# Question about TM Apostol's Calculus Method of Exhaustion. Exercise 1.1.4 1(a)

I am trying to solve the exercises after the explanation of the Method of Exhaustion in Apostol's Calculus I book. The question of interest is question 1(e) of exercise I 1.4 and it goes thus:

If the ordinate at each $$x$$ is $$ax^2 + c$$, calculate the area. For context, below is a figure of the method of exhaustion from the book. The ordinate at each $$x$$ in this figure is $$x^2$$.

Here is my attempt at the problem: Area of a single rectangle is: $$\frac{b}{n}\cdot\left(a\cdot \left(\frac{kb}{n}\right)^2 + c \right)$$

So the sum $$S_n$$ of the area of the outer rectangles for $$k = 1 ... n$$ is $$S_n = \frac{b}{n}\left(a\left(\frac{b}{n}\right)^2 + c \right) + \frac{b}{n}\left(a \left(\frac{2b}{n}\right)^2 + c \right)+ \cdots + \frac{b}{n}\left(a \left(\frac{nb}{n}\right)^2 + c \right),$$ from which we get $$S_n = a\left(\frac{b^3}{n^3}\right)(1^2 + \cdots +n^2) + \left(\frac{b}{n}\right)nc$$

Similarly, for the sum $$s_n$$ of the inner rectangles, with $$k = 1 \ldots (n-1)$$ inner rectangles, we get

$$s_n = a\left(\frac{b^3}{n^3}\right)(1^2 + \cdots +(n-1)^2) +bc-\frac{bc}{n}.$$

Now we know that

$$1^2 + \cdots + (n-1)^2 < \frac{n^3}{3}<1^2 + \cdots + n^2$$ Multiplying the inequality above by $$\frac{ab^3}{n^3}$$ and adding $$bc$$, we get the inequality:

$$\frac{ab^3}{n^3}\left[1^2 + \cdots + (n-1)^2\right] + bc < \frac{ab^3}{3}+bc<\frac{ab^3}{n^3}\left[1^2 + \cdots + n^2\right]+bc$$

From the LHS of the ineq above, we have that

$$\left[\frac{ab^3}{n^3}\left[1^2 + \cdots + (n-1)^2\right] + bc - \frac{bc}{n}\right] < \left[\frac{ab^3}{n^3}\left[1^2 + \cdots + (n-1)^2\right] + bc \right] < \left[\frac{ab^3}{3}+bc\right]\ \ \ \ \ (*)$$

And so we have that $$\left[\frac{ab^3}{n^3}\left[1^2 + \cdots + (n-1)^2\right] + bc - \frac{bc}{n}\right] < \left[\frac{ab^3}{3}+bc\right] < \left[\frac{ab^3}{n^3}\left[1^2 + \cdots + n^2\right]+bc\right]$$

i.e we have that $$s_n < \left[\frac{ab^3}{3}+bc\right] < S_n$$.

Now the rest of the proof is to show that $$\frac{ab^3}{3}+bc$$ is the ONLY number between $$s_n$$ and $$S_n$$, i.e if A is a number such that $$s_n, then $$A = \frac{ab^3}{3}+bc$$ And this can easily be shown using law of trichotomy and contradiction to show that $$A > \frac{ab^3}{3}+bc$$ and $$A < \frac{ab^3}{3}+bc$$ are both FALSE, hence $$A = \frac{ab^3}{3}+bc$$. I won't post that here for brevity.

My confusion is then: If we are able to show that $$\frac{ab^3}{3}+bc$$ is the only number between $$s_n$$ and $$S_n$$, what about the number

$$\frac{ab^3}{n^3}\left[1^2 + \cdots + (n-1)^2\right] + bc$$

from equation (*) above which is obviously between $$s_n$$ and $$S_n$$?

Am I doing something wrong with my attempt?

Let $$A_n:= \frac{ab^3}{n^3}[1^2 + \cdots + (n-1)^2] + bc.$$ Then it is true that $$s_n < A_n < S_n,$$ but $$A_n$$ depends on $$n$$. It isn't a unique number, contrary to $$\frac{ab^3}{3}+bc.$$ You may notice that, as expected, $$A_n \to \frac{ab^3}{3}+bc \,\,\,\,\,\, (n \to \infty)$$