Let $(f: \mathbb{R}^2 \rightarrow \mathbb{R}^2)$ be defined as $(f(x, y) = (y, x)^t)$, and let $(\gamma: [0, 1] \rightarrow \mathbb{R}^2)$ be a continuously differentiable curve with $(\gamma(0) = \gamma(1))$. Then it holds that
$ \int_{0}^{1} \langle f(\gamma(x)), D_x\gamma(x) \rangle \, dx = 2\pi. $
To verify the given statement, let's consider the integral again:
$ \int_{0}^{1} \langle f(\gamma(x)), \frac{{d\gamma(x)}}{{dx}} \rangle \, dx. $
According to the definition of $(f(x, y) = (y, x))$, we have $(f(\gamma(x)) = (\gamma_2(x), \gamma_1(x))).$
$(Dx\gamma(x))$ denotes the derivative of $(\gamma(x))$ with respect to $(x)$. The integral can be rewritten as:
$ \int_{0}^{1} (\gamma_2(x) \frac{{d\gamma_1(x)}}{{dx}} + \gamma_1(x) \frac{{d\gamma_2(x)}}{{dx}}) \, dx. $
This can be further simplified as:
$ \int_{0}^{1} \frac{{d(\gamma_2(x) \gamma_1(x))}}{{dx}} \, dx. $
By applying the Fundamental Theorem of Calculus, we get:
$ (\gamma_2(1) \gamma_1(1)) - (\gamma_2(0) \gamma_1(0)). $
Since $(\gamma(0) = \gamma(1))$, we have $(\gamma_2(1) = \gamma_2(0))$ and $(\gamma_1(1) = \gamma_1(0))$. Thus, the expression becomes:
$ \gamma_2(1) \gamma_1(1) - \gamma_2(0) \gamma_1(0) = \gamma_2(0) \gamma_1(0) - \gamma_2(0) \gamma_1(0) = 0. $
Hence, the integral $(\int_{0}^{1} \langle f(\gamma(x)), \frac{{d\gamma(x)}}{{dx}} \rangle \, dx)$ evaluates to $0$.Which means that the statement is false.