Calculus: Area Under Curve with Infinite Rectangles

[minase]

In calculus class when we were learning how definite integral derived from. We added up infinite # of rectangles under the curve. As n number of interval increased length of the base of the rectangles approached 0. Can you multiply 0 by infinity and get back the area you had.

 

[matt grime]

You are not multiplying 0 by infinity. I can think of no arithmetic where this is allowed.

You are not adding up an infinite number of rectangles. At each stage you are adding a strictly finite number of rectangles, each of strictly positive width. Your integral exists if the limit of these sums exists in the appropriate sense (as the largest width of a rectangle tends to zero) and is the same for all choices of sequences of partitions.

You are *not* multuplying zero by infinity.

 

[ssd]

Can you multiply 0 by infinity and get back the area you had.

You may note that x-->0 implies that x is not exactly equal to 0.

 

[Gib Z]

Its just like getting a better approximation over and over. As the base of these rectangles get smaller, the approximation is better. Think, if there were an infinite number, the approximation would be exact. This integral takes this in the limit, as in, it gets closer and closer to base zero, but not exact.

 

[Hurkyl]

You may note that x-->0 implies that x is not exactly equal to 0.

Just to make it clear, x is not any particular number here: it is a variable. It's like the i in \sum_{i=0}^10} i^2, or the x you see in f(x) = x^2.

Its just like getting a better approximation over and over. As the base of these rectangles get smaller, the approximation is better.

This is a good way to think of it; there is a quantity, called the area under your curve. You know how to approximate the area by adding up the area of finitely many rectangles. That the integral exists is simply saying that you can force this approximation to be as good as you like simply by setting an upper bound on the width of the rectangles.

Think, if there were an infinite number, the approximation would be exact.

This is not a good way to think of it. In standard analysis, there aren't an infinite number. Even if you were to use nonstandard analysis, an infinite number of rectangles simply means that you are infinitessimally close to the area under the curve.

This integral takes this in the limit, as in, it gets closer and closer to base zero, but not exact.

I'm not clear precisely what you meant, so I will say something just in case -- the integral is exactly the area under the curve. It is the approximations of the integral by Riemann sums that are approximate.

 

[slider142]

As was said before, the most important idea is that you're looking for a certain bound.
Going back to the definition of an infinite series, such a series actually represents a sequence of partial sums, each of which has a finite number of terms, and thus each partial sum is some number. In the real number system, if we have an increasing sequence (each term is greater than or equal to the one before it), and we can find some real number larger than every term in the sequence. we say the sequence has an upper bound. But in the real number system, if we have such a sequence that is bounded from above, but is always increasing, then there must be some real number L that is the smallest real number larger than every term in the sequence, called the least upper bound. It is this number that we call the sum of the infinite series.
In the case of your Riemann sum, taking the limit as n approaches infinity is really looking for the least upper bound of the set of real sums where n is finite; that is, if your integral is for a positive function. I'm sure you can generalize this idea to series for which we look for greatest lower bounds and so on.

 

[Gib Z]

Hey Hurkyl, from your last quote of me, I seem to have spoken ambiguously, sorry. I meant that the base of the rectangles would approach zero, but not be exactly zero, hence his original thought on 0*infinity=Area. Of course the integral is exact :)