# Inexact differential-path function

From Wikipedia:

an inexact differential cannot be expressed in terms of its antiderivative for the purpose of integral calculations

What's the mathematical reason that inexact differential cannot be expressed in terms of its antiderivative? Not enough information? Too complicated? $$\mathrm{d}U$$ is an exact differential. If you integrate $$\mathrm{d}U$$, what will you get then?

• This question would be better off in a part of stack exchange devoted to maths – Nuclear Chemist Jan 5 at 13:13

## 1 Answer

I think you answer comes from the part of the sentence that you left of the quote from Wikipedia:

... i.e. it's value can't be inferred by looking just at the initial and final states of the given system.

Inexact differential are the result of integrating nonstate functions (functions which aren't path independent). The fundamental theorem of calculus, which allows the integration of a function over a range by determining the antiderivative at the end points of that range, requires path independence in order to be true. So it won't work for inexact differentials that are path dependent.

As for integrating over $$dU$$, $$U$$ is a state function and so you can apply the fundamental theorem and solve the integral using the antiderivatives at the end points.