This question is motivated by a calculation in Section 2.2.4 of Griffiths' book on an Introduction to Electrodynamics, in which he shows that the field of a point charge is curl-free.
The field of a point charge at the origin is given in spherical coordinates by $$\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ \hat{\mathbf{r}}.$$
Here $\epsilon_0$ is the permittivity of free space, $q$ is the magnitude of the charge, $r$ is the radius from the origin, and $\hat{\mathbf{r}}$ is radial spherical basis vector.
The line integral of interest is $$\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{E}\cdot d\mathbf{l},$$ and is taken over an arbitrary path in $\mathbb{R}^3$ (that presumably doesn't include the origin). This integral is calculated in spherical coordinates, so the infinitesimal displacement is given by $$d\mathbf{l}=dr\ \hat{\mathbf{r}}+r\ d\theta\ \hat{\boldsymbol\theta}+r\ \sin\theta\ d\phi\ \hat{\boldsymbol\phi}.$$
In Griffiths' notation $\theta$ is the azimuthal angle, and $\phi$ is the polar angle. To calculate the integral he first observes $$\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ dr.$$ The integral is then evaluated as follows $$\int_\mathbf{a}^\mathbf{b}\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\int_\mathbf{a}^\mathbf{b}\frac{q}{r^2}\ dr=\frac{-1}{4\pi\epsilon_0}\frac{q}{r}\bigg|^{r_b}_{r_a}=\frac{1}{4\pi\epsilon_0}\bigg(\frac{q}{r_a}-\frac{q}{r_b}\bigg).$$ Here $r_a$ and $r_b$ are the radii associated with $\mathbf{a}$ and $\mathbf{b}$. It is then argued that the integral around any closed path is zero, so that $\mathbf{E}$ is curl-free by Stokes theorem. I cannot understand how he got from $$\frac{1}{4\pi\epsilon_0}\int_\mathbf{a}^\mathbf{b}\frac{q}{r^2}\ dr$$ to $$\frac{-1}{4\pi\epsilon_0}\frac{q}{r}\bigg|^{r_b}_{r_a}.$$ Doesn't this integral need to be parameterized? If he is integrating over a contour, how is it legitimate to just take the antiderivative? I would think that if he is integrating over a path, he would need to express $r=r(\theta,\phi)$ and find a corresponding formula for $dr$.
I am obviously missing something, but I can't seem to figure out what it is. I would be grateful to anyone who would be able to help.
Edit: I asking how he got from $\int_\mathbf{a}^\mathbf{b}\frac{1}{\mathbf{r}^2(x)}d(\mathbf{x})$ to $\int_{r(\mathbf{a})}^{r(\mathbf{b})}\frac{1}{r^2}dr$. Thanks to @Alex Burdin for the suggestion.
$$\vec{E}=E_r\vec{r}+0\vec{\theta}+0\vec{\phi}$$ thus
$$\vec{E}.\vec{dl}=E_rdr=\frac{q}{4\pi\epsilon_0}\frac{dr}{r^2}$$ with $$\int_{r_a}^{r_b}\frac{1}{r^2}dr=\Bigl[\frac{-1}{r}\Bigr]_{r_a}^{r_b}$$
$$-\frac{1}{r_b}-(-\frac{1}{r_a})=\frac{1}{r_a}-\frac{1}{r_b}$$