To determine if the statement is true or false, let's analyze the integrals given in the problem.

We have:

• $C$, a circle of radius 1 centered at $(0, 0)$ and traversed counterclockwise.
• $-C$, the same circle but traversed clockwise.

For any continuous real-valued function $f(x, y)$, the claim is that: $\int_{-C} f \, ds = -\int_{C} f \, ds.$

Explanation:

In line integrals along a closed curve, reversing the direction of traversal changes the sign of the integral. This is because the orientation of the curve affects the direction of $ds$ (the differential of arc length along the path). When you reverse the orientation of $C$, you effectively multiply $ds$ by $-1$, so: $\int_{-C} f \, ds = -\int_{C} f \, ds.$

This property holds for any continuous function $f(x, y)$ along the path $C$.

Conclusion:

The statement is true.

Would you like further details or have any questions?

Here are some related questions to explore this concept further:

1. How does reversing the orientation of a curve affect vector line integrals?
2. What is the impact of curve orientation in Green's Theorem applications?
3. Can this result be generalized for curves other than circles?
4. How does arc length $ds$ relate to the orientation of a path?
5. How does this concept apply to surface integrals in higher dimensions?

Tip: Reversing the path of integration in a line integral simply changes the sign of the integral value.