2.1 The Area Under a Power Function

Let $a$ be a positive number, let $r$ be a positive number, and let $S_a^r$ be the set of points $(x,y)$ in $\mbox{{\bf R}}^2$ such that $0\leq x\leq a$ and $0\leq y\leq x^r$. In this section we will begin an investigation of the area of $S_a^r$.

Our discussion will not apply to negative values of $r$, since we make frequent use of the fact that for all non-negative numbers $x$ and $t$

$$(x \leq t) \mbox{ implies that } (x^r \leq t^r).$$

Also $0^r$ is not defined when $r$ is negative.

The figures for the argument given below are for the case $r=2$, but you should observe that the proof does not depend on the pictures.

Let $n$ be a positive integer, and for $0\leq i \leq n$, let $${x_i={{ia}\over n}}$$.

Then $${x_i-x_{i-1}={a\over n}}$$ for $1\leq i\leq n$, so the points $x_i$ divide the interval $[0,a]$ into $n$ equal subintervals. For $1\leq i\leq n$, let

$$\begin{eqnarray*} I_i&=&B(x_{i-1},x_i\colon 0,x_{i-1}^r) \\ O_i&=&B(x_{i-1},x_i\colon 0,x_i^r). \end{eqnarray*}$$

If $(x,y)\in S_a^r$, then $x_{i-1}\leq x\leq x_i$ for some index $i$, and $0\leq y\leq x^r\leq x_i^r$, so

$$(x,y)\in B(x_{i-1},\; x_i\colon 0,\; x_i^r)=O_i \mbox{ for some } i\in \{1,\cdots ,n\}.$$

Hence we have

$$S_a^r\subset \bigcup_{i=1}^n O_i,$$

and thus

$$\mbox{\rm area}(S_a^r)\leq\mbox{\rm area}(\bigcup_{i=1}^n O_i).$$ (2.1)

If $(x,y)\in I_i$, then $0\leq x_{i-1}\leq x\leq x_i\leq a$ and $0\leq y\leq x_{i-1}^r\leq x^r$ so $(x,y)\in S_a^r$. Hence, $I_i\subset S_a^r$ for all $i$, and hence

$$\bigcup_{i=1}^n I_i\subset S_a^r,$$

so that

$$\mbox{\rm area}(\bigcup_{i=1}^n I_i)\leq\mbox{\rm area}(S_a^r).$$ (2.2)

Now

$$\begin{eqnarray*} \mbox{\rm area}(I_i)&=&\mbox{\rm area}\left(B(x_{i-1},x_i\colo... ...t( {{(i-1)a}\over n}\right)^r={{a^{r+1}}\over {n^{r+1}}}(i-1)^r, \end{eqnarray*}$$

and

$$\begin{eqnarray*} \mbox{\rm area}(O_i)&=&\mbox{\rm area}\left(B(x_{i-1},x_i\colo... ...r n}\left( {{ia}\over n}\right)^r={{a^{r+1}}\over {n^{r+1}}}i^r. \end{eqnarray*}$$

Since the boxes $I_i$ intersect only along their boundaries, we have

$$\mbox{\rm area}(\bigcup_{i=1}^nI_i) = \mbox{\rm area}(I_1)+\mbox{\rm area}(I_2)+\cdots +\mbox{\rm area}(I_n)\mbox{{}}$$

$$= {{a^{r+1}}\over {n^{r+1}}}0^r + {{a^{r+1}}\over {n^{r+1}}}1^r+\cdots + {{a^{r+1}}\over {n^{r+1}}}(n-1)^r\mbox{{}}$$

$$= {{a^{r+1}}\over {n^{r+1}}}(1^r+2^r+\cdots +(n-1)^r),$$ (2.3)

and similarly

$$\begin{eqnarray*} \mbox{\rm area}(\bigcup_{i=1}^n O_i)&=&\mbox{\rm area}(O_1)+\m... ...1}}}n^r \\ &=& {{a^{r+1}}\over {n^{r+1}}}(1^r+2^r+\cdots +n^r). \end{eqnarray*}$$

Thus it follows from equations (2.1) and (2.2) that

$${{a^{r+1}}\over {n^{r+1}}}(1^r+2^r+ \cdots +(n-1)^r)\leq\mbox{\rm area}(S_a^r)\leq {{a^{r+1}}\over {n^{r+1}}}(1^r+2^r+\cdots +n^r).$$ (2.4)

The geometrical question of finding the area of $S_a^r$ has led us to the numerical problem of finding the sum

$$1^r + 2^r + \cdots + n^r.$$

We will study this problem in the next section.

2.5   Definition (Circumscribed box.) Let $\mbox{\rm cir}(S_a^r)$ be the smallest box containing $(S_a^r)$. i.e.

$$\mbox{\rm cir}(S_a^r) = B(0,a;0,a^r) \hspace{3em}(r \geq 0).$$

Notice that $\mbox{\rm area}(\mbox{\rm cir}(S_a^r)) = a\cdot a^r = a^{r+1}$. Thus equation (2.4) can be written as

$${(1^r+2^r+ \cdots +(n-1)^r) \over n^{r+1}} \leq {\mbox{\rm a... ...{\rm cir}(S_a^r))} \leq {(1^r+2^r+\cdots +n^r) \over n^{r+1}}.$$ (2.6)

Observe that the outside terms in (2.6) do not depend on $a$.

Now we will specialize to the case when $r=2$. A direct calculation shows that

$$1^2 = 1,\mbox{{}}$$

$$1^2+2^2 = 5,\mbox{{}}$$

$$1^2+2^2+3^2 = 14,\mbox{{}}$$

$$1^2+2^2+3^2+4^2 = 30,\mbox{{}}$$

$$1^2+2^2+3^2+4^2+5^2 = 55.$$ (2.7)

There is a simple (?) formula for $1^2+2^2+\cdots +n^2$, but it is not particularly easy to guess this formula on the basis of these calculations. With the help of my computer, I checked that

$$1^2+\cdots +10^2=385 \mbox{ so } {{1^2+\cdots +10^2}\over {10^3}}=.385$$

$$1^2+\cdots +100^2=338350 \mbox{ so } {{1^2+\cdots +100^2}\over {100^3}}=.33835$$

$$1^2+\cdots +1000^2=333833500 \mbox{ so } {{1^2+\cdots +1000^2}\over {1000^3}}=.3338335$$

Also

$$\begin{eqnarray*} {{1^2+\cdots +999^2}\over {1000^3}} &=& {{1^2+\cdots +1000^2}\... ...1000^3}}-{{1000^2}\over {1000^3}}=.3338335-.001\\ &=& .3328335. \end{eqnarray*}$$

Thus by taking $n=1000$ in equation (2.6) we see that

$$.332\leq {\mbox{\rm area}(S_a^2) \over \mbox{\rm area}(\mbox{\rm cir}(S_a^2))} \leq .3339 .$$

On the basis of the computer evidence it is very tempting to guess that

$$\mbox{\rm area}(S_a^2)= {1\over 3} \mbox{\rm area}(\mbox{\rm cir}(S_a^2)) = {1\over 3}a^3.$$