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Total Change

The Evaluation Theorem says that if f is continuous on [a,b], then $$${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$$ where F is any antiderivative of f. This means that F'=f and equation can be rewritten as $$${\int_{{a}}^{{b}}}{F}'{\left({x}\right)}{d}{x}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$$.

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: $$${\int_{{a}}^{{b}}}{F}'{\left({x}\right)}{d}{x}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$$.

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, $$${\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{V}'{\left({t}\right)}{d}{t}={V}{\left({t}_{{2}}\right)}-{V}{\left({t}_{{1}}\right)}$$$ is the change in the amount of water in the reservoir between time $$${t}_{{1}}$$$ and $$${t}_{{2}}$$$.
  2. If $$$\frac{{{d}{n}}}{{{d}{t}}}$$$ is the rate of growth of a population, then$$${\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}\frac{{{d}{n}}}{{{d}{t}}}{d}{t}={n}{\left({t}_{{2}}\right)}-{n}{\left({t}_{{1}}\right)}$$$ 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 $$${v}{\left({t}\right)}={s}'{\left({t}\right)}$$$. So, $$${\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{v}{\left({t}\right)}{d}{t}={s}{\left({t}_{{2}}\right)}-{s}{\left({t}_{{1}}\right)}$$$ 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 $$${v}{\left({t}\right)}\ge{0}$$$ (the particle moves to the right) and also the intervals when $$${v}{\left({t}\right)}\le{0}$$$ (the particle moves to the left). In both cases the distance is computed by integrating $$${\left|{v}{\left({t}\right)}\right|}$$$, the speed.
  4. The acceleration of the object is $$${a}{\left({t}\right)}={v}'{\left({t}\right)}$$$, so $$${\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{a}{\left({t}\right)}{d}{t}={v}{\left({t}_{{2}}\right)}-{v}{\left({t}_{{1}}\right)}$$$ 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 $$${v}{\left({t}\right)}={{t}}^{{2}}-{5}{t}+{6}$$$ (in meters per second).

  1. Find the displacement of the particle during the time period $$${1}\le{t}\le{4}$$$.
  2. Find the distance traveled during the time period $$${1}\le{t}\le{4}$$$.

Solution.

  1. The displacement is $$${s}{\left({4}\right)}-{s}{\left({1}\right)}={\int_{{1}}^{{4}}}{v}{\left({t}\right)}{d}{t}={\int_{{1}}^{{4}}}{\left({{t}}^{{2}}-{5}{t}+{6}\right)}{d}{t}={\left(\frac{{1}}{{3}}{{t}}^{{3}}-\frac{{5}}{{2}}{{t}}^{{2}}+{6}{t}\right)}{{\mid}_{{1}}^{{4}}}=$$$

    $$$={\left(\frac{{1}}{{3}}\cdot{{4}}^{{3}}-\frac{{5}}{{2}}\cdot{{4}}^{{2}}+{6}\cdot{4}\right)}-{\left(\frac{{1}}{{3}}\cdot{{1}}^{{3}}-\frac{{5}}{{2}}\cdot{{1}}^{{2}}+{6}\cdot{1}\right)}=\frac{{3}}{{2}}$$$. 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 $$${v}{\left({t}\right)}={\left({{t}}^{{2}}-{5}{t}+{6}\right)}={\left({t}-{2}\right)}{\left({t}-{3}\right)}$$$ and so $$${v}{\left({t}\right)}\le{0}$$$ on the interval [2,3] and $$${v}{\left({t}\right)}\ge{0}$$$ on [1,2] and [3,4]. Therefore, distance travelled is $$${\int_{{1}}^{{4}}}{\left|{v}{\left({t}\right)}\right|}{d}{t}={\int_{{1}}^{{2}}}{v}{\left({t}\right)}{d}{t}+{\int_{{2}}^{{3}}}-{v}{\left({t}\right)}{d}{t}+{\int_{{3}}^{{4}}}{v}{\left({t}\right)}{d}{t}=$$$

    $$$={\int_{{1}}^{{2}}}{\left({{t}}^{{2}}-{5}{t}+{6}\right)}{d}{t}-{\int_{{2}}^{{3}}}{\left({{t}}^{{2}}-{5}{t}+{6}\right)}{d}{t}+{\int_{{3}}^{{4}}}{\left({{t}}^{{2}}-{5}{t}+{6}\right)}{d}{t}=$$$

    $$$={\left(\frac{{1}}{{3}}{{t}}^{{3}}-\frac{{5}}{{2}}{{t}}^{{2}}+{6}{t}\right)}{{\left|_{{1}}^{{2}}-{\left(\frac{{1}}{{3}}{{t}}^{{3}}-\frac{{5}}{{2}}{{t}}^{{2}}+{6}{t}\right)}\right|}_{{2}}^{{3}}}+{\left(\frac{{1}}{{3}}{{t}}^{{3}}-\frac{{5}}{{2}}{{t}}^{{2}}+{6}{t}\right)}{{\mid}_{{3}}^{{4}}}={\left(\frac{{1}}{{3}}\cdot{{2}}^{{3}}-\frac{{5}}{{2}}\cdot{{2}}^{{2}}+{6}\cdot{2}\right)}-$$$

    $$$-{\left(\frac{{1}}{{3}}\cdot{{1}}^{{3}}-\frac{{5}}{{2}}\cdot{{1}}^{{2}}+{6}\cdot{1}\right)}-{\left(\frac{{1}}{{3}}\cdot{{3}}^{{3}}-\frac{{5}}{{2}}\cdot{{3}}^{{2}}+{6}\cdot{3}\right)}+{\left(\frac{{1}}{{3}}\cdot{{2}}^{{3}}-\frac{{5}}{{2}}\cdot{{2}}^{{2}}+{6}\cdot{2}\right)}+$$$

    $$$+{\left(\frac{{1}}{{3}}\cdot{{4}}^{{3}}-\frac{{5}}{{2}}\cdot{{4}}^{{2}}+{6}\cdot{4}\right)}-{\left(\frac{{1}}{{3}}\cdot{{3}}^{{3}}-\frac{{5}}{{2}}\cdot{{3}}^{{2}}+{6}\cdot{3}\right)}=\frac{{11}}{{6}}\approx{1.83}$$$ meters.