Deriving E(ave)=kT: Is it Correct?

[resurgance2001]

I have attached a photo of a page from a book that I am studying. The author is showing a derivation of E(ave) = kT

However, I can't follow this derivation. He says that you need to use integration by substitution and I am OK with that. But when he actually does the substitution it looks wrong. Where there is uv - int vdu , he seems to have left out E in the first term. He also seems to have left out a negative sign. Then the E suddenly pops up again in the second term, int v du and then he says limit e^-E/kT = 0, as E becomes 0. Is that correct? I thought it became 1, or am I completely off the mark? Thanks

 

[axmls]

What is the name of this book? You may want to check if there's an errata sheet out there, because unless we're both losing our minds or don't see some major steps, then there's something off here.

 

[D H]

The expression after "We then have ..." is just wrong. The end result is correct.

There's a much easier way to do this. Let x=E/(kT). Then E=kTx, dE=kT\,dx, resulting in
\int_0^\infty E e^{-E/kT} dE = (kT)^2 \int_0^\infty xe^{-x}dx = (kT)^2
That \int_0^\infty x e^{-x}dx = 1 is one of those definite integrals you should just know. In fact, you should know that \int_0^\infty x^n e^{-x}dx = n! for all non-negative integers n.