Sorry for clouding up things if I did :). You are correct, not all of them have a discontinuity in them. Your integral is an improper Riemann integral of the second kind: http://en.wikipedia.org/wiki/Improper_Riemann_integral. In the right hand corner there are some examples of improper...
Yes, since for stationary/critical/etc... points to exist, your function's derivative has to have points in which its value is 0. Since your function can never have 0 values, you're correct.
The graphical interpretation is also quite neat. Try these in Mathematica, it'll all be clear in a...
You've basically said the definition yourself. An improper integral is an integral in which an infinite discontinuity appears in one of the limit points, or inside the interval. There are numerous methods of solving these integrals (they constitute a series of 14 courses and seminars, each 2...
\frac{ds}{dt}=v(t)
\frac{dv}{dt}=a(t)
Where s(t) is the displacement vector, v(t) is the velocity vector and
a(t) is the acceleration vector.
If you look at it in any given point, so I'm guessing that it's correct.