Integrating differential form. Integral split with same integration margins.

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In my textbook of physical chemistry is example of integration of total differential. I'll paraphrase it as I understood it.

By integrating $$1) dU=\delta q-pdV$$ we get $$2) \int_{i,L}^f dU=\int_{i,L}^f (\delta q-pdV )$$ where path is p=const $$3) \int_{i,L}^f dU=\int_{i,L}^f \delta q- \int_{i,L}^fpdV $$ then $$4) U_f-U_i= q_p- p(V_f-V_i)$$

I don't understant transition from 2 to 3 step. Why is is possible to divide path integral, where path is p=const to two integrals with same integration margins?

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There is well known definite integral propertie: $$\int_i^f(adx+bdx)=\int_i^f a(x)dx+\int_i^f b(x)dx$$ Proof of these property is in many textbooks of calculus. Line integral has analogous property. It is because line integral is limit of sum. $$ \int_{AB}=\lim_{n\to \infty} \sum_{i=1}^n (P(x,y) \Delta x_i +Q(x,y) \Delta y_i )=\lim_{n\to \infty}\sum_{i=1}^n P(x,y)\Delta x_i +\lim_{n\to \infty}\sum_{i=1}^n Q(x,y)\Delta y_i =\int_{AB} P(x,y)d x+\int_{AB} Q(x,y)dy $$