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Finding an explicit formula

  1. Apr 4, 2005 #1
    How do u find an explicit formula when given an integral of a function.
    For example, the integral from 0 to x of tg(t)dt=x+x^2, how do u find the forumla for g(t)????
     
  2. jcsd
  3. Apr 5, 2005 #2
    This involves a portion of the Fundamental Theorem of Calculus which says:

    The function:

    [tex]F(x) = \int_{0}^{x} f(t) dt[/tex]

    is an indefinite integral or antiderivative of f. That is:

    [tex]F'(x) = f(x)[/tex]

    Explicit form is simply in terms of a function F.
     
  4. Apr 5, 2005 #3

    dextercioby

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    Evaluate that integral by parts...(You could have written t*g(t),the way you did,it can be mistaken with "tangent" of 't').

    Daniel.
     
  5. Apr 5, 2005 #4

    HallsofIvy

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    There is no general method. We do integration by using the "anti-derivative" of a function f: a function F whose derivative is f- and that often is a matter of "remembering" a correct f. There are ways of altering a function whose anti-derivative you do not immediately "remember" to a simpler function with a related anti-derivative- but those are often "ad hoc" and can be used only for certain situations. One of annoying things about learning (and teaching) "Calculus II" (generally "methods of integration" is that you have to learn many unrelated "methods" (tricks) that work only in limited situations.
     
    Last edited: Apr 5, 2005
  6. Apr 5, 2005 #5
    Speaking about the Fundamental thm of calculus, i was wondering why is it that for F(x)= int from 0 to x for f(t)dt, the function F is the constant function 0?
     
  7. Apr 5, 2005 #6
    Here, matrix_204, is the is the simplest answer you could possibly find:

    Consider a function f(x). If this function is smooth, we can naturally associate with it another function A(x), defined as "the area between f(x) and the x-axis counted from the point x = 0 to x". We do not have a formula for A(x), but we know that it is a function because for each value x there is only one area A.

    Now, consider breaking up the area under the curve f in to many infinitesimally thin rectangles (the same type rectangles we use in a Reimann sum), each of which has an infinitesimal area dA. This is a very graphical argument, so I hope you are picturing these little rectangles dA. Now, how can we express the area of a little rectangle dA in terms of its height and width?

    The height of the rectangle at point x is f(x) and the width of the rectangle is dx. So we have established the fundamental theorem:

    dA = f(x) dx

    dA/dx = f(x)

    Now we can find a formula for A(x), it is the function whose derivative is f(x).

    I am curious to see what anyone thinks of this derivation (which I made up, but do not expect to be unique to me). Obviously, it is about the loosest thing this side of Newton's fluxions, but in a certain real sense it works.
     
  8. Apr 5, 2005 #7

    dextercioby

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    It isn't,unless the integrand is identically zero

    [tex] f(t) \equiv 0 [/tex].

    Daniel.
     
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