Show that F is conservative assuming it's values depend only on endpoints.

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Assuming that the value of \int\F \bulletdr, then F is a conservative function.

The class is Calc 3. My professor went through this proof in class but it was the one proof in this section that I didn't fully comprehend. Usually I try to find the proof from another source, such as online, to solidify my understanding. I'm unable to find this proof, though.

Can anybody link me to where this proof is recited, or even if it has a formal name?
Thanks.
 
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PS. Sorry for the crappy latex formatting. There should be a "C" underneath the integral sign, and that's supposed to be the vector valued function F dotted with dr, the parametrization of curve C. Thank you.
 
The proof depends stongly on the precise definition of "conservative". Some texts use "the integral depends only on the endpoints" as the definition of "conservative" in which case there is nothing to prove! What definition does your textbook use?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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