If two parametrization have the same endpoints, do they have the same path integral? Let $c(t)=(\cos t,\sin t)$, for $0\leq t\leq 2\pi$, and let $p(t)=(\cos t,−\sin t)$, for $0\leq t\leq 2\pi$. Is it true that $\displaystyle \int \limits _c​f\,ds=\int \limits _p f\,ds$ for every continuous (real-valued) function $f$ on $\mathbb{R}^2$?
Answer choices are:
a. True for all functions $f$.
b. True for some functions $f$ and false for others.
c. False for all functions $f$.
d. It's complicated.
I tried plugging in the endpoint and see that $c(0)=(1,0)$, $c(2\pi )=(1,0)$, $p(0)=(1,0)$, $p(2\pi )=(1,0)$. Is this sufficient to say it's true for all functions $f$?
Again, thanks and happy math doing!
 A: You are right that the integrals are the same, but not for the reason you provide.
An integral of the form $\int_C f \, ds$ can be interpreted as representing the signed area of a ``ribbon'' with height $f$ over a base curve $C$, as shown in the image below.  (In the image, the curve $C$ is shown in red in the $xy$-plane.)

(Parenthetical note: I say that $\int_C f \, ds$ is the signed area of the ribbon because, just as in single-variable calculus, we count the area as negative where the function $f$ is negative.  That doesn't affect the considerations here, but is important for understanding the full situation.)
When it's described this way, it should be clear that the area depends only on two things:  the path, and the height function.  Using two different parametrizations of the same path shouldn't make a difference -- it's still the same ribbon!
To return to the example in your question:  $c(t)$ and $p(t)$ are two different parametrizations of the unit circle (one oriented clockwise, one oriented counter-clockwise), so the integrals $\int_c f \, ds$ and $\int_p f \, ds$ are descriptions of the exact same area.
But note that it's not enough to verify that the two paths have the same endpoints!  Consider the image below, which shows two different paths joining the two endpoints $A = (1,1)$ and $B = (-1,-11)$. The first path (in red) is a line segment; the second path (in blue) is a semicircle.  Over each paths we construct a ``ribbon'' using the same height function (in this example, $f(x,y) = 2 + x^2 y$). As the image suggests, the ribbons have quite different areas, despite the fact that the paths begin and end at the same point.

Perhaps the most basic example of this is when we consider the function $f(x,y) = 1$.  Then $\int_c f \, ds$ is the length of the path $c$, and $\int_p f \, ds$ is the length of the path $p$.  Just because two paths begin and end at the same point doesn't automatically mean that they have the same length!  However, if the paths are the same, the integrals will be the same.
(You may find it helpful to think of two different parametrizations of a path as being like two different cars driving down the same road but at different speeds.  The length of the road is the same, regardless of how quickly you drive along it.)
A: I think it is true for all f. Because
$\int_{0}^{2\pi}f(cost,-sint)dt=-\int_{0}^{-2\pi}f(cosu,sinu)du=\int_{-2\pi}^{0}f(cosu,sinu)du=\int_{-2\pi}^{0}f(cos(u+2\pi),sin(u+2\pi))du=\int_{0}^{2\pi}f(cost,sint)dt$
