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2.5 Area Under the Curve $y={1\over {x^2}}$

The following argument is due to Pierre de Fermat (1601-1665) [19, pages 219-222]. Later we will use Fermat's method to find the area under the curve $y=x^\alpha$ for all $\alpha$ in $\mbox{{\bf R}}\setminus \{-1\}$.

Let $a$ be a real number with $a > 1$, and let $S_a$ be the set of points $(x,y)$ in $\mbox{{\bf R}}^2$ such that $1\leq x\leq a$ and $\displaystyle {0\leq y\leq {1\over {x^2}}}$. I want to find the area of $S_a$.


Let $n$ be a positive integer. Note that since $a > 1$, we have

\begin{displaymath}1<a^{{1\over n}} <a^{{2\over n}} <\cdots < a^{{n\over n}}=a.\end{displaymath}

Let $O_j$ be the box

\begin{displaymath}O_j=B\left( a^{ {j-1}\over n},a^{{j\over n}}\colon 0,
{1\over {\left( a^{ {j-1}\over n}\right)^2}} \right).\end{displaymath}

Thus the upper left corner of $O_j$ lies on the curve $\displaystyle {y={1\over {x^2}}}$.

To simplify the notation, I will write

b = a^{1\over n}.


\begin{displaymath}O_j = B\left(b^{j-1},b^j: 0,{1\over b^{2(j-1)}}\right),\end{displaymath}


\mbox{\rm area}(O_j) & = &{ b^{j}-b^ {j-1} \over
= { (b-1)b^{ j-1} \over b^{2(j-1)} }
= {( b-1) \over b^{(j-1)}}.


\mbox{\rm area}\left(\bigcup_{j=1}^n O_j\right) &=& \mbox{\rm ...
...ight)\left(1+{1\over b}+\cdots
+ { 1\over b^{(n-1)}}\right). \\

Observe that we have here a finite geometric series, so

$\displaystyle \mbox{\rm area}\left( \displaystyle {\bigcup_{j=1}^n}O_j\right)$ $\textstyle =$ $\displaystyle \left( b-1 \right)
\left( {1-{1\over b^{n}}\over 1-{1\over b}} \right)$ (2.30)
  $\textstyle =$ $\displaystyle b\left(1-{1\over b}\right)\left( {1-{1\over b^{n}}\over 1-{1\over b}} \right
= b\left(1-{1\over b^{n}}\right).$ (2.31)

S_a\subset\bigcup_{j=1}^n O_j
\end{displaymath} (2.32)

\mbox{\rm area}(S_a)\leq\mbox{\rm area}(\displaystyle {\bigcup_{j=1}^n}O_j)=b \left(1-{1\over b^{n}}\right).
\end{displaymath} (2.33)

Let $I_j$ be the box

\begin{displaymath}I_j=B\left(a^{ {j-1}\over n},a^{j\over n}:0,{1\over a^{{2j\over n}}}\right)
= B\left(b^{j-1},b^j: 0 ,{1\over b^{2j}}\right)\end{displaymath}


so that the upper right corner of $I_j$ lies on the curve $y={1\over x^2}$ and $I_j$ lies underneath the curve $y={1\over x^2}$. Then

\mbox{\rm area}(I_j) &=&\left({b^{j}-b^{j-1}\over b^{2j}}\righ...
...)}} = {(b-1)\over b^2b^{j-1}} = {\mbox{\rm area}(O_j)\over b^2}.


\mbox{\rm area}\left(\displaystyle {\bigcup_{j=1}^n}I_j\right)...
={1\over b^2}\cdot b(1-{1\over b^n})
= b^{-1}(1-b^{-n}).


\begin{displaymath}\bigcup_{j=1}^nI_j\subset S_a,\end{displaymath}

we have

\begin{displaymath}\mbox{\rm area}\left(\displaystyle {\bigcup_{j=1}^n}I_j\right)\leq\mbox{\rm area}(S_a);\end{displaymath}


\begin{displaymath}b^{-1} (1-b^{-n}) \leq\mbox{\rm area}(S_a).\end{displaymath}

By combining this result with (2.33), we get

b^{-1}(1-b^{-n})\leq\mbox{\rm area}(S_a)\leq b(1-b^{-n}) \mbox{ for
all }
n\in\mbox{{\bf Z}}^+.

Since $\displaystyle {b = a^{1\over n}}$, we can rewrite this as
a^{-{1\over n}}(1 - a^{-1}) \leq \mbox{\rm area}(S_a) \leq a^{1\over n}(1-a^{-1}).
\end{displaymath} (2.34)

2.35   Exercise. What do you think the area of $S_a$ should be? Explain your answer. If you have no ideas, take $a=2$ in (2.34), take large values of $n$, and by using a calculator, estimate $\mbox{\rm area}(S_a)$ to three or four decimal places of accuracy.

2.36   Exercise. A Let $a$ be a real number with $0<a<1$, and let $N$ be a positive integer. Then

\begin{displaymath}a=a^{N\over N}<a^{{N-1}\over N}<\cdots <a^{2\over N}<a^{1\over N}<1.\end{displaymath}

Let $T_a$ be the set of points $(x,y)$ such that $a\leq x\leq 1$ and $0\leq
y\leq\displaystyle { {1\over {x^2}}}$. Draw a sketch of $T_a$, and show that

\begin{displaymath}a^{1\over N}(a^{-1}-1)\leq\mbox{\rm area}(T_a)\leq a^{-{1\over N}}(a^{-1}-1).\end{displaymath}

The calculation of $\mbox{\rm area}(T_a)$ is very similar to the calculation of $\mbox{\rm area}(S_a)$.

What do you think the area of $T_a$ should be?

2.37   Exercise. Using the inequalities (2.6), and the results of Bernoulli's table in section 2.2, try to guess what the area of $S_a^r$ is for an arbitrary positive integer $r$. Explain the basis for your guess. ( The correct formula for $\mbox{\rm area}(S_a^r)$ for positive integers $r$ was stated by Bonaventura Cavalieri in 1647[6, 122 ff]. Cavalieri also found a method for computing general positive integer power sums.)

next up previous index
Next: 2.6 Area of a Up: 2. Some Area Calculations Previous: 2.4 Finite Geometric Series   Index
Ray Mayer 2007-09-07