Definite Integral of Definite Integral

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[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?
 
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SteamKing
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[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]? Would [itex]\int_0^uf(t)dt[/itex] be a separate function and not just some single value?
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"
 
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Yes, but I just want to clarify that what does ##g(x)## and ##f(x)## in the link correspond to in this problem?
 
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SteamKing
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Yes, but I just want to clarify that what does ##g(x)## and ##f(x)## in the link correspond to in this problem?
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.
 
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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.
What does OP stand for?
 
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WWGD
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(O)riginal (P)ost(er).
 
  • #7
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f(t) in Paul's notes corresponds to ∫u0f(t)dt \int_0^uf(t)dt in the OP.
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).
 
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SteamKing
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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).
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.
 
  • #9
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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.
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?
 
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SteamKing
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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?
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.
 
  • #11
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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?
 
  • #12
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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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