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Next: 2.2 Some Summation Formulas Up: 2. Some Area Calculations Previous: 2. Some Area Calculations   Index

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 $t$

\begin{displaymath}(x \leq t) \mbox{ implies that } (x^r \leq t^r). \end{displaymath}

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 $\displaystyle {x_i={{ia}\over n}}$.

Then $\displaystyle {x_i-x_{i-1}={a\over n}}$ 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

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).

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

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

Hence we have

\begin{displaymath}S_a^r\subset \bigcup_{i=1}^n O_i,\end{displaymath}

and thus
\mbox{\rm area}(S_a^r)\leq\mbox{\rm area}(\bigcup_{i=1}^n O_i).
\end{displaymath} (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

\begin{displaymath}\bigcup_{i=1}^n I_i\subset S_a^r,\end{displaymath}

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


\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,


\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

Since the boxes $I_i$ intersect only along their boundaries, we have
$\displaystyle \mbox{\rm area}(\bigcup_{i=1}^nI_i)$ $\textstyle =$ $\displaystyle \mbox{\rm area}(I_1)+\mbox{\rm area}(I_2)+\cdots +\mbox{\rm area}(I_n)\mbox{{}}$  
  $\textstyle =$ $\displaystyle {{a^{r+1}}\over {n^{r+1}}}0^r + {{a^{r+1}}\over {n^{r+1}}}1^r+\cdots + {{a^{r+1}}\over
  $\textstyle =$ $\displaystyle {{a^{r+1}}\over {n^{r+1}}}(1^r+2^r+\cdots +(n-1)^r),$ (2.3)

and similarly

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

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

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

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

\begin{displaymath}1^r + 2^r + \cdots + n^r. \end{displaymath}

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.

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


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}}.
\end{displaymath} (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

$\displaystyle 1^2$ $\textstyle =$ $\displaystyle 1,\mbox{{}}$  
$\displaystyle 1^2+2^2$ $\textstyle =$ $\displaystyle 5,\mbox{{}}$  
$\displaystyle 1^2+2^2+3^2$ $\textstyle =$ $\displaystyle 14,\mbox{{}}$  
$\displaystyle 1^2+2^2+3^2+4^2$ $\textstyle =$ $\displaystyle 30,\mbox{{}}$  
$\displaystyle 1^2+2^2+3^2+4^2+5^2$ $\textstyle =$ $\displaystyle 55.$ (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

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

\begin{displaymath}1^2+\cdots +100^2=338350 \mbox{ so } {{1^2+\cdots +100^2}\over

\begin{displaymath}1^2+\cdots +1000^2=333833500 \mbox{ so } {{1^2+\cdots +1000^2}\over


{{1^2+\cdots +999^2}\over {1000^3}} &=& {{1^2+\cdots +1000^2}\...
...1000^3}}-{{1000^2}\over {1000^3}}=.3338335-.001\\
&=& .3328335.

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

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

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

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

next up previous index
Next: 2.2 Some Summation Formulas Up: 2. Some Area Calculations Previous: 2. Some Area Calculations   Index
Ray Mayer 2007-09-07