Definite Integral of Definite Integral

[andyrk]

[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?

 

[SteamKing]

[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"

 

[andyrk]

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

 

[SteamKing]

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.

 

[andyrk]

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]

(O)riginal (P)ost(er).

 

[andyrk]

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).

 

[SteamKing]

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.

 

[andyrk]

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?

 

[SteamKing]

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.

 

[andyrk]

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 [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]