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 $${0\leq y\leq {1\over {x^2}}}$$. I want to find the area of $S_a$.

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

$$1<a^{{1\over n}} <a^{{2\over n}} <\cdots < a^{{n\over n}}=a.$$

Let $O_j$ be the box

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

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

To simplify the notation, I will write

$$b = a^{1\over n}.$$

Then

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

and

$$\begin{eqnarray*} \mbox{\rm area}(O_j) & = &{ b^{j}-b^ {j-1} \over b^{2(j-1)}} = { (b-1)b^{ j-1} \over b^{2(j-1)} } = {( b-1) \over b^{(j-1)}}. \end{eqnarray*}$$

Hence

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

Observe that we have here a finite geometric series, so

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

$$= b\left(1-{1\over b}\right)\left( {1-{1\over b^{n}}\over 1-{1\over b}} \right ) = b\left(1-{1\over b^{n}}\right).$$ (2.31)

Now

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

so

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

Let $I_j$ be the box

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

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

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

Hence,

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

Since

$$\bigcup_{j=1}^nI_j\subset S_a,$$

we have

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

i.e.,

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

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 $${b = a^{1\over n}}$$, we can rewrite this as

$$a^{-{1\over n}}(1 - a^{-1}) \leq \mbox{\rm area}(S_a) \leq a^{1\over n}(1-a^{-1}).$$ (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

$$a=a^{N\over N}<a^{{N-1}\over N}<\cdots <a^{2\over N}<a^{1\over N}<1.$$

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

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

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.)