Thanks to the OP for posting this great question. Indeed in mathematical physics  it is more expedient to manipulate integrals and derivatives using the approach of the [infinitesimal calculus](https://encyclopediaofmath.org/wiki/Infinitesimal_calculus) or the [nonstandard calculus](https://en.wikipedia.org/wiki/Nonstandard_calculus). However, mathematical rigor can often provide clarity and avoid *differentiability* issues which lead to erroneous calculations.

Addressing the first part of the question requires application of the [change of variables theorem](https://mathworld.wolfram.com/ChangeofVariablesTheorem.html). We denote the twice differentiable $C^2(\mathbb{R})$ (for convenience) scalar functions $x: t \mapsto x(t), v: t \mapsto \frac{dx}{dt}(t)$, $a: t \mapsto \frac{d^2x}{dt^2}(t)$ and $F_c + F_{nc}: x \mapsto F_c(x) + F_{nc}(x)$. The definite integral manipulation corresponding to the mathematically accurate definition of the mass normalized work ($\frac{1}{m} \int_{x_1}^{x_2} (F_c + F_{nc})(x) \; dx := \int_{t_1}^{t_2} a(x(t)) \frac{dx}{dt}(t) \; dt$), which uses the change of variable $x = x(t)$, given as 
$$\frac{1}{m} \int_{x_1}^{x_2} (F_c + F_{nc})(x) \; dx = \int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} a(x(t)) \frac{dx}{dt}(t) \; dt,$$ where $x(t_1) = x_1$ and $x(t_2) = x_2$ implies that 
$$\int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} \frac{dv}{dt}(t) v(t) \; dt,$$ while the change of variable $v = v(t)$ (applied *in reverse*, to state the manipulation in words) implies that 
$$\int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} \frac{dv}{dt}(t) v(t)  \; dv = \int_{v_1}^{v_2} v(v) \; dv = {\Huge[} \frac{v^2}{2}{\Huge]}{\Huge|}_{v_1}^{v_2},$$
where $v(t_1) = v_1$ and $v(t_2) = v_2$. The work-energy theorem in the one dimensional case follows directly from this result. This completes the first part of the answer.

The vector dot product $(a_x \hat{i}_{x} + a_y \hat{i}_{y} + a_z \hat{i}_{z}) \cdot d\vec{r} = a_x d x + a_y dy + a_z dz$ so that, using the previous result, $$\int_{C} (a_x \hat{x} + a_y \hat{y} + a_z \hat{z}) \cdot d\vec{r} = \int_{(x_1,y_1,z_1)}^{(x_2,y_2,z_2)} (a_x \hat{x} + a_y \hat{y} + a_z \hat{z}) \cdot d\vec{r} \\ = \int_{x_1}^{x_2} a_x dx + \int_{y_1}^{y_2} a_y dy + \int_{z_1}^{z_2} a_z dz = \frac{1}{2}({\large[}v_x^2{\large]}{\Huge|}_{v_{x1}}^{v_{x2}} + {\large[}v_y^2{\large]}{\Huge|}_{v_{y1}}^{v_{y2}} + {\large[}v_z^2{\large]}){\Huge|}_{v_{z1}}^{v_{z2}} = \frac{1}{2} [\|\vec{v}\|^2]{\Huge|}_{\vec{v}_1}^{\vec{v}_2},$$ which completes the second part of the answer.


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Finally, defining $\Delta W_{c} := \int_{x_1}^{x_2} \vec{F}_{c} \cdot d \vec{r} := \int_{x_1}^{x_2} \vec{F}_{c} \cdot \frac{d \vec{v}}{dt}(t) \; dt$, $\nabla V(\vec{r}) := - \vec{F}_c(\vec{r})$ (such a potential function $V \in C^1(\mathbb{R})$ exists due to the definition of [conservative forces (which are not necessarily central forces but have a form related to the central forces)](https://physics.stackexchange.com/q/213061/254821)) $\vec{F}_c(\vec{r}) := \|\vec{F}_c (\vec{r}) \| \frac{\vec{r}}{\|\vec{r}\|}$, $\Delta W_{nc} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot d \vec{r} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot \frac{d \vec{v}}{dt}(t) \; dt$ and $T(\vec{v}) := \frac{1}{2} m \|\vec{v}\|^2$, we can state the work-energy theorem for a material particle on which conservative and non-conservative forces act as 
$$\Delta W_{c} + \Delta W_{nc} = V(\vec{r}_1) - V(\vec{r}_2) + \Delta W_{nc} = T(\vec{v}_2) - T(\vec{v}_1),$$
with the *conservation form* given as
$$V(\vec{r}_1) + T(\vec{v}_1) = V(\vec{r}_2) + T(\vec{v}_2) - \Delta W_{nc}.$$.