# Definite Integral of Definite Integral

1. Apr 12, 2015

### andyrk

$h(x)= \int_0^x (\int_0^uf(t)dt). du$, 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?

Last edited: Apr 12, 2015
2. Apr 12, 2015

### SteamKing

Staff Emeritus
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"

3. Apr 12, 2015

### andyrk

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

4. Apr 12, 2015

### SteamKing

Staff Emeritus
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.

5. Apr 13, 2015

### andyrk

What does OP stand for?

6. Apr 13, 2015

### WWGD

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

7. Apr 13, 2015

### andyrk

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
8. Apr 13, 2015

### SteamKing

Staff Emeritus
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. Apr 13, 2015

### andyrk

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
10. Apr 13, 2015

### SteamKing

Staff Emeritus
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. Apr 13, 2015

### 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?

12. Apr 14, 2015

### HallsofIvy

Staff Emeritus
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 $F'(x)= f(x)= \int_0^x h(t)dt$