This entry is a general assist for my class working on a project. Suppose you knew the rate at which births are occurring (call it a function of time, b(t)) and you knew the rate at which deaths are occurring (a function d(t)). If the only way the population changes is through births and deaths, then if P(t) is the function describing the size of the population in time, then P'(t) = b(t) - d(t). (It is still your job to explain why this makes biological sense.)
Okay, now for the general principle. Anytime you know the rate of change of a quantity, you can always get back to the original quantity through a definite integral (assuming the rate of change is continuous, anyway). This is the heart of the 2nd Fundamental Theorem of Calculus. Not using P and t as variables (so that you have at least something to translate), here is the basic idea.
Suppose you know f '(x). Then A(x) = ∫0x f '(z) dz is an antiderivative of f '. But so is f(x) since that is where f '(x) comes from. That is f(x) = A(x) + C for some constant. In particular, A(0) = 0, so C=f(0). That is, f(x) = f(0) + ∫0x f '(z) dz.
This will always work, even if I don't start the integral at 0: f(x) = f(a) + ∫ax f '(z) dz. Written another way, it looks like the first Fundamental Theorem of Calculus: f(x) - f(a) = ∫ax f '(z) dz.
In other words, the 2nd FTC implies that every function is its starting value plus the integral of its rate of change.
Now, for our population problem, we don't actually know the rate of change completely; we only know the value at specific points. So instead of computing an integral (to get an exact value), we will approximate the integral using a Riemann sum. We are restricted to using the table data, so Δt=2 is forced upon us. For example, ∫02 b(t) dt can only be estimated with a single rectangle while ∫04 b(t) dt would involve two rectangles. The idea of the Riemann sum is that we choose b(tk*) as one of our data points (either on the left or right).
More specifically, on the interval [0,2] (k=1), we can either use t1*=0 so that b(t1*)=100 or use t1*=2 so that b(t1*)=135. In the first case, the rectangle for k=1 contributes b(t1*)Δt = 200, while the second case leads to a contribution of b(t1*)Δt = 270. The average (midpoint) of these two values (200+270)/2 = 235 is the estimate that would come from using the trapezoid sum. We do this for each of the 8 intervals between our data points, for both the births and the deaths.
By considering our estimates for the number of births and deaths in each of the intervals ([0,2], [2,4], [4,6], etc.), we can produce an estimate of the new population at each of the times (2, 4, 6, 8, etc.). By thinking about what estimates lead to the largest predicted population, we get an upper limit (i.e. bound) for our estimate --- no population consistent with this data can ever go above that value. Similarly, we can choose those estimates to create a lower bound. The true population will be somewhere in between.