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I'm conscious this question has been already asked here:

For continuous function $f$, prove: $\int_{0}^{x} \; \left[\int_{0}^{t}f(u) \;du \right] \;dt=\int_{0}^{x} f(u)(x-u)du$

Prove: $\int_{0}^{x} f(u)(x-u) du = \int_{0}^{x} \left(\int_{0}^{u} f(t)dt\right)du$

But I still have doubts. Let me explain. I did this in the right side:

$F(x)=\int_{0}^{x}\left ( \int_{0}^{u}f(s)ds \right )du=x\int_{0}^{x}f(u)du-\int_{0}^{x}uf(u)du$

$\implies F'(x)=xf(x)+\int_{0}^{x}f(u)du-xf(x)=\int_{0}^{x}f(u)du\tag1$

Then, by the left side we have:

$G(x)=\int_{0}^{x}\left ( \int_{0}^{u}f(s)ds \right )du=\int_{0}^{x}\mathfrak{F}(u)du $

with $\mathfrak{F}(u)=\int_{0}^{x}f(s)ds$

$\Rightarrow G'(x)=\mathfrak{F}(x)=\int_{0}^{x}f(s)ds\tag2$

So $(1)=(2)$, but that doesn't imply that $F(x)=G(x)$. An example of that are the functions $h(x)=3x^2$ and $k(x)=3x^2+5$, i.e., $h'(x)=k'(x)$ but $h(x)\neq k(x)$

So, how can I conclude that $F(x)=G(x)$?

In the other posts I saw that they argue that $(1)=(2)=0$ if $x=0$, and that's true, but I don't understand why they conclude with that. Can you help me?

Aiden Chow
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luisegf
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    You have $F'(x) = G'(x)$ for all $x$, and $F(0) = G(0)$. So what can you tell about the difference function $h = F-G$? – Martin R Sep 03 '20 at 18:39
  • That, $h'(x)=0$ because $F'(x)=G'(x)$ and h(0)=0. So, when two functions have the same derivative and those functions evaluated in zero, are equal to zero, does that mean always that those functions are the same? – luisegf Sep 03 '20 at 18:49
  • $h'(x) = 0$ means that $h$ is constant. Therefore ... – Martin R Sep 03 '20 at 18:50
  • But, for an example: Let be $m(x)=3x^2$ and $n(x)=3x^2+5$, then $h(x)=n(x)-m(x)$. We have that $h'(x)=0$ always but $n(x) \neq m(x)$. Those functions have the same derivative, but are not the same function. So, I understand that it is a necessary condition that both functions evaluated in zero, are equal to zero. Is it true? – luisegf Sep 03 '20 at 19:01
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    Yes, but it seems that you are thinking too complicated. Two functions $F$, $G$ are identical if their difference $h = F-G$ is identically zero. That is equivalent to $h'(x) = $ for all $x$ and $h(x_0) = 0$ for some $x_0$. – Martin R Sep 03 '20 at 19:06
  • I got it! Thank you very much!! – luisegf Sep 03 '20 at 19:42

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