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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 or the 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. 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 $$\int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} \frac{dv}{dt}(t) v(t) \; dt = = \int_{t_1}^{t_2} v(v(t)) \frac{dv}{dt}(t) \; dt = \int_{v_1}^{v_2} v(v) \; dv = {\Huge[} \frac{v^2}{2}{\Huge]}{\Huge|}_{v_1}^{v_2},$$ 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.

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) $\vec{F}_c(\vec{r}) := \|\vec{F}_c (\vec{r}) \| \frac{\vec{r}}{\|\vec{r}\|}$), $\Delta W_{nc} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot d \vec{r} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot \frac{d \vec{v}}{dt}(t) \; dt$ 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 $$V(\vec{r}_1) + T(\vec{v}_1) = V(\vec{r}_2) + T(\vec{v}_2) - \Delta W_{nc}.$$.

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 or the 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. 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 $$\int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} \frac{dv}{dt}(t) v(t) \; dt = \int_{t_1}^{t_2} v(v(t)) \frac{dv}{dt}(t) \; dt = \int_{v_1}^{v_2} v(v) \; dv = {\Huge[} \frac{v^2}{2}{\Huge]}{\Huge|}_{v_1}^{v_2},$$ 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.

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) $\vec{F}_c(\vec{r}) := \|\vec{F}_c (\vec{r}) \| \frac{\vec{r}}{\|\vec{r}\|}$), $\Delta W_{nc} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot d \vec{r} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot \frac{d \vec{v}}{dt}(t) \; dt$ 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 $$V(\vec{r}_1) + T(\vec{v}_1) = V(\vec{r}_2) + T(\vec{v}_2) - \Delta W_{nc}.$$.

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 or the 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. 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 $$\int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} \frac{dv}{dt}(t) v(t) \; dt = \int_{v_1}^{v_2} v(v) \; dv = {\Huge[} \frac{v^2}{2}{\Huge]}{\Huge|}_{v_1}^{v_2},$$ 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.

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) $\vec{F}_c(\vec{r}) := \|\vec{F}_c (\vec{r}) \| \frac{\vec{r}}{\|\vec{r}\|}$), $\Delta W_{nc} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot d \vec{r} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot \frac{d \vec{v}}{dt}(t) \; dt$ 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 $$V(\vec{r}_1) + T(\vec{v}_1) = V(\vec{r}_2) + T(\vec{v}_2) - \Delta W_{nc}.$$.

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 or the 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. 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 $$\int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} \frac{dv}{dt}(t) v(t) \; dt = = \int_{t_1}^{t_2} v(v(t)) \frac{dv}{dt}(t) \; dt = \int_{v_1}^{v_2} v(v) \; dv = {\Huge[} \frac{v^2}{2}{\Huge]}{\Huge|}_{v_1}^{v_2},$$ 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.

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) $\vec{F}_c(\vec{r}) := \|\vec{F}_c (\vec{r}) \| \frac{\vec{r}}{\|\vec{r}\|}$), $\Delta W_{nc} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot d \vec{r} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot \frac{d \vec{v}}{dt}(t) \; dt$ 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 $$V(\vec{r}_1) + T(\vec{v}_1) = V(\vec{r}_2) + T(\vec{v}_2) - \Delta W_{nc}.$$.

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 or the 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. 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.

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) $\vec{F}_c(\vec{r}) := \|\vec{F}_c (\vec{r}) \| \frac{\vec{r}}{\|\vec{r}\|}$), $\Delta W_{nc} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot d \vec{r} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot \frac{d \vec{v}}{dt}(t) \; dt$ 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 $$V(\vec{r}_1) + T(\vec{v}_1) = V(\vec{r}_2) + T(\vec{v}_2) - \Delta W_{nc}.$$.

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 or the 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. 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 $$\int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} \frac{dv}{dt}(t) v(t) \; dt = \int_{v_1}^{v_2} v(v) \; dv = {\Huge[} \frac{v^2}{2}{\Huge]}{\Huge|}_{v_1}^{v_2},$$ 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.

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) $\vec{F}_c(\vec{r}) := \|\vec{F}_c (\vec{r}) \| \frac{\vec{r}}{\|\vec{r}\|}$), $\Delta W_{nc} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot d \vec{r} := \int_{x_1}^{x_2} \vec{F}_{nc} \cdot \frac{d \vec{v}}{dt}(t) \; dt$ 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 $$V(\vec{r}_1) + T(\vec{v}_1) = V(\vec{r}_2) + T(\vec{v}_2) - \Delta W_{nc}.$$.