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Definite Integral of Definite Integral

  1. Apr 12, 2015 #1
    [itex]h(x)= \int_0^x (\int_0^uf(t)dt). du[/itex], then why is [itex]h'(x) = \int_0^uf(t)dt[/itex]? Shouldn't it be ##
    h(x) - h(0)## in the first equation? where ##h(x)## is the antiderivative of [itex]\int_0^uf(t)dt[/itex]? But wait, isn't antiderivative of a function without limits on it? Like for [itex]\int_a^bf(x)dx[/itex] we would say, let ##F(x)## be the antiderivative of ##f(x)##, i.e. [itex]F(x) = ∫f(x)dx[/itex]. And then we apply limits on ##F(x)## do evaluate the definite integral. So what does ##h(x)## mean in the beginning? Does it mean that ##h(x)## is the antiderivative of [itex]\int_0^uf(t)dt[/itex], i.e. [itex]h(x) = ∫(\int_0^uf(t)dt).dx[/itex] and then we apply the limits 0 and x on it? Would [itex]\int_0^uf(t)dt[/itex] be a separate function and not just some single value?
     
    Last edited: Apr 12, 2015
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  3. Apr 12, 2015 #2

    SteamKing

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    This is an example of applying the Fundamental Theorem of the Calculus:

    http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfDefiniteIntegral.aspx

    Scroll down to near the bottom of the page to the section called "Fundamental Theorem of Calculus, Part I"
     
  4. Apr 12, 2015 #3
    Yes, but I just want to clarify that what does ##g(x)## and ##f(x)## in the link correspond to in this problem?
     
  5. Apr 12, 2015 #4

    SteamKing

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    g(x) in Paul's notes corresponds to h(x) in the OP.

    f(t) in Paul's notes corresponds to ## \int_0^uf(t)dt## in the OP.
     
  6. Apr 13, 2015 #5
    What does OP stand for?
     
  7. Apr 13, 2015 #6

    WWGD

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    (O)riginal (P)ost(er).
     
  8. Apr 13, 2015 #7
    But the limits in this are from 0 to u not 0 to x. So shouldn't we be equating h'(u) rather than h'(x) to the integrand because the integrand would be a function of u (say g(u)) and not x. So it wouldn't make any sense to say h'(x) = g(u).
     
    Last edited: Apr 13, 2015
  9. Apr 13, 2015 #8

    SteamKing

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    u is only a dummy variable. The function h(x) is defined as ##h(x)= \int_0^x (\int_0^uf(t)dt). du##, and you wish to find h'(x) = dh(x) / dx.
     
  10. Apr 13, 2015 #9
    Do you mean to say that we can substitute u = x in the inner definite integral then? But wouldn't that mean that u and x are the same whereas they should have been different?
     
    Last edited: Apr 13, 2015
  11. Apr 13, 2015 #10

    SteamKing

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    No, all I'm saying is that u is used in the inner integral to avoid confusion with the limit x in the outer integral. It's more of a symbol thing.
     
  12. Apr 13, 2015 #11
    What's bothering me is that if ##h'(x)=\int_0^uf(t)dt##, RHS is a function of u and LHS is a function of x. So how can they be related?
     
  13. Apr 14, 2015 #12

    HallsofIvy

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    If [itex]F(x)= \int_a^x f(u)du[/itex] then [itex]F'(x)= f(x)[/itex], not f(u). If [itex]F(x)= \int_0^x f(u)du[/itex] with [itex]f(u)= \int_0^u h(t) dt[/itex], Then [itex]F'(x)= f(x)= \int_0^x h(t)dt[/itex]
     
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