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Call me stupid, but I would like to know whether my understanding is okay:

$$\frac{d}{dx}\left(\int_0^x f(s)ds\right)=\frac{d}{dx}F(x)=f(x)$$

user91500
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luka5z
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    Fundamental theorem of analyis. – gndz Feb 04 '14 at 16:20
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    This is correct (assuming that $f$ is sufficiently smooth, such as being continuous). The intuition is that, roughly, $F(x+\delta) = \int_0^{x+\delta} f = \int_0^x f + \int_x^{x+\delta} f = F(x) + \int_x^{x+\delta} f$, and for small $\delta$, $\int_x^{x+\delta} f \approx f(x) \delta$. – copper.hat Feb 04 '14 at 16:21

2 Answers2

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I didn't find this at all intuitive until very recently when I worked through the following to get happy with it. Start with definition of a derivative

$$\frac{dg}{dx} = \lim_{h\to0} \frac{g(x+h)-g(h)}{h} $$

then

$$\frac{d}{dx}\int_0^xf(s)\;ds =\lim_{h\to0} \;\left[\frac{\int_0^{x+h}f(s)\;ds\;-\;\int_0^{x}f(s)\;ds}{h}\right]$$

$$= \lim_{h\to0} \;\left[\frac{\int_x^{x+h}f(s)\;ds}{h}\right]$$

$$= \lim_{h\to0} \;\left[\frac{f(x) h + O(h^2)}{h}\right]$$

$$= \lim_{h\to0}\; [f(x) + O(h)]$$

$$=f(x)$$

This is the Fundamental Theorem of Calculus. There is a sequence of Khan Academy videos on this starting here.

TooTone
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    What the intuition behind the $O(h^2)$ term in the third line of the second equation? – David Simmons Feb 04 '14 at 16:55
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    @DaveS It means $\int_x^{x+h}f(s)ds = f(x)h$ to first order. – TooTone Feb 04 '14 at 17:01
  • can you explain that part more – Muhammad Umer Mar 17 '14 at 00:34
  • @MuhammadUmer it means the error term is small so you can ignore it. I don't think what I've written here would count as a proof. For that see, e.g., http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Proof_of_the_first_part. – TooTone Mar 17 '14 at 00:36
  • i have read that page, but that has problem. theorem proven using assumption. Can you explain how did you do integral..and then where did h go – Muhammad Umer Mar 17 '14 at 00:48
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This is precisely the Fundamental Theorem of Calculus. And $f(x)$ should be continuous on some closed interval.

michek
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