Interpreting an Integral Whose Integrand Is a Rate of Change

1. The function \(P=P(t) \) relates the population \(P\) (in billions of persons) as a function of the time \(t\) (in years). Suppose \(\int_{0}^{10}P^\prime (t)\, dt=3.\) Since \(P^{\prime} (t)\) is the rate of change of the population with respect to time, then the change in population from \(t=0\) to \(t=10\) is \(3\) billion persons.
2. The function \(v=v(t) \) is the speed \(v\) (in meters per second) of an object at a time \(t\) (in seconds). If \(\int_{0}^{5}v(t) dt=8,\) then the object travels 8 m during the interval \(0\leq t\leq 5.\) Do you see why? The speed \(v=v(t) \) is the rate of change of distance \(s\) with respect to time \(t.\) That is, \(v=\dfrac{ds}{dt}.\)