# 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.

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The vertical distance is $x^2$, because that is what the parabola signifies: Exactly the points that are $x$ away from the origin horizontally and at the same time $x^2$ away from the origin in the vertical direction. It is oftentimes written as $y = x^2$ where $y$ signifies the vertical distance and $x$ the horizontal distance. – Arthur Dec 12 '11 at 11:21
Ah..right.How about figuring out the area with less mathematical rigour.Too much usage of these inequalities,induction and things like that confuses me!!Anyways i'll give it a try,please go for it – alok Dec 12 '11 at 11:26
The best I can do is tell you to imagine a pyramid with square base, height $b$ and side length $b$. At distance $x$ from the top, the area of the cross section is $x^2$, and the total volume is $\frac{b^3}{3}$. But again, there is some rigor showing the connection between that volume and the area you want to compute. Proving integrals without any rigor is really not possible. – Arthur Dec 12 '11 at 11:35
Ok go for it arthur.I'll put it in my effort to understand your way of computing the area.Let's see if it's different from that of apostol's. – alok Dec 12 '11 at 11:43

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.