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Showing why an integral is the area under a curve...

by Jd0g33
Tags: curve, integral, showing
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Sep6-13, 11:43 PM
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So I've been spending a lot of time lately trying to figure out why an integral will give you the area under the curve. I asked the forum and got some great answers, but all were in terms of infinite sums, dx, and infinite rectangles. I think I've come upon a more fundamental answer that I haven't heard from anybody yet (although I'm sure its obvious to most people.) :)

[itex]y(x)[/itex] is a curve on a graph. Now as [itex]y(x)[/itex] gets larger, the area under the curve [itex]A(x)[/itex] gets larger. In fact, say [itex]y(x)=3[/itex]. Then the area at [itex]x=1[/itex] would be 3. Its rate of change at that instant would be 3.

Which means the rate of change of the area under the curve is equal to y(x): [itex]A'(x)=y(x)[/itex].

That being said, the actual area under the curve would equal the anti-deriviative of y(x): [itex]A(x)=∫y(x)[/itex].

Is this correct? If so, why do we use the confusing dx when it's really not even a number that you could multiply by to get the area of a rectangle (other than showing the integral is with respect to x). I guess I just don't like the [itex]\frac{dy}{dx}[/itex] notation.

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Sep6-13, 11:59 PM
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Definite integrals are the "area under the curve" by construction - that is why you were referred to the Riemann sum.

The connection between the definite integral and the anti-derivative is the "fundamental theorem of calculus" - Newton was the first to discover this connection; both derivatives and definite integrals were known and used prior to Newton (e.g., Archimedes, Descartes just to name the most famous).

It is the fundamental theorem of calculus that tells us that the anti-derivative can be viewed as an "indefinite integral".

The proof is fairly simple today:

The notation was originated by Leibniz; it is very suggestive of "infinitesimal relations" which we now carry out via the method of limits, which is very rigorous, but not very suggestive. The "dx" in the integral identifies the integration variable - it is a handy notation. You will find that a good notation is very useful in mathematics because it reminds you of important things.
Sep7-13, 03:11 PM
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If f is a step function (For our purposes here constant except at a finite number of jump continuities) it is obvious that the integral is the area. If f is not a step function, but is well behaved it is the limit of a sequence of step function, and the integral is the limit of the integrals of the step functions. Too you do not like the dx notation it is widely used, and not likely to be changed based on your say so.

Sep8-13, 11:56 AM
P: 362
Showing why an integral is the area under a curve...

An integral doesn't calculate the area under the curve, the integral defines the area under the curve. Integrals are definitions, not ways of calculating.

As you move on in math, the Riemann integral becomes unwieldy and not that useful (note, it is the only integral most physicists and engineers ever see or even hear of). But when you get into graduate school or professional mathematics, you really work with the Lebesgue integral. This integral allows you to integrate FAR more functions than the Riemann. For example you can integrate the function

f(x)= 1 for rational
0 for irrational

That function is wild, and not Riemann integrable. It is discontinuous EVERYWHERE (note that just because a function is discontinuous at a point, or multiple, or even infinitely many points does NOT mean that is not Riemann integrable). However you CAN integrate it using the Lebesgue integral.

To answer your question. Your question is flawed. Integrals define, they don't calculate.
Sep8-13, 12:09 PM
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In the "naive" Riemann sum approach, the integral can ONLY be defined in so far as the Least Upper Sum and the Greatest Lower Sum converge to the SAME value.
That value is given the name the "integral".

Unless such a value exists, it is meaningless within the Riemann approach to say that we can measure something called the "area" beneath the curve.

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