2
$\begingroup$

Let's suppose we have a curve $y = x^2$

Now when we take two points on the curve, we aren't completely sure how to connect them(will elaborate more later), compared to two points on a straight line which we are sure how to connect(shortest distance).

My argument is that the only way we know how to connect two points (precisely) is by marking the shortest distance between them (please do tell if I'm wrong here and for the sake of the question do assume this for a second) now, to connect two points on the curve we need more points between the two original points and then connect all of those points by marking the shortest distance between them. But then the same initial argument applies to any two adjacent points - that we aren't sure how to connect them again (compared to a straight line). So now we need to consider something like infinity too - basically keep increasing the resolution of the points on the curve to a non finite degree to connect the points and form the curve precisely.

Now comes my question - in this backdrop, how is it that by integrating we can find the precise value of the area under the curve?, since we were unable to form the curve in the first place precisely unless by using the concept of infinity.

Is there any literature which explains my question simply?

$\endgroup$
10
  • $\begingroup$ Thats the whole idea of calculus. You make the distance so small, divide the interval into infinitely many parts, so that the error which you are talking about approaches zero. $\endgroup$
    – Shailesh
    Dec 28 '15 at 14:10
  • $\begingroup$ What does "connect two points" mean for you here? How does the matter that the two points you want to "connect" happen to lie on a curve? What does all that have to do with integrals? $\endgroup$ Dec 28 '15 at 14:11
  • $\begingroup$ You say it approaches zero, but that's what I want to know -how? And a proof that there is no error would help a ton. Plus it wouldn't be possible unless we use the concept of infinity - what I thought was that integrating a curve should always give an approximate value but it doesn't it gives a precise value $\endgroup$
    – novice
    Dec 28 '15 at 14:14
  • $\begingroup$ @henning makholm , well what I'm trying to convey is that it was very hard to precisely form the curve in the first place, but somehow we are easily able to obtain the area under the curve please do give me a little time to convey myself better here though $\endgroup$
    – novice
    Dec 28 '15 at 14:19
  • $\begingroup$ Any hints please? Is the question clear now? What can I read up.. This has been bothering me for a while now $\endgroup$
    – novice
    Dec 28 '15 at 14:49
1
$\begingroup$

Yes, there is an error. If you are talking about a curve like $x^2$, the error in measurement of area is proportional to h say. where h is the gap between 2 consecutive divisions you are splitting your curve to assume them to be rectangles.

Now since you are saying I'll reduce the gap between the lines to a value as close to 0 as possible, the error (since proportional) also reduces to a value as close to 0 as possible.

Recall that 10+h as Limit $h\to 0$ is $10$.

$\endgroup$
0
$\begingroup$

In real analysis the integral is usually defined using Darboux sums.
You define partition of the interval and then you let n go to infinity.
Then the sum of the (signed) areas of the rectangles (as n goes to infinity)
is defined/said to be the integral (the value of integral).

Check this e.g.

Darboux Integral

$\endgroup$
3
  • $\begingroup$ Not exactly what I'm looking for mate. $\endgroup$
    – novice
    Dec 28 '15 at 14:43
  • $\begingroup$ OK, I read your comments. I think I now understand what you're asking. It's more philosophical than mathematical I think, but you cannot state it quite well, it seems :) Unfortunately I cannot answer it so well in English (as I want to), or it will be very difficult for me to say the least. $\endgroup$ Dec 28 '15 at 15:25
  • $\begingroup$ Can you point me to some resources I can read up then? Also if you could point out the ambiguity on the question would help a lot - so I can make it clearer $\endgroup$
    – novice
    Dec 28 '15 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.