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If s(t) is a position function with rate of change v(t) = s'(t), then
because both of these quantities describe the same thing:
s(b) – s(a)
is the difference between the starting position at t = a and the ending position at t = b, while
is the weighted distance travelled, taking into account that distance travelled in opposite directions cancels out. Either of these quantities gets us the net change in s from a to b.
There's no particular reason that s(t) needs to describe position and v(t) velocity. As far as the mathematics is concerned, the important thing is that the function inside the integral
is the rate of change, or derivative, of the function outside the integral
s(b) – s(a).
We can generalize from the special case with position and velocity to get the following.
The Fundamental Theorem of Calculus (FTC): If f is continuous on the interval [a, b] and is the derivative of F on [a, b], then
You should memorize the FTC, but you should also understand what it means. Being able to write down
from memory is a good start. Remember that
1) f must be continuous on [a, b]
2) f must equal F'
3) on the right-hand side, we use the upper limit of integration first:
There are other ways to write the FTC. Since we can change the variable of integration, we could write
Or, since the important part of this equation is that the function being integrated is the rate of change of the function outside the integral, we could also write
If you're getting mixed up with all the f's and F's, you can think of the FTC as
Just as we integrate velocity to find the change in position, we integrate the rate of change of f to find the change in f.