My Calculus book defines a smooth parametric curve as
- $(x(t),y(t))$ and $(x'(t), y'(t))$ are continuous
- $(x'(t), y'(t))$ do not both equal $0$ for some $t$.
I understand why the first one is important and why the second one is important for calculating the length of a parametric curve, but when calculating line integrals why is it important the derivatives still not be zero at any point along it? Even if our velocity vector is zero and it starts going "backwards" or doing other "crazy" stuff along the curve, as long as we're adding and subtracting along those various infinitesimal displacement vectors the work done by the vector field, won't the vector field have still done the same net work at the end?
Because it is not the parametrization that the authors are defining as "smooth", but the curve itself (that is, the set of points).
"Curve" is a notoriously difficult concept to define, except by way of parametrization, so it is usually defined to mean the image of a continuous map of an interval into the space. But if you define "curve" itself by means of parametrization, then you also have to define the properties of curves by parametrization.
The property here is that a curve is "smooth" if it has a parametrization of the form you have given. This rules out sharp corners such as you see in triangles, squares, or more general polygonal curves. The only way a parametrization can get around such a corner is either for the derivative to be discontinuous, or to be $0$. If you think of the parametrization as the path of a "bug" travelling along the curve, before it can turn the corner the bug has to come to a complete stop. Otherwise its momentum would carry it beyond the corner.
Note that it only requires a single parametrization meeting the conditions to qualify a curve as smooth. Such curves will have plenty of other parametrizations that do not meet the conditions. A circle is a smooth curve, yet one can imagine a bug setting out upon it, only to stop and turn around after its glasses before heading out again (e.g., the parametrization $(\cos(t^3 - t), \sin(t^3 -t))$). Such a parametrization does not disqualify the circle as smooth, since the classical $(\cos t, \sin t)$ parmetrization works.
The authors are not saying that this is the only type of curve usable for line integration. Integrations over polygons are done all the time. What they are saying is that this is the type of curve they are discussing at this point in time. Doubtless after covering the properties of these curves, they will broaden the scope to more general curves - at least to piecewise smooth curves.