^x^/^3}$$ + $$\frac{3}{e^x^/^3}$$ ) x is from 0 to infinity. So for $$\frac{x}{e^x^/^3}$$ 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.
$$\int_0^{\infty} xe^{-x/3}dx~=~\lim_{b \to \infty}\int_0^b xe^{-x/3}dx$$

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.