[SOLVED] [Calculus]--Help with surface area of revolution please

*Problem*: The curve y=e^(-x), x__>__0 is revolvd about the xaxis. Does the resulting surface have finite or infinite area? [Remember tht you can sometimes decide whether improper integral converges w/out calculating it exactly]

*Solution Attempt*:

Surface area over [a,b]= 2(pi)*integral[f(x)*sqrt(1+f'(x)^2)dx]

Comparison test for improper integrals assuming f(x)>g(x)>0 for x>a: if integral[f(x)dx] on [a,infinity] converges, then integral[g(x)dx] on [a,infinity] also converges.

So,

Surface area= 2(pi)*integral[e^(-x)*sqrt(1+e^(-2x))dx].. are the bounds [0,infinity]? What do I do after that to find out if it's finite or not? Do I use the comparison theorem? I'll appreciate any help. Thanks in advance