Mean values of x or y in functions of the form $f(x) = g(y)$ ?

[NormsBreaker]

Suppose I have a function of the form[MATH] f(x) = g(y) [/MATH]? Then how am I supposed to find the mean values of x or y ?

I have the knowledge to find mean values of functions of the form[MATH]y=f(x)[/MATH]using the area under the curve but not more than that.

Although I am more interested in a generalized solution, currently I am stuck with a function of the form
[MATH] y=\frac{kx^3}{e^{mx} -1} [/MATH]
Where k and m are constants.
And I am interested in the mean value of x i. e.[MATH] <x>. x \in (0, \infty) y \in (0, 1.248 \times 10^{-26}] [/MATH]So far, I've tried to somehow represent the above function in the form of[MATH]x=f(y)[/MATH], but to no success. Even after using the McLaurin series expansion, I couldn't do it. It always translates to the form :
[MATH] f(x) = g(y) ? [/MATH]

 

[Steven G]

You say that you tried ... but to no success. Not even with McLaurin's series. Can we see your work so that we can help you? It is quite hard to help you if we do not see you work. For the record if you followed the posting rules you would have received help by now.

 

[NormsBreaker]

@Jomo !
Thanks for your time!
Here's what I have tried.
Using McLaurin Series Expansion,
[MATH]mx + (mx) ^2 /2 + (mx) ^3/6)/kx^3=1/y [/MATH][MATH]=> m/(kx^2)+(m^2)/2kx + (m^3)/6k=1/y[/MATH]This gets stuck to a form [MATH]f(x) =g(y) [/MATH] even after series expansion.
I will very much appreciate any help with this although if there is some other way to find out [MATH]<x>[/MATH] exactly(series expansion is bound to give erraneous results, since the series is being curtailed ) , of which I am not aware of, that would be the best.