Total Change
The Evaluation Theorem says that if f is continuous on [a,b], then `int_a^bf(x)dx=F(b)F(a)` where F is any antiderivative of f. This means that F'=f and equation can be rewritten as `int_a^bF'(x)dx=F(b)F(a)`.
We know that F'(x) represents the rate of change of y=F(x) with respect to x and F(b)F(a) is the change in y when x changes from a to b. So we can reformulate the Evaluation Theorem in words as follows.
Total Change Theorem. The integral of a rate of change is the total change: `int_a^bF'(x)dx=F(b)F(a)`.
This principle can be applied to all of the rates of change in the natural and social sciences.
 If V(t) is the volume of water in a reservoir at time t, then its derivative is the rate at which water flows into the reservoir at time t. So, `int_(t_1)^(t_2)V'(t)dt=V(t_2)V(t_1)` is the change in the amount of water in the reservoir between time `t_1` and `t_2`.
 If `(dn)/(dt)` is the rate of growth of a population, then`int_(t_1)^(t_2)(dn)/(dt)dt=n(t_2)n(t_1)` is the increase in population during time period from `t_1` to `t_2`.
 If an object moves along a straight line with position function s(t), then its velocity is `v(t)=s'(t)`. So, `int_(t_1)^(t_2)v(t)dt=s(t_2)s(t_1)` is the change of position, or displacement, of the particle during the time period from `t_1` to `t_2`.
CAUTION! If we want to calculate the distance traveled during the time interval, we have to consider the intervals when `v(t)>=0` (the particle moves to the right) and also the intervals when `v(t)<=0` (the particle moves to the left). In both cases the distance is computed by integrating `v(t)`, the speed.  The acceleration of the object is `a(t)=v'(t)`, so `int_(t_1)^(t_2) a(t)dt=v(t_2)v(t_1)` is the change in velocity from time `t_1` to time `t_2`.
Example. A particle moves along a line so that its velocity at time t is `v(t)=t^25t+6` (in meters per second).
 Find the displacement of the particle during the time period `1<=t<=4`.
 Find the distance traveled during the time period `1<=t<=4`.
Solution.

The displacement is `s(4)s(1)=int_1^4 v(t)dt=int_1^4 (t^25t+6)dt=(1/3t^35/2t^2+6t)_1^4=`
`=(1/3*4^35/2*4^2+6*4)(1/3*1^35/2*1^2+6*1)=3/2`. This means that the particle’s position at time t=4 is 1.5 m to the right of its position at the start of the time period.

Note, that `v(t)=(t^25t+6)=(t2)(t3)` and so `v(t)<=0` on the interval [2,3] and `v(t)>=0` on [1,2] and [3,4]. Therefore, distance travelled is `int_1^4 v(t)dt=int_1^2 v(t)dt+int_2^3 v(t)dt+int_3^4 v(t)dt=`
`=int_1^2 (t^25t+6)dtint_2^3 (t^25t+6)dt+int_3^4 (t^25t+6)dt=`
`=(1/3t^35/2t^2+6t)_1^2(1/3t^35/2t^2+6t)_2^3+(1/3t^35/2t^2+6t)_3^4=(1/3*2^35/2*2^2+6*2)`
`(1/3*1^35/2*1^2+6*1)(1/3*3^35/2*3^2+6*3)+(1/3*2^35/2*2^2+6*2)+`
`+(1/3*4^35/2*4^2+6*4)(1/3*3^35/2*3^2+6*3)=11/6~~1.83` meters.