# Apostol Calculus, Method of Exhaustion

In Apostol's Calculus, he goes through the method of exhaustion to find the area under a parabola from $$0 \ to\ b$$. Using the fact that,

\begin{align} &1^2+2^2+...+(n-1)^2 < \frac{n^3}{3} < 1^2+2^2+...+n^2\label{1} \\ &\Rightarrow \frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2) < \frac{b^3}{3} < \frac{b^3}{n^3}(1^2+2^2+...+n^2) \nonumber \\ &\Rightarrow s_{n} < \frac{b^3}{3} < S_{n} \nonumber \end{align}

where $$s_{n}$$ and $$S_{n}$$ are the lower and upper approximation (rectangles), respectively, for the area under the parabola. We then have to show that $$A=\frac{b^3}{3}$$ is the only number that satisfies

$$$$s_{n}

for every $$n\geq1$$. Using the left-most side of the first inequality, he adds $$n^2$$ and then multiplies both sides by $$\frac{b^3}{n^3}$$,

\begin{align*} &\frac{b^3}{n^3}(1^2+2^2+...+n^2)<\frac{b^3}{3}+\frac{b^3}{n^2} \\ &\Rightarrow S_{n}<\frac{b^3}{3}+\frac{b^3}{n^2} \end{align*}

for the right-most side of the inequality, he subtracts $$n^2$$ and multiplies by $$\frac{b^3}{n^3}$$,

\begin{align*} &\frac{b^3}{3}-\frac{b^3}{n^2}<\frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2)\\ &\Rightarrow \frac{b^3}{3}-\frac{b^3}{n^3}

this implies,

\begin{align*} \frac{b^3}{3}-\frac{b^3}{n}

This is where it gets confusing. He says the only possibilities are:

$$\begin{array}{ccc} A>\frac{b^3}{3},& A<\frac{b^3}{3},& A=\frac{b^3}{3} \end{array}$$

and proceeds to show that $$A=\frac{b^3}{3}$$ via contradictions for the first two cases. This is fine, but what about $$\frac{b^3}{n}$$ in the inequality? why don't we have to consider the possible relationships between $$A$$ and $$\frac{b^3}{n}$$?

• You have a confusing typo. The far LHS and far RHS of the inequality above "This is where it get confusing" are identical. Please edit. Commented Nov 14, 2018 at 10:58
• I see! I have fixed it, thank you very much. Commented Nov 14, 2018 at 18:00
• For $b\ne 0$ we have $|b^{-3}(A-b^3/3|<1/n^2$ for $all$ $n\in \Bbb Z^+$, which is not possible unless $b^{-3}(A-b^3/3)=0$, which is not possible unless $A=b^3/3$. Note that $A$ and $b$ have no relation to $n$. Commented Nov 15, 2018 at 6:12
• What about $A<\frac{b^3}{3}$, wouldn't that satisfy the inequality in your response? Commented Nov 15, 2018 at 18:39

$$\frac{b^3}{3} - \frac{b^3}{n} < A < \frac{b^3}{3} + \frac{b^3}{n}$$
Picking it up from here. Note that the Apostol's ultimate goal in this proof is to somehow show that $$A = \frac{b^3}{3}$$. He basically just applied the law of trichotomy which says (I write informally) that given any two arbitrary numbers $$x$$ and $$y$$, exactly one of the following is true: $$x < y$$ or $$x > y$$ or $$x = y$$.
Applying that, we have $$A$$ which is an arbitrary number and we have $$\frac{b^3}{3}$$ also a number. We do not have any idea about the relationship between $$A$$ and $$\frac{b^3}{3}$$. However, we know from the law of Trichotomy that exactly one of the following (relations) is true: $$A < \frac{b^3}{3}$$ or $$A > \frac{b^3}{3}$$ or $$A = \frac{b^3}{3}$$.