When a particle moves back and forth in a straight line it is important to
distinguish between velocity (which includes direction) and speed (which does
not include direction). The speed of the particle is simply the scalar part of
the velocity. If the particle is restricted to a straight line then the speed is
given by |v(t)| which removes any distinction being made about the direction of
the particle. If we think of the definite integral as an "accumulation" then the
distinction between velocity and speed leads directly to the distinction between
Total Distance and Net Displacement:
Total Distance =
Net Displacement = 
If I run down a road for 5 miles turn around and run back 4miles, how far have I gone? (One mile or Nine miles?) The question is not clear. If it is asking for total distance traveled, then we must add up all of the distance irrespective of direction. (Nine miles) On the other hand, if you are after net displacement, then the distance coming back is subtracted from the distance out and the answer becomes one mile.
In the discussion above it is important to think of v(t)dt as an infinitely small distance (or a differential of the distance). The definite integral simply adds up all of those small distances from t=a up to t=b. When finding Net Displacement v(t)dt can be negative at some points and the accumulated sum can thus be zero or even negative. The Total Distance by comparison can only increase.
Another way to view this phenomena is break up the motions into positive and negative velocities. If we find the t values where the particles changes direction (i.e. velocity goes through a sign change) then we can determine the positions at those times. By subtracting the proper positions we can determine either the total distance or the net distance traveled. The disadvantage of this method is that you must determine an equation for the position of the particle which will include a constant of integration. Please realize however that any constant will cancel when the subtraction is carried out and the two methods presented here are truly identical.
The above paragraph should sound very familiar to those of you who just learned the Fundamental Theorem of Calculus:
where F(x) is the antiderivative
of f(x).
Since we learned earlier that the velocity v(t) is simply the derivative of the position x(t):
where x(t) is the antiderivative
of v(t)!!
Example:
To complete the introduction, we shall revisit a simple example. Consider a
ball being thrown upward with an initial velocity of 96 ft/sec. The acceleration
on earth is downward at 32 ft/sec. We will ask the following questions:
A. What is v(t)?
B. What is |v(t)|?
C. Find the Net Displacement from t=1 to t=5.
D. Find the Total Distance traveled from t=1 to t=5.
Solution:
A) 
We have arbitrarily chosen downward to be negative. We also used the fact
that v(0)=96.
B) Since v(t) changes sign at t=3 we find that:
C) Net Displacement = 
D) Total Distance = 
Realize that the ball is simply moving upward until time=3sec at which time it
comes back down to the same place that it was at t=1sec. Please compare the area
under the curves:
The area under the "speed" curve is always positive and will add throughout
the integral. On the "velocity" curve the area is under the axis for half of the
time and this cancels the area above the curve.
Alternate Solution: If we assume
that x(0)=0 then:
We realize that v(t) changes sign at t=3 then we calculate:
x(1)=80
x(3)=144
x(5)=80
144-80 = 64ft going up and another 64ft coming back down.
The graph below shows the Height (x) as a function of the time (t)
