Path Integral

Let $[a,b] \subset \mathbb{R}$ where $a \neq b$.

We partition that interval using points $x_0, x_1, \dots, x_n$ where $x_0 = a$ and $x_n = b$.

We want an expression for the total length covered by summing all the partition pieces. Each piece contributes $x_{i+1} - x_i$, so the total is

$$\sum_{i=0}^{n-1} (x_{i+1} - x_i)$$

Do you agree that this sum telescopes? Every intermediate point appears once as a left endpoint and once as a right endpoint with opposite sign, so everything cancels except the outermost terms:

$$\sum_{i=0}^{n-1} (x_{i+1} - x_i) = b - a$$

I can choose variable step size $x_{i+1} - x_i)$ and even choose a constant one but only the endpoints will matter.

By the del or nabla operator article, we understand that for a scalar function $h(x,y)$ the gradient $\nabla h$ encodes how $h$ changes in every direction. Dotting the gradient with a small displacement $d\vec{r}$ gives the infinitesimal change in $h$ along that displacement:

$$\nabla h \cdot d\vec{r} = \frac{\partial h}{\partial x}\,dx + \frac{\partial h}{\partial y}\,dy$$

This is exactly how $h$ changes over the small step $d\vec{r}$.

Now suppose we have a path $C$ through space. We can sum all those infinitesimal changes along the entire path:

$$\int_C \nabla h \cdot d\vec{r}$$

Just as with our partition, each infinitesimal contribution is a difference of $h$ at two neighbouring points. Summing them all, the intermediate values telescope and we are left with

$$\int_C \nabla h \cdot d\vec{r} = h(\vec{r}_{\text{end}}) - h(\vec{r}_{\text{start}})$$

The path taken does not matter. Only the endpoints do. This is the gradient theorem, the multivariable analogue of the fundamental theorem of calculus.