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Integral as area under the curve.

  1. Jun 2, 2005 #1
    For showing how the integral is the area under the curve, i know how to drive the relation [tex] \lim_{x_k\rightarrow 0} \sum_{k=1}^\infty f(x_k)dx_k[/tex]. But i don't know how this is the integral itself. I know that i can try specific examples, use the sum, take a finite interval and show that it is approximately similar to its integral. But is there anyway to prove it generally? Thanks alot.
     
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  3. Jun 3, 2005 #2

    Galileo

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    The area under the graph of a function is defined as the Riemann sum/integral.
     
  4. Jun 3, 2005 #3
    hmmm, defined? is there no way to prove it then?

    How do i show that the Riemann sum is equal to the Riemann integral?
     
  5. Jun 3, 2005 #4

    Galileo

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    How is a (Riemann) integral defined in your book? Isn't it just the limit of a Riemann sum?
     
  6. Jun 3, 2005 #5

    HallsofIvy

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    What can be done is show that the limit of the Riemann sums has the properties we want from "area". Those properties are (1) if U is a subset of V then the area of U is less than or equal to the area of V, (2) if U and V do not overlap or overlap only on their boundaries, then area of U union V equal area of U plus area of V, (3) the area of a rectangle with length l and width w is lw.

    Given f(x) such that graph of y= f(x) is always above the x-axis for x between a and b, partition the x-axis between a and b into n intervals. In each (closed) interval, define x* to be the value of x that makes f(x*) smallest in that interval (f continuous is sufficient for that to exist-though not necessary) and define x* to be the value of x that makes f(x*) largest in that interval.
    It's clear that the Riemann sum, using the x* values, is the area of the "outer" rectangles, that the Riemann sum, using the x* values, is the area of the "inner" rectangles and that the area under the curve lies between them.
    If the limits, as n goes to infinity, are the same (that is, if f is Riemann integrable) by the "pinching theorem", the common value must be the area under the curve.

    I'm wondering if misogynistic feminist was asking about showing that that is the difference F(b)- F(a) where F is the anti-derivative of f. That's different from talking about Riemann sums.
     
  7. Jun 5, 2005 #6
    hmmm yeah, my question was actually how the Riemann sum equals to F(b)-F(a). Because i want to show that the integral is a sum, and connect it to the Riemann sum. So far, these things are disparate pieces to me. Sorry for being vague...
     
  8. Jun 5, 2005 #7
    Recall that the Reimann Sum is a sum of individual strips of height f(x) and width delta x. The Reimann integral's definition using the Riemann sum is

    [tex] \lim_{\Delta x \rightarrow 0} \sum_{k=1}^\infty f(x_k)\Delta x[/tex]

    as you showed, but in that expression, the [itex] \Delta x_k [/tex] term as you may recall represents the small change (dx) in the x parameter, or translates algebraically to

    [tex] \Delta x_k = x_n - x_{n-1} [/tex] for a strip of x-width 'n' . If you substitute that into the Reimann expression,

    [tex] \lim_{x_k\rightarrow 0} \sum_{k=1}^\infty f(x_k)\left( x_k - x_{k-1}\right)[/tex]

    Lets spit out the first few terms of the reimann sum:

    [tex]\sum_{k=1}^\infty f(x_i)\left( x_k - x_{k-1}\right) = (f(x_i)x_1 - f(x_i)x_0) + (f(x_i)x_2 - f(x_i)x_1) + (f(x_i)x_3 - f(x_i)x_2) \ ... [/tex]

    What do you notice about the terms? Note in the first term, you have added f(x)x_1 but then for the second term you have subtracted it. In the second term you add f(x)x_2 but in the third term you subtract it. The only terms that dont get subtracted are the very first term, f(x)x_a, and the very last term, f(x)x_b. Its from this effect that the reimann integral simplifies to F(b) - F(a).


    This is off the top of my head, and I'm sure its got a mistake or two (or six hundred) so please correct me if I'm wrong.
     
  9. Jun 5, 2005 #8
    hey ! i'm somewhat getting the picture. I can see that grouping the pluses and minuses together, i get something like [tex] f(x_i)x_1 +f (x_1)x_2 +....- f(x_i)x_0-f (x_i)x_1-... [/tex] And this suggests [tex] F(b)-F(a) [/tex]. But right now, all i see are sums and not integrals. I can't really see how [tex] f(x_i)x_1 +f (x_1)x_2 +....=F(b)[/tex]. Sorry if i'm a little persistent, but i can't seem to make any connection here.
     
  10. Jun 5, 2005 #9
    No, I totally understand if my description isn't adequate to quench your thirst, but I think I should leave it to a mathematician to give you the full description just in case I made an error, we wouldn't want to build on faulty grounds.

    edit (very rough cliffnotes): As you add more strips, the width of the strips decrease, taking the limit to infinity of the number of strips, the width of each strip approaches zero. This is where the sum turns into the integral. since were taking the sum of an (approaching) infinite amount of strips but each strip kinda subtracts the strip before it (like we just saw), and what you end up with are the anti derivative's value at B minus the one at A.

    sorry these notes suck.
     
    Last edited: Jun 5, 2005
  11. Jun 5, 2005 #10

    HallsofIvy

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    And F(x) is the anti-derivative of f(x). Yes, that's what I said before that I thought you meant.

    You are using the word "integral" to mean anti-derivative and think that's a bit confusing in this context. As I said before, you can show that the area is the limit of the Riemann sums (not the Riemann sums themselves.

    Now, given f(x) positive for x> a and DEFINE F(t) to be "the area bounded by the graph of y= f(x), y= 0, x= a and x= t.

    Then, for any h (small enough that t-h is still larger than a), F(t+h) is the area bounded by the graph of y= f(x), y= 0, x= a, and x= t+h. (You might want to do the cases h>0, h< 0 separately.)

    F(t+h)- F(t)= area bounded by the graph of y= f(x), y= 0, x= t, and x= t+h.

    Taking x* to be the value of x in [t,t+h] such that f(x* is maximum and x* to be the value such that f(x*) is minimum. Then that area lies between f(x*)h and f(x*)h. That is, f(x*)h< F(t+h)- F(t)< f(x*)h so f(x*)< (F(t+h)- F(t))/h< f(x*). Of course, these x* and x*depend upon h but it is easy to see that, as h-> 0 x* and x*, trapped between t and t+h, both converge to t so that
    lim(h->0) (F(t+h)-F(t))/h= dF/dt= f(t).

    That is, f is the derivative of this area function and so the area function is an anti-derivative of f. "An" anti-derivative because the choice of a is arbitrary.
     
    Last edited: Jun 5, 2005
  12. Jun 8, 2005 #11
    Misogynist + Feminist = 0 ?

    In your question I don't hear that you need a proof of the fundamental Theorem of Calculus. Right? Instead, I hear you asking the intuitive: What would have prepared us intuitively to know that summing up a bunch of f()s would get us a single F()?

    If that is your question, following is my answer (I am going to answer you in English, but I suggest that you might want to sit and ponder it for a while):

    In that sum you gave, what you are doing is adding up a bunch of n-dimensional objects and getting a single n+1 dimensional object! It's like stacking a bunch of squares (2-D) to get a pyramid (3-D) -- where the squares are getting bigger and bigger.

    The f() function is the n-dimensional object, and F() is the next higher dimension. One single F() object will hold, contain, consist-of a bunch of f() layers -- when you sum all the f()'s, you are building an F().

    This is the "why" to your question, and it is really quite intuitive once grasped. If you need more, I can say it a different way. Then again, I may not have addressed your Q at all.


    Steve Rives
     
    Last edited: Jun 8, 2005
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