Prove that integration can be used to find area under curve?

Question

I understand Riemann sums but dont see the connection to integral calculus

Answer 1

If you understand Riemann sums, then you must understand that by any reasonable definition of area, that as the mesh of the partitions approaches zero, the limit of the Riemann sums will approach the signed area under the curve. Now, the integral is BY DEFINITION the limit of these Riemann sums, which means, in effect, that the integral is defined to be the signed area under the curve.

What you may be confused on, is that in practice, evaluation of integrals is performed not by working directly with the Riemann sums, but rather by finding an antiderivative of the function and then evaluating it at two points. So if you mean to ask what is the connection between this process of finding antiderivatives and finding the area under the curve, then the proper question is "prove that _antidifferentiation_ may be used to find the area under the curve." The theorem that states this is possible is called the Fundamental Theorem of Calculus. Since the proof requires work, I'll just link you to the chapter on integrals in my analysis text where the theorem is proven -- the proof you are looking for starts on page 16. http://www.math.utah.edu/~taylor/5_Integral.pdf

Answer 2

As the "slabs" of the Riemann sum get thinner and thinner, the area under a curve gets closer and closer to the sum of the area in those slabs -- there's less and less area under the curve that's omitted or over the curve that's included in the Riemann sum.

The integral is the limit of that process -- it's the value that the sum approaches as the slabs continue to get narrower.

I know that's not a real "proof," because it's really a definition of an integral. But the mathematical proof is far beyond me...

Answer 3

dude

riemann sum IS an integral

it is the sum of the areas of rectangles under the curve so the number of rects approches infinity

so if u take the integral of a certain function between two values a and b, you are taking the riemann sum

Answer 4

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