Actually, no. Let the curve be c(t)=(a(t),b(t)). In the expression int_c f(x,y)dx, you have x= the i-coordinate of the curve= a(t). As for ds, it equals sqrt{a'(t)^2+b'(t)^2}dt, with a little help from differential geometry. So,Am I right in thinking that line integrals of the form int_c f(x,y)dx arise by approximating the arc length delta s by the line segment delta x so that as delta x tends to 0 delta s / delta x = 1?

ds/dx->1 cannot hold.

The general setting for the curvilinear integral int_c f(x,y)ds, is calculating the signed area of the surface, formed by the curve c and the graph of f. Let's see this (half-vigorously); Consider a partition of the curve, that is divide it into minor segments of length ds. For a random point (x,y) on the minor segment, f(x,y)ds is the signed area of the parallelogram of sides ds and f(x,y). Adding these up, we get Sum{f(x,y)ds}, which is actually a riemannian sum, converging to int_c f(x,y)ds.I can undertsand int_c f(x,y)ds as dividing the curve into smaller and smaller arcs. Is this similar to resolving vecors into i components and j components (for int_c f(x,y)dy) .

For practical calculation, if c(t)=(a(t),b(t)), then

int_c f(x,y)ds=int f(a(t),b(t)) sqrt{(a'(t))^2+(b'(t))^2}dt.

Hope this helps.