# Proving a theorem in line integrals

1. Feb 13, 2016

### anhtu2907

At the bottom of the picture, I couldn't understand why differentiating with respect to x gives the first integral at the right-hand side 0. Thanks for reading.

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2. Feb 13, 2016

### HallsofIvy

In my opinion, that is simply wrong. Rather than taking $C_1$ to be "any path from (a, b) to $(x_1, y)$" we must choose $C_1$ to be the vertical line from (a, b) to $(a, y)$ then take $C_2$ to be the horizontal line from $(a, y)$ to $(x, y)$.

3. Feb 13, 2016

### Gianmarco

I'm no math pro, but my guess would be that since you have the hypothesis that your integral is path independent, then
$$\int_{C_1} F\cdot dr = \int^{(x1,y)}_{(a,b)} F\cdot dr = f(x1,y) - f(a,b)$$
which differentiated w.r.t. x gives 0 and w.r.t. y does not (since the point (x,y) is arbitrary, y is arbitrary but x1,a,b are fixed)