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Limit of an integrand

  1. Apr 26, 2010 #1
    Hi All,

    i got a short question. if i have an integral an let the integrand approach zero, how do i handle that?

    [int][tex]\int_a^b \! f(c+x) \, dx.[/tex]

    what happens if lim(x->0) ?
    Maple still computes the integral as if nothing has happened...

    thanks all

    edit: cant get the integral sign to work....
  2. jcsd
  3. Apr 26, 2010 #2


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    There is no variable x to take the limit of.
    Note that in
    \int_a^b \! f(c+x) \, dx,
    x is a dummy variable. When you evaluate the integral, you will get something which doesn't depend on x anymore, just on - in this case - a, b and c.

    It's like asking, what happens when you take the limit of x -> 0 of 1/2, and wondering why Maple still gives 1/2.
  4. Jul 25, 2010 #3
    Hi again,

    im still a bit puzzled by this. what happens when
    \int f(c+x)h(x)dx is the according integral? Consider f and h to be pdfs.

    thanks again
  5. Jul 25, 2010 #4
    more detailed i got a probability:

    P=[tex]\int_{a}^{b}f(c+x)h(x)dx/(\int f(c+x)h(x)dx)[/tex]

    I was now told that as x approaches zero the denominator approaches one because we can eliminate the resulting f(c)...

    or is it just a misstatement?
  6. Jul 26, 2010 #5
    Note that [tex] x [/tex] is a dummy variable in the integral.

    Equivalently, writing as follows may make it clearer:
    [tex] \lim_{x \rightarrow 0} \int_{a}^{b} f(c + x) dx = \lim_{x \rightarrow 0} \int_{a}^{b} f(c + y) dy [/tex]

    That's why "there is no [tex] x [/tex] for you to take limit".

    Or, loosely speaking, after performing integration, you will "remove the dependence of the integrator variable, leaving the integral depending on the upper and lower bound", i.e. [tex] \int_{a}^{b} f(x)dx = g(a,b) [/tex].

    Hope this helps.
    Last edited: Jul 26, 2010
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