Apostol's method of exhaustion to find area under x^2 I'm a high school student currently going through Apostol's calculus. I'm not that familiar with proofs, but I learned Calculus in school up to partial fractions, but we focused more on problems instead of the concept/proof, so please bear with me. I'm stuck in the part where he used the method of exhaustion to prove that the area of $b^2$ is $\frac{b^3}{3}$. After some inequalities, we find that there are 3 possibilities for the area: $A>\frac{b^3}{3}$, $A<\frac{b^3}{3}$, and $A=\frac{b^3}{3}$
To proof that $A=\frac{b^3}{3}$, we can do this by contradiction. I can prove by contradiction that $A>\frac{b^3}{3}$ is not possible, namely through the following method (please correct me if this is wrong):
$$A<\frac{b^3}{3}+\frac{b^3}{n}$$ for all $n>=1$
$$A-\frac{b^3}{3}<\frac{b^3}{n}$$
Since we assume that $A>\frac{b^3}{3}$, then $A-\frac{b^3}{3}$ >0, so we can divide from both sides and multiply both sides by n
$$n<\frac{b^3}{A-\frac{b^3}{3}}$$
$$\frac{b^3}{A-\frac{b^3}{3}}>0$$
$$\frac{b^3}{A-\frac{b^3}{3}}+1>1$$
Since $\frac{b^3}{A-\frac{b^3}{3}}+1$ is more than one it could be a value of n since $n>=1$, therefore it contradicts $n<\frac{b^3}{A-\frac{b^3}{3}}$. I tried to use the same method for the other possibility, $A<\frac{b^3}{3}$, but I can't get it to work. I'm also a bit confused, for instance, which inequality should I use, from?
$$\frac{b^3}{3}-\frac{b^3}{n}<A<\frac{b^3}{3}+\frac{b^3}{n}$$
Apostol chose the latter, but I'm not sure why. I tried to use both to contradict the second inequality, but I fail to contradict it. Please help.

You should use the left-hand inequality
$$
\frac{b^3}3-\frac{b^3}n<A\qquad\text{for all $n\ge1$}\tag1
$$
to prove that $A\ge\frac{b^3}3$.  The argument by contradiction follows the same lines as the proof you've given: If $A<\frac{b^3}3$, then you can use algebra to rearrange (1) into the statement
$$
n<\frac{b^3}{\frac{b^3}3-A}\qquad\text{for all $n\ge1$}\tag2
$$
which is an impossibility since (2) is violated for any $n$ that exceeds $\frac{b^3}{\frac{b^3}3-A}$.
Why did we choose the left-hand inequality (1) to prove that $A\ge\frac{b^3}3$? Inequality (1) is saying that $A$ is larger than $\frac{b^3}3$ minus a small positive quantity (namely $\frac{b^3}n$). As $n$ gets larger, the quantity $\frac{b^3}n$ gets smaller and smaller and the expression $\frac{b^3}3-\frac{b^3}n$ gets closer and closer to $\frac{b^3}3$ from below. But we know $A$ is larger than the expression $\frac{b^3}3-\frac{b^3}n$. Thinking this way, your intuition should be that it's impossible for $A$ to then be strictly less than $\frac{b^3}3$, i.e. you should be able to contradict the assertion that $A<\frac{b^3}3$.
