Cubing functions

The analysis here is entirely parallel to the analysis for squaring functions. When f(x) = kx³ the graph appears to be an upward curving arc. Again it makes no sense to ask about slope directly. As before, we look at chords first. Pick a point on the graph of y = f(x), call it (a, ka³) and another different point, (a+h, k(a+h)³) also on the graph.

The chord joining these points has slope

$$\begin{array} {rcl} {\displaystyle\frac{k(a+h)^3 - ka^3}{a+h... ...)a + a^2)}{a+h-h}}\\ & = & k((a+h)^2 + (a+h)a + a^2)\end{array}$$

As h approaches , the slopes of the chords approach 3ka² and it is reasonable to define the slope of the graph to be 3ka² at the point (a, ka³).

For example, the line y = ka³ + 3ka²(x-a) touches the graph of y = f(x) only at (a,ka²) but does not cross it:

$$kx^3 - (ka^3 + 3ka^2(x-a)) = k(x^3 - a^3 - 3a^2(x-a)) = k(x-a)^2(x+2a) \geq 0$$

since $x\geq 0$.

The area problem is solved in the same fashion as with the squaring functions. Again pick a positive number a and divide the segment on the horizontal axis from (0,0) to (a,0) into N subintervals of equal width. The endpoints of these subintervals first coordinates $0, a/N, 2a/N, \dots, (N-1)a/N$, and each subinterval has width a/N. It is clear from the diagram below that the area between our segment of the horizontal axis and the graph of y = f(x) is less than or equal to the sum of the areas of the circumscribed rectangles, which is given by

$$\begin{array} {rcl} U_N & := & (a/N)k(a/N)^3 + (a/N)k(2a/N)^... ...\\ & = & k(a^3/N^3)(1^3 + 2^3 + 3^3 + \cdots + N^3),\end{array}$$

and the area is larger than or equal to the sum of the areas of the inscribed rectangles:

$$\begin{array} {rcl} L_N & := & (a/N)k(0a/N)^3 + (a/N)k(a/N)^... ...(a^3/N^3)(0^3 + 1^3 + 2^3 + 3^3 + \cdots + (N-1)^3).\end{array}$$

In other words,

$$L_N \leq \;{\rm Area}\;\leq U_N.$$

Now we can apply our observation about differences of powers of consecutive integers:

$$\begin{array} {rcl} L_N & = & k(a^3/N^3)(0^3 + 1^3 + +2^3 +... ...{4}\right)}\\ & = & k(a^4/N^4)(N^4/3)\\ & = & ka^4/4\end{array}$$

and

$$\begin{array} {rcl} U_N & = & k(a^4/N^4)(1^3 + +2^3 + 3^3 +... ...4}\right)}\\ & = & k(a^4/N^4)(N^4/4)\\ & = & ka^4/4.\end{array}$$

In other words,

$$L_N \leq ka^4/4 \leq U_N$$

Therefore

$$\vert{\rm Area} - ka^4/4\vert \leq U_N - L_N \leq k(a^4/N^4)N^3 = ka^4/N$$

Therefore, the difference between the area and ka⁴/4 is smaller than any positive number, meaning ${\rm Area} = ka^4/4$.