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Part of my paper is asking me to conceptualize how the definite integral is used to find the area under a curve and show the relation to the antiderivative. I do well at the beginning, but I always end up lost(I think because I'm not very good at connecting the two). I said that the area can be found by making lots of rectangles under a curve, and the more rectangles, the more accurate your area. And if we have change in x approaching 0, we can solve for actual area. This is represented by the Riemann Sum. Is this the definite integral? So far I just have that the definite integral is the area under a curve between two points, a and b.and what part does A= Sum f(x)dx play? Is that the anti derivative? We got assigned this before we've really covered alot of this, so consider me a newbie(very, very new)

Sara
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  • The definite integral is just the Riemann sum in the limit as the norm of your chosen partition goes to $0$ (assuming said sum exists) -- basically the limit as each of your $\Delta x_i$'s go to $0$. The relation to the antiderivative is the fundamental theorem of calculus -- to really get a hold on that I'd suggest going through the proof. –  Dec 11 '14 at 02:28
  • And... For some basic information about writing math at this site see e.g. here, here, here and here. –  Dec 11 '14 at 02:29
  • so would it be accurate to say that the Riemann sum is the sum of the products(rectangles) and the definite integral is the Riemann sum as the change in x approaches 0(as the base of the rectangles approaches 0)? – Sara Dec 11 '14 at 02:32
  • That's basically it. There can be some slight complications -- for instance functions can be made up whose upper and lower Riemann sums don't converge to the same limit -- in which case the function is said to NOT be integrable. But you've got the basic idea down. –  Dec 11 '14 at 02:34
  • Yes! and thanks for the links. My teacher gave us that the equation and function for the area are related by the antiderivative. I tried looking around and I think I could get by but I really want to understand it. When i was looking around, I saw the A= Sum f(x)dx and that if you take the antiderivative of f, you get A. I'm just stuck on connecting the definite integral with that idea (if it's right) – Sara Dec 11 '14 at 02:41
  • Is it that the definite integral as the limit of a riemann sum is found by taking the anti derivative of f? – Sara Dec 11 '14 at 02:44
  • $\sum_i^n f(x_i)\Delta x_i$ IS the Riemann sum. The fundamental theorem of calculus tells us that if we have $\int_a^b f'(x)dx$ then this is the same as $f(b)-f(a)$ -- subject to some conditions. $f(x)$ is called the "antiderivative" of $f'(x)$. It is just the class of functions, all of whose derivative is $f'(x)$. For example, looking at $\int_0^1 e^{2x}dx$. We know that any function of the form $\dfrac {e^{2x}}{2} + C$, where $C$ is a constant, will have a derivative of $e^{2x}$. So the FTC tells us that $\int_0^1 e^{2x}dx = (\dfrac {e^{2(1)}}{2} + C) - (\dfrac {e^{2(0)}}{2} + C)$ –  Dec 11 '14 at 02:46
  • ... $=\dfrac {e^2}{2} - \dfrac 12$. That's a big help because this way, no Riemann sums have to be calculated. –  Dec 11 '14 at 02:51
  • If you'd like more info, check out this other question. The first answer is really good. –  Dec 11 '14 at 03:06

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