I ran into a problem that, initially, I thought was a typo.
$$\int_C\ e^xdx $$
where C is the arc of the curve $x=y^3$ from $(-1,-1)$ to $(1,1)$.
I have only encountered line integrals with $ds$ before, not $dx$ (or $dy$, for that matter). At first, I thought the $dx$ was supposed to be a $ds$, but that led to an unsolvable integral.
Unfortunately, my book does not cover this topic very well, and the online answers I have found are rather vague. I tried plugging in $x=y^3$ to get $$ \int_{-1}^{1}\ e^{y^3}3y^2dy $$
which evaluates to $e – \frac{1}{e}$. This is also what I get when I do $ \int_{-1}^{1}\ e^{x}dx $, so I am inclined to believe it is correct. However, I'm not sure, and I'd like to know for certain if my intuition is valid.
Best Answer
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.