Squaring functions

When f(x) = kx² the graph appears to be an upward curving arc. Notice that it makes no sense to ask about slope directly as we have only discussed slopes for line segments. We take a round about tack. Pick a point on the graph of y = f(x), call it (a, ka²) and another different point, (a+h, k(a+h)²) also on the graph. The chord joining these points has slope

$$\frac{k(a+h)^2 - ka^2}{a+h -a} = k\frac{(a+h)^2 - a^2}{h} = k\frac{(a+h-h)(a+h+a)}{a+h-h} = k(2a+h)$$

As h approaches , the slopes of the chords approach 2ka and it is reasonable to define the slope of the graph to be 2ka at the point (a, ka²). For example, the line y = ka² + 2ka(x-a) touches the graph of y = f(x) only at (a,ka²) but does not cross it:

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

The area problem is solved in similar fashion: we approximate the area, and then try to see what we have approximated. 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

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

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

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

In other words,

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

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

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

and

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

In other words,

$$L_N \leq ka^3/3 \leq U_N$$

Therefore

$$\vert{\rm Area} - ka^3/3\vert \leq U_N - L_N \leq k(a^3/N^3)N^2 = ka^3/N$$

Therefore, the difference between the area and ka³/3 is smaller than any positive number, meaning ${\rm Area} = ka^3/3$. Pretty slick.