# Definite Integral of Definite Integral

$h(x)= \int_0^x (\int_0^uf(t)dt). du$, then why is $h'(x) = \int_0^uf(t)dt$? Shouldn't it be ##
h(x) - h(0)## in the first equation? where ##h(x)## is the antiderivative of $\int_0^uf(t)dt$? But wait, isn't antiderivative of a function without limits on it? Like for $\int_a^bf(x)dx$ we would say, let ##F(x)## be the antiderivative of ##f(x)##, i.e. $F(x) = ∫f(x)dx$. 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 $\int_0^uf(t)dt$, i.e. $h(x) = ∫(\int_0^uf(t)dt).dx$ and then we apply the limits 0 and x on it? Would $\int_0^uf(t)dt$ be a separate function and not just some single value?

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SteamKing
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$h(x)= \int_0^x (\int_0^uf(t)dt). du$, then why is $h'(x) = \int_0^uf(t)dt$? Shouldn't it be ##
h(x) - h(0)## in the first equation? where ##h(x)## is the antiderivative of $\int_0^uf(t)dt$? But wait, isn't antiderivative of a function without limits on it? Like for $\int_a^bf(x)dx$ we would say, let ##F(x)## be the antiderivative of ##f(x)##, i.e. $F(x) = ∫f(x)dx$. 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 $\int_0^uf(t)dt$, i.e. $h(x) = ∫(\int_0^uf(t)dt).dx$? Would $\int_0^uf(t)dt$ 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"

Yes, but I just want to clarify that what does ##g(x)## and ##f(x)## in the link correspond to in this problem?

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.

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?

WWGD
Gold Member
(O)riginal (P)ost(er).

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.

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.

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?

HallsofIvy
If $F(x)= \int_a^x f(u)du$ then $F'(x)= f(x)$, not f(u). If $F(x)= \int_0^x f(u)du$ with $f(u)= \int_0^u h(t) dt$, Then $F'(x)= f(x)= \int_0^x h(t)dt$