Archimedes Method To Find The Area Under A Curve I'am reading Tom Apostol's Calculus volume-1 text (page 3 and 4),where he talks about calculating the area under a curve which eventually leads to the concept of the definite integral.In the below figure he chooses an arbitrary point on the base and denote's it's distance from $0$ by $x$.
How can the vertical distance be $x^2$?In particular,if the length of the base itself is $b$,then how come the altitude is $b^2$?
He then winds up by concluding that the area $A = \frac{b^3}{3}$ by considering approximations from above and below.
I have a problem in understanding his derivation due to unfamiliarity with mathematical induction and other rigour.Is there an informal way (i mean less rigour) of computing the area? I'll be very happy if i understand this cause i want to think and solve this problem the archimedes way!!
I'am aware of possible duplicate thread however this is purely based on apostol's text.I even went through this Area under a curve is an integral but couldn't understand it.

 A: Ok,I decided to take a shot.I would be finding the area bounded by $f(x) = x^2$ between $x = 0$ and $x = 2$.If we partition the interval into little sub intervals of length $\Delta x$,which is the difference of the end points $b$ and $a$.That is $\Delta x = \frac{b  -  a}{n}$.In my case it's $\Delta x = \frac{2-0}{n} = \frac{2}{n}$.
I'am imagining $n$ sub-intervals each of length $\frac{2}{n}$.Now i'll draw vertical lines hitting the curve from those subintervals as shown below.

I'am wondering if it can be called as "circumscribed rectangles"!!(let me know if that's the case).So the next subinterval is $\frac{4}{n}$ and so on.We're gonna add up the areas of these rectangles for any value of $n$(In general).I'am gonna form what we call the "upper sum",it's the sum that's larger than the area under the parabola.It's the summation of the areas of those rectangles.Here's how it is.$$s(n) = \sum_{i = 1}^{n}f(\frac{2i}{n})\Delta x$$ 
$$ = \sum_{i = 1}^{n}f(\frac{2i}{n})^2(\frac{2}{n})$$
$$ = \sum_{i = 1}^{n}(\frac{8}{n^3})i^2$$
$$ = \frac{8}{n^3}[\frac{n(n+1)(2n+1)}{6}]$$
$$ = \frac{4}{3n^2}(2n^2 + 3n +1)$$
$$ =\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^2}$$
Now if i take the limit as $n$ goes to $\infty$
$$ = \lim_{n \to \infty } \frac{8}{3}+\frac{4}{n}+\frac{4}{3n^2} = \frac{8}{3}$$.
here $\Delta x $ is the base of each rectangles.$f(\frac{2i}{n})$ is the function for height as $i$ goes from $1$ to $n$.
Now If we inscribe these rectangles and follow the same procedure we might get an other expression but the area will still be the limit i.e $\frac{8}{3}$.
SO that's my derivation without much mathematical rigour!!.Please let me know if i have made a mistake or if it can be done in a much better way.
A: $x\cdot x^2=x^3$, thus, the area of a rectangle.
Archimedes tells us that the area of the hyperbola is two thirds of the triangle formed by its Tangents, thus
$$\frac{x^3}{2 (2/3)} = \frac{x^3}{3}.$$
