# 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.

1. 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 .
2. If (dn)/(dt) is the rate of growth of a population, thenint_(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 .
3. 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.
4. 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^2-5t+6 (in meters per second).

1. Find the displacement of the particle during the time period 1<=t<=4 .
2. Find the distance traveled during the time period 1<=t<=4 .

Solution.

1. The displacement is s(4)-s(1)=int_1^4 v(t)dt=int_1^4 (t^2-5t+6)dt=(1/3t^3-5/2t^2+6t)|_1^4=

=(1/3*4^3-5/2*4^2+6*4)-(1/3*1^3-5/2*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.

2. Note, that v(t)=(t^2-5t+6)=(t-2)(t-3) 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^2-5t+6)dt-int_2^3 (t^2-5t+6)dt+int_3^4 (t^2-5t+6)dt=

=(1/3t^3-5/2t^2+6t)|_1^2-(1/3t^3-5/2t^2+6t)|_2^3+(1/3t^3-5/2t^2+6t)|_3^4=(1/3*2^3-5/2*2^2+6*2)-

-(1/3*1^3-5/2*1^2+6*1)-(1/3*3^3-5/2*3^2+6*3)+(1/3*2^3-5/2*2^2+6*2)+

+(1/3*4^3-5/2*4^2+6*4)-(1/3*3^3-5/2*3^2+6*3)=11/6~~1.83 meters.