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f(t) is continuous function.So I know that $\int _0^x {f(t) dt}=$ "The net signed area between $t=0, y=0, t=x$, and $y=f(t)$"

And I can find the same result with endless small rectangulars areas method.

"The net signed area between $t=0, y=0, t=x$, and $y=f(t)$"=$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})$

Therefore,

$$\int _0^x {f(t) dt}=\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})$$

Is there any other analytical method (such as transform methods or any other one) to prove that the equation is correct?

EDIT:During my attempts to proof the equality, I have noticed that the equality can be proved via power series expression of $f(x)$. I did not think that way before asking the question. I would like to share it with you. It is also welcome your comments about my approach and waiting for your different methods to show a proof via another method as well.
$ f(x) =f(0)+\frac{f'(0)x}{1!}+\frac{f''(0)x^2}{2!}+.....=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $

$$\int _0^x {f(t) dt}=\int _0^x(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} t^n)dt=\sum_{n=0}^{\infty} (\frac{f^{(n)}(0)}{n!}\int _0^x t^n dt)=\sum_{n=0}^{\infty} (\frac{f^{(n)}(0)}{n!}\frac{x^{n+1}}{n+1})$$ $$\int _0^x {f(t) dt}=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^{n+1}}{(n+1)!}$$ $$(1)$$


$$f(\frac{kx}{n})=\sum_{m=0}^{\infty} \frac{f^{(m)}(0)}{m!} (\frac{kx}{n})^m$$ $$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $a_j$ are constants. More information about summation http://en.wikipedia.org/wiki/Summation

$$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})=\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n \sum_{m=0}^{\infty} \frac{f^{(m)}(0)}{m!} (\frac{kx}{n})^m=\lim_{n\to\infty} \frac{x}{n}\sum_{m=0}^{\infty} \frac{x^m}{n^m} \frac{f^{(m)}(0)}{m!} \sum \limits_{k=1}^n k^m=\lim_{n\to\infty} \frac{x}{n}[f(0)n+\frac{f'(0)x}{n 1!}(\frac{n^2}{2}+\frac{n}{2})+ \frac{f''(0)x^2}{n^2 2!}(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})+\frac{f'''(0)x^3}{n^3 3!}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+\frac{f^{(4)}(0)x^4}{n^4 4!}(\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30})+...... ]= \lim_{n\to\infty} [f(0)x+\frac{f'(0)x^2}{n^2 1!}(\frac{n^2}{2}+\frac{n}{2})+ \frac{f''(0)x^3}{n^3 2!}(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})+\frac{f'''(0)x^4}{n^4 3!}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+\frac{f^{(4)}(0)x^5}{n^5 4!}(\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30})+...... ]= [f(0)x+\frac{f'(0)x^2}{ 2!}+ \frac{f''(0)x^3}{ 3!}+\frac{f'''(0)x^4}{ 4!}+\frac{f^{(4)}(0)x^5}{ 5!}+...... ]$$

$$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})=\sum_{m=0}^{\infty} \frac{f^{(m)}(0)x^{m+1}}{(m+1)!}$$ $$(2)$$

Equation $(1)$ and equation $(2)$ are equal to each other. The proof is completed.

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Actually, $\int_0^xf(t)\,dt$ is the net signed area between $t=0$, $y=0$, $t=x$, and $y=f(t)$. –  Arturo Magidin Mar 21 '12 at 21:15
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Note the issue of the variable, too: the integral you have is the area under the graph of $y=f(t)$ up to $t=x$. $x$ is playing the role of a limit of the integral, not of the variable. "The net signed are under $f(x)$" implies the integral is with respect to $x$, which yours is not. –  Arturo Magidin Mar 21 '12 at 21:21
    
I edited . Thanks for your comment –  Mathlover Mar 21 '12 at 21:21
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What you mean by "any other" and what is your definition of integral? For a Riemann integral I doubt there will be one, as you have to use definition somewhere, so all arguments will be similar. For a Lebesgue integral there might be, but this is another story with much more theory involved. –  dtldarek Apr 17 '12 at 23:14
    
Other ways mean maybe derivatives n times both side and goes limits or delta dirac function using or some integral transform methods (such as en.wikipedia.org/wiki/Integral_transform) . I tried some of them but till now I could not show they are equal each other. –  Mathlover Apr 17 '12 at 23:21
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I suppose you're assuming $f$ is continuous. The statement is easily seen to be true for $f(t)=t$, and by subtracting a multiple of this we can assume $f(0) = f(x)$. By the Stone-Weierstrass theorem, trigonometric polynomials with period $x$ are uniformly dense in the continuous functions with period $x$, and so it suffices to prove the result for $f(t) = \cos(2 \pi j t/x)$ and $\sin(2 \pi j t/x)$ for each integer $j$. But for those it can be done explicitly: $\sum_{k=1}^n \sin(2 \pi j k/n) = 0$ for all integers $j$, $\sum_{k=1}^n \cos(2 \pi j k/n) = n$ if $j$ is a multiple of $n$, $0$ otherwise...

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