The value of a scalar line integral does not depend on the choice of parametrization (because it is defined without reference to a parametrization). Prove this directly. That is, suppose that \(\mathbf{r}_{1}(t)\) and \(\mathbf{r}(t)\) are two parametrizations such that \(\mathbf{r}_{1}(t)=\mathbf{r}(\varphi(t))\), where \(\varphi(t)\) is an increasing function. Use the Change of Variables Formula to verify that
\[
\int_{c}^{d} f\left(\mathbf{r}_{1}(t)ight)\left\|\mathbf{r}_{1}^{\prime}(t)ight\| d t=\int_{a}^{b} f(\mathbf{r}(t))\left\|\mathbf{r}^{\prime}(t)ight\| d t
\]
where \(a=\varphi(c)\) and \(b=\varphi(d)\).