The general case

There are no new ideas here, we just have to chase through the algebra for a general positive integer power p. The pictures look just like the special cases of p=2 and p=3.

First the slope problem:

$$\begin{array} {rcl} {\displaystyle\frac{f(a+h)-f(a)}{a+h-a}}... ...& = & k((a+h)^{p-1} + (a+h)^{p-2}a +\cdots+ a^{p-1})\end{array}$$

As h approaches , the slopes of the chords approach kpa^p-1 and it is reasonable to define the slope of the graph to be kpa^p-1 at the point (a, ka^p). For example, the line y = ka^p + kpa^p-1(x-a) touches the graph of y = f(x) only at (a,ka²) but does not cross it:

$$\begin{array} {rcl} f(x) - (kpa^3 + kpa^{p-1}(x-a)) & = & kx... ... \cdots + (p-2)xa^{p-3} + (p-1)a^{p-2})\\ & \geq & 0\end{array}$$

since $x\geq 0$.

As for the area problem,

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

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

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

In other words,

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

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

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

and

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

In other words,

$$L_N \leq ka^{p+1}/(p+1) \leq U_N$$

Therefore

$$\vert{\rm Area} - ka^{p+1}/(p+1)\vert \leq U_N - L_N \leq k(a^{p+1}/N^{p+1})N^{p} = ka^{p+1}/N$$

Therefore, the difference between the area and ka^p+1/(p+1) is smaller than any positive number, meaning ${\rm Area} = ka^{p+1}/(p+1)$. Wow, such a simple formula!