Line integral with respect to arc length

Let $C$ be any curve in $\mathbb{R}^d$. A parametrization of $C$ is a map $\gamma: [a,b] \to \mathbb{R}^d$ which trace along the points on $C$ in a specific order. For any two functions $f$, $g$ defined on the set of points belong to $C$, we define the line integral over $C$ by

$$\int_C f dg \stackrel{def}{=} \int_\gamma f dg \stackrel{def}{=}\int_a^b f(\gamma(t))\,(g \circ \gamma)'(t) dt$$

i.e. the line integral is defined through a integral over a specific parametrization of the curve. The key is the value of the integral on the right is independent of the choice of parametrization. For clarity, one can drop the explicit parameter $t$ from the expression.

For your case, $\int e^{x} dx$ really means $\int e^{x(\gamma(t))}\,(x\circ\gamma)'(t) dt$ for whatever parametrization you choose to evaluate the integral. The $s$ you usually see stands for the arc length parametrization, it is only one possible choice of parametrization. You don't need to use it if it make your life harder.

Define $\vec{r}(t) = <t^3,t>$. Define $\vec{F}(t)= <e^{t^3}, 0>$. Then you want $\int_C\vec{F}\cdot d\vec{r}$, which can now be integrated along the $t$ line from -1 to 1. But this is looking at it as a vector field not a scalar field. But it is consistent with the notation.

It is fairly standard to write $\int_C P (x,y)dx+Q (x,y)dy $ for the line integral $\int_C\vec F\cdot d\vec r $, where $\vec F (x,y)=(P (x,y),Q (x,y))$. So in this case $\vec F (x,y)=(e^x ,0)$.

Here one can solve as you did; or you can notice that $\vec F $ is conservative, with potential function $f (x,y)=e^x $. Thus $$\int_C e^x\,dx=f (1,1)-f (-1,-1)=e-\frac1e. $$

It is important to note all this works because $P $ depends only on $x $.