TITLE: Line integral with respect to arc length
QUESTION [13 upvotes]: 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.

REPLY [2 votes]: 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 $.