Proof that the upper and lower Darboux sums approach the same limit.

Question

While finding the area under the curve we break the curve into infinite numbers of strips and then find the area of individual strips. First we take circumscribed strips and then inscribed strips. They lead to an overestimation and underestimation of the actual area, respectively. But in limiting condition both the areas comes out to be equal. And that is equal to integral of function. Why this area is equal to integral of the function over the given interval? How can we proof this fundamental theorem of calculus?

Answer 1

Given a function $f(x)$, one can break down any interval $\left( 0,b \right)$ into rectangles of smaller and smaller widths. In the limiting case, you achieve the exact area between the curve and the base.

Hopefully this provides you with some idea of how it works:

By splitting the interval into smaller sub-intervals of length $\frac{b}{n}$, we can multiply them by the function's value at rightmost (or leftmost, resulting in the lesser and greater approximate areas, respectively) points. These points are as follows -

$$\frac{b}{n}, \frac{2b}{n}, \cdots, \frac{\left(n - 1\right)b}{n},\frac{nb}{n} = b$$

Their heights at each of these points are $f\left(\frac{kb}{n} \right)$, where $k \in \{ 1,2, \cdots , n \}$.

We can now multiply the bases and the heights, and add these rectangles together to achieve an approximation that improves as $n$ increases -

$$\sum_{k=1}^n \left[ \left( \frac{b}{n} \right) f \left( \frac {kb}{n}\right) \right]$$

Using the idea of a limit, we can achieve the exact area -

$$\lim_{n\to \infty} \sum_{k=1}^n \left[ f \left( \frac {kb}{n}\right) \left( \frac{b}{n} \right) \right] = \int_{0}^b f(x)dx$$

Note: the same is true if you take the heights from the leftmost side, only difference being that $\sum_{i=1}^n$ becomes $\sum_{i=1}^{n-1}$.

Not a wholly rigorous proof, but might be of use intuitively. As the rectangles shrink in width, the accuracy as a whole improves.

As to why this limiting case area is equal to the integral, I'd suppose it's by definition.

Answer 2

You're mixing up the logic of how things proceed. You ask:

They lead to an overestimation and underestimation of the actual area, respectively. But in limiting condition both the areas comes out to be equal. And that is equal to integral of function. Why this area is equal to integral of the function over the given interval?

The answer to your question is that this happens by definition of "Riemann-integrable" and the definition of the Riemann integral. Let me briefly outline the integration theory goes:

1. Given a bounded function $f:[a,b] \to \Bbb{R}$, and a partition $P = \{t_0, \dots, t_n\}$ of $[a,b]$ (which just means $a=t_0 < \dots < t_n = b$), we define the upper and lower sums of $f$ with respect to $P$ by \begin{align} U(f,P):= \sum_{i=1}^n M_i (t_i - t_{i-1}) \quad \text{and} \quad L(f,P):= \sum_{i=1}^n m_i(t_i - t_{i-1}) \end{align} where $M_i := \sup\{f(x)| \, \, t_{i-1}\leq x \leq t_i\}$ and $m_i := \inf\{f(x)| \, \, t_{i-1}\leq x \leq t_i\}$. Intuitively, these are over and under approximations to "the area under the graph of $f$".

2. Next, a slightly technical lemma shows that for any partitions $P,P'$ of $[a,b]$, we have $L(f,P) \leq U(f,P')$. This implies \begin{align} \underbrace{\sup\{L(f,P)| \, \, \text{$P$ a partition of $[a,b]$}\}}_{:= \underline{\int_a^b}f} \leq \underbrace{\inf\{U(f,P)| \, \, \text{$P$ a partition of $[a,b]$}\}}_{:= \overline{\int_a^b}f}. \tag{$*$} \end{align} Intuitively what this says is that "the limit of under-approximations is smaller than the limit of over-approximations". The symbols $\underline{\int_a^b}f$ and $\overline{\int_a^b}f$ are respectively called the Lower and Upper integrals of $f$ over the interval $[a,b]$. So, what this result says is that "the lower integral is smaller than the upper integral".

3. So far, we have only talked about partitions, upper and lower sums, and upper and lower integrals, but we haven't even defined what it means for a function to be Riemann-integrable. Well, like you worry about, it is very possible that the upper and lower integrals are distinct from each other (i.e it is sometimes possible to have a strict inequality $<$ in $(*)$). Well, if the upper and lower integrals do not agree, then it seems that the function is "pretty bad" (as you can later prove, every continuous, or monotone function, or many other "nice functions" all will have upper and lower integrals being equal).

So, what we do at this stage is make a definition.

We define a $f:[a,b] \to \Bbb{R}$ to be Riemann-integrable if it is bounded and the upper and lower integrals of $f$ are equal; i.e if \begin{align} \underline{\int_a^b}f = \overline{\int_a^b}f \end{align} In this case, we define the Riemann integral of $f$ over $[a,b]$ to be this common number; \begin{align} \int_a^bf &:= \underline{\int_a^b}f = \overline{\int_a^b}f \end{align}

So, what this definition is trying to do is to pick out a certain class of "nice functions". These "nice functions" are called "Riemann-integrable". What is the special property of these functions? Well they have the defining property that their upper and lower integrals are equal. Not every function has this property. So, the answer to your question is literally that it is true by definition.

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Finally, you ask:

How can we proof this fundamental theorem of calculus?

At this stage there is nothing to prove. Everything is just a definition (properly defining things in integration is quite a technical matter, which is why almost any high-school introduction/ most elementary math/physics introductions to integration are not $100\%$ rigorous, and they're only meant to convey the broad ideas). The Fundamental Theorem of Calculus is actually a completely different thing, and it relates the notion of integration and differentiation (but I guess this isn't what you intended, and rather you just used the term "fundamental" in a more colloquial sense).