**Quote:**

Suppose $C$ is a smooth curve given by $\vec{r}(t)$, $a \leq t \leq b$. Also suppose that $\Phi$ is a function whose gradient vector, $\nabla \Phi=f$, is continuous on $C$. Then

$$

\int_C f \cdot \,\mathrm{d}\vec{r}

= \Phi(\vec{r}(b))-\Phi(\vec{r}(a)).

$$

To prove this, we start by rewriting the integral using the parameterization of $C$. So

$$

\int_C f \cdot \,\mathrm{d}\vec{r}

= \int_a^bf(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t

$$

Since $\Phi$ is the potential function of $f$,

$$

\int_a^b f(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t

= \int_a^b \nabla \Phi(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t,

$$

and with the substitution $\omega=r(t)$, $\mathrm{d}\omega=r^{\prime}(t)\,\mathrm{d}t$,

$$

\int_a^b \nabla \Phi(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t

= \int_{\omega_1}^{\omega_2}\bigl[\Phi(\omega)\bigr]'\,\mathrm{d}\omega,

$$

and since $\omega_1=\vec{r}(a)$ and $\omega_2=\vec{r}(b)$, the fundamental theorem of calculus gives

$$

\int_C f \cdot \mathrm{d}\vec{r}

= \Phi(\omega_2) - \Phi(\omega_1)

= \Phi(\vec{r}(b)) - \Phi(\vec{r}(a)). \quad\Box

$$

Is this proof complete? Can you explain why the equality

$$

\int_a^b \nabla \Phi(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t

= \int_{\omega_1}^{\omega_2}\bigl[\Phi(\omega)\bigr]'\,\mathrm{d}\omega,

$$

holds? Though I can see it intuitively, I seek a more mathematically formal reasoning.