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For many years now I have thought about this but have not been able to get a clear answer. We all know that $\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ gives us a function we call as the derivative that gives the gradient of the function at $x$. To get the anti-derivative, we can use the $\int$ of the derivative and get back the original $f(x)$.

This part of $\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ has been explained to me many times since high school and is crystal clear. By using this I obtain that derivative of $\sin$ is $\cos$ and derivative of $x^n$ is $nx^{n-1}$. What has never been explained to me is the proof that $\int$ of a function in fact gives the area under $f(x)$ and that also using the Riemann sum. I have always wondered about this and have not been able to get the answer. What is the proof of this?

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please don't try to prove anything using f(x)=x since that is not a good example. I need a proof here for continuos functions. – quantum231 Apr 30 '13 at 20:17
@quantum231 I think the problem here is that you want to prove the riemann sums or the integral are equal to the area below graph. But the thing is, the area below the graph is just an intuitive non mathematical concept.You can't talk about it mathematically, therefore you can't prove the integral will give the area below the graph. If what I just said doesn't apply to you, then how do you define area below the graph? – Git Gud Apr 30 '13 at 20:18
Let $s(t)$ be the area under $f(x) |_{[0,t]}$, consider $\frac{1}{h}(s(t+h)-s(t))$ as $h\to 0$. – Ma Ming Apr 30 '13 at 20:20
@GitGud I'm not sure I would say the area below the graph is non-mathematical. Signed area under the graph of a function seems like a perfectly well-founded geometric idea. I always thought of Riemann integrability as a definition for the signed area under a curve. – rschwieb Apr 30 '13 at 20:38

For simplicity, assume that $f(x)\ge0$ for all $x\in[a,b]$. Riemann sums give the sum of areas of certain pairwise disjoint rectangles (overlapping boundary lines don't count), which by any reasonable concept of area is the area of the union of these rectangle. And if we can give a meaning to "area under the curve" then this area is "obviously" between the Riemann lower sums (which describe areas of subsets of the area under the curve) and Riemann upper sums (same with supersets). If the function $f$ is Riemann integrable on $[1,b]$ (which just says that lower and upper Riemann sums converge to a fixed number $c$ as the corresponding partitions become finer and finer (the rectangles become thinner and thinner), then we say $\int_a^bf(x)\,\mathrm dx=c$ and as this integral is between all lower and upper Riemann sums, the same number $c$ must be what we call area under the curve.

If you consider $F(x):=\int_0^xf(t)\,\mathrm dt$ for a continuous function $f$, then you will note that $F(x+h)-F(x)=\int_x^{x+h}f(t)\,\mathrm dt$ can be estimated by a simple Riemann sum with a single rectangle. The lower sum is a rectangle of width $h$ and heght $\min\{ f(t)\mid x\le t\le x+h\}$ and the upper Riemann sum is the same with $\max$. Therefore the difference quotient $\frac{F(x+h)-F(x)}h$ is a number between the min and max of $f$ on $[x,x+h]$. As $h$ becomes smaller, both min and max converge to $f(x)$ if - as we required - $f$ is continuous. In other words, the derivative of $F$ is $f$.

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