doubt in standard proof - if $\int_C F. dr$ is path-indenpendent show that $F=\nabla \phi$

Question

if $\int_C F. dr$ is path-indenpendent show that $F=\nabla \phi$

It is a standard proof given in text. But i did not understand how he concludes if path-dependent setting $ " \phi (x,y,x)" = \int_C F. dr$ [IN QUOTATIONS]

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Answer 1

Assuming $\phi(x,y,z)$ is a generic function for the result of that line integral, integrating with respect to the differential of vector r, if the integral is path-independent, would be the same as integrating with respect to the differential of each component of the vector. Recall that the gradient is the sum of the partial derivatives of the function, so the sum of the integrals with respect to the differential of each of the vector components is equal to integrating the gradient along a path.

Answer 2

The claim is that if the integral is path-independent, then there exists some $\phi$ having $F = \nabla\phi$. To demonstrate this claim, the book constructs and defines a function $\phi$ that does this.

The book defines $\phi$ by the integral. Since the integral is path-independent, the only thing matters is the endpoint $(x, y, z)$ and so $\phi$ is well-defined.