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How to prove that $$\int^a_bf(x)dx=\lim_{\Delta x\rightarrow0}\sum_{x=a}^bf(x)\Delta x$$ (which $\int^a_bf(x)dx=g(a)-g(b)$ where $\frac{d}{dx}g(x)=f(x))$?

I know this is a very basic thing of integration but it seems that I can't its proof anywhere. Please help me... Thank you.


Maybe I should say it in this way: prove that $\int^a_bf(x)dx$ is the area between the curve y=f(x) and x-axis in the interval [b,a].

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How do you define $\int_a^b f(x)dx$? –  Chris Janjigian Sep 9 '12 at 13:24
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What is $$\sum_{x=a}^b$$ anyway?? Are $\,a,b\,$ integers so that the variable $\,x\,$ can run between them? I think there's a serious mistake/typo in this question. –  DonAntonio Sep 10 '12 at 3:01
    
That is the definition of the integral in most of the calculus books. Well, except for a few typos. –  Keivan Sep 11 '12 at 3:04
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Please see my question similar to yours. math.stackexchange.com/questions/123048/… –  Mathlover Sep 11 '12 at 8:36
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1 Answer 1

up vote 1 down vote accepted

If you mean you want a proof that shows that if

$$F'(x)=f(x)$$

then

$$F(b)-F(a)=\lim_{n \to \infty} \Delta x\sum_{x=0}^{n-1} f(x+n\Delta x)$$

where

$$\Delta x=\frac{b-a}{n}$$

I suggest you look in to the fundamental theorem of calculus.

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