I checked it by choosing a value of moment ( say 375) and there are three possible curvature values that could produce this moment value.will be correct.
I see, so your inverse is, mathematically spoken, not a function but just a general relation which is, as you pointed ot, not unique. Functional operators like integrals are defined for functions only - the outcome has to be unique for every abscissa value, which is not true for the inverse of M(Phi). So there is not much you could do analytically with Phi(M).
What should Phi(375) yield? A vector with three values?
How would you define integration over Phi(M)?
What should be the outcome of the definite integral int(Phi(M),M,350,400)?
This wouldn't all make much sense.
The only thing you can do is to turn your inverse into a function by restricting the Phi-range like your pocket calculator does if you ask for the inverse of sinus. arcsin(0.5) yields pi/6, but of course arcsin/0.5) has an infinite number of values (pi/6 + k*2*pi and 5*pi/6 + k*2*pi, with k=integer). But your calculator (and also Mathcad) restricts the angle to the range -pi/2 to pi/2 which makes the inverse a function with unique function values and that way analytical operation like integrals are valid.
You can do the same with your Phi(M). Restrict phi to the range 0 to 0.005 (approx.) and you have a function you are able to integrate. From your graph I see you can create three different functions Phi(M). You are able to integrate each of them frim 350 to 400. But what will you do with the three results?
I am not sure if this would be of help to you. And to automatically determine the number of possible inverse functions and determinig the exact phi-ranges could be a bit cumbersome.
So whatever you have to calculate by the integral of Phi(M), either it is sufficient to use the Phi-range up to the first maximum only or there may be another way to obtain it?
