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Why defined integral is figure area?

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Why figure area is defined as defined integral? i have signature
    [tex]P=\int\limits^b_af^{\prime}(x)\mbox{d}x=f(b)-f(a)[/tex]
    what is a prove of this signature? I have other signature:
    [tex]P=\lim_{n\to\infty}\sum\limits^n_1\left(x_{i+1}-x_i\right)f\left(x_i\right)[/tex]
    how it can be proved that these signatures are equivalent? why cant I define figure area under [tex]f(x)[/tex] in a and b limits as ex. [tex]P=f^{\prime}(b)-f^{\prime}(a)[/tex]?
     
  2. jcsd
  3. Jan 15, 2010 #2

    HallsofIvy

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    Because it would give the wrong answer? I have no idea why you would think the area would be equal to f'(b_- f'(a)! Suppose f is a constant function: f(x)= h for all x. f'(x)= 0 for all x so your "[itex]P= f'(b)- f'(a)= 0[/itex]. But the area under the graph, from x= a to x= b, would be a rectangle with height h and width b-a: h(b-a) is not 0.
    On the other hand, the integral of the constant function, f(x)= h is [itex]\int h dx= hx. That, evaluated from a to be is hb- ha)= h(b-a).

    the complete explanation of why the integral is equal to the area is the proof of the "Fundamental theorem of Calculus" (part 1) which is rather complicated but given in any Calculus text. Basically, the idea is to partition the region under the curve into rectangles. You could, for example, divide the x-axis from a to b into "n" equal length regions, then construct rectangles having as height, say, the lowest value of f(x) in each region. The resulting rectangles will cover an area that is clearly beneath the curve and so has area less than the actual area under the curve. Now do the same thing except use the largest value of f(x) in each region. Those rectangles will will extend above the curve and so have area more than the actual area of the region.

    That is, the actual area lies between the two sums. Now take the limit as n goes to infinity. If the two sums have a limit and the limit is the same (and if that is NOT true, the integral does not exist and the formula you give is NOT true), then, since the true area is always between the two sums, it must be equal to the joint value- the integral.

    By the way, the correct English word is "formula", not "signature".
     
  4. Jan 16, 2010 #3
    thanks friend, I going to think about this;]
     
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