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, 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` .
  3. If an object moves along a straight line with position function s(t), then its velocity is `v(t)=s'(t)` . So, displacement and distance`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.