Some extensions of our results

Since our operations of finding area and finding slope involved only adding or subtracting values of the given function, it is clear that the results we obtained will hold for polynomial functions defined on the non-negative real numbers. For example, if $f:[0,\infty)\rightarrow[0,\infty)$, f(x) = 3x³ + 4x + 2 then

$$\begin{array} {rcl} f'(a) & = & 9a^2 + 4\\ \int_0^a f(u) du & = & (3/4)a^4 + 2a^2 + 2a\end{array}$$

as we see from applying our results in the previous sections to the functions f(x) = 3x³, f(x) = 4x and f(x) = 2 separately and then adding the results.

Next, one might wonder about negative constants. These pose no problem, as the result on the graphs is to invert the positive and negative portions of the vertical axes. In the area problem, this results in our having to declare that area below the horizontal axis should be counted as negative quantity. Thus

$$\begin{array} {rcl} \int_0^{\pi}-2u^2\;du & = & -2\int_0^{\pi}u^2\;du \\ & = & -2(a^3/3)\end{array}$$

Why adopt such a convention? The reason is that in many cases we want to consider sums of the form

$$\frac{a}{N}\left(f(a/N) + f(2a/N) + \cdots + f(Na/N)\right)$$

without necessarily regarding them as area. For example, f(x) might be the temperature at hour x during the day, a = 24 and N = 48. Our sum would then represent the average temperature during the day, if we sampled the temperature every half an hour, multiplied by 24 hours. This would be an estimate of the quantity called degree-hours. The procedure we used to calculate area will apply equally well to this problem, even if f takes on negative values. For example, if f(x) = x²-12x then the number of degree-hours would be

$$\int_0^{24}u^2 - 12u du = 24^3/3 - 12(24)^2/2 = 24^3/12 = 2(24)^2$$

Furthermore, f'(a) would represent the rate of change of the temperature at hour a. For example, f'(1) = 2(1) - 12 = -10 means that the temperature is falling at a rate of 10 degrees per hour at one o'clock (AM), and f'(7) = 2(7)-12 = 2 means that the temperature is rising at the rate of 2 degrees per hour at seven AM. If this is in degrees Fahrenheit, f is not very realistic.

We also need not restrict ourselves to polnomial functions whose domain are just the non-negative real numbers, but allow all real numbers. The results for f' change not at all, nor do their interpretations as slopes. However, we lose the tangent line not crossing the graph property, something we will have to account for in some way or another. The integration properties don't change either if we interpret them correctly. By using symmetry it can be shown that if a < b and

$$\int_a^b u^p\;du$$

is supposed to represent the area over the segment joining (a,0) and (b,0) and if area below the horizontal axis should count as negative, then

$$\int_a^b u^{p-1}\;du = b^{p}/p - a^p/p$$

is what we get. Of all the results we have stated so far, this is the hardest to see.

If a > 0 then this just says that the area we want is the area over the interval from (0,0) to (0,b) minus the area over the interval from (0,0) to (0,a). For the other cases the area analogy breaks down and one has to rely on the algebra to pull us through and show that the formula works.