How do we arrive at the definite integral to find area approximated by a sum of rectangles?

Question

The area enclosed by a one variable function from a to b can be approximated by $n$ rectangles$$A \approx \sum_{i=1}^{n} f(x_i)(x_i-x_{i-1})$$ and if we let $n \rightarrow \infty$ we get $$A = \sum_{i=1}^{\infty} f(x_i)\Delta x$$ which is evaluated using an integral $$\int_a^b f(x) dx = F(a) - F(b)$$ this last step seems like a bit of a jump and nobody has really explained this to me before, I was wondering if anyone could provide some insight as to how this works and why an infinite sum can evaluated in this way. I understand that it obviously works and using the idea of displacement and velocity I understand how this would provide the area under the graph as the area under a velocity graph is simply the change in position, but this summation notation seems like a leap and I see this used in many derivations of other formulas so I would really like to fully understand this concept. Thanks for any help!

Answer 1

To make the expression $A = \sum_{i=1}^{\infty} f(x_i)\Delta x$ rigorous (and understandable) you can do the following: Fix the interval $[a,b]$ for which you have a function $f:[a,b]\rightarrow \mathbb R$ and for which you want to define $\int_a^b f(x) dx$.

Some necessary terms: Any finite sequence $(x_0,x_1,\ldots,x_n)$ with $a=x_0<x_1<x_2\ldots <x_n=b$ we call partition of interval. The value $\max\{x_i-x_{i-1}:i\in\{1,2,\ldots n\}\}$ is called the norm of the partition of interval, i.e. it is the maximal length of a step $x_i \rightarrow x_{i+1}$. Let be $P_\tau([a,b])$ the set of all partitions of interval for $[a,b]$ with norm less or equal to $\tau$. Now you can define $$S_\tau = \sup\left\{\sum_{i=1}^{n} f(x_i)(x_i-x_{i-1}): (x_0,x_1,\ldots,x_n) \in P_\tau([a,b]) \right\}$$ and $$s_\tau = \inf\left\{\sum_{i=1}^{n} f(x_i)(x_i-x_{i-1}): (x_0,x_1,\ldots,x_n) \in P_\tau([a,b]) \right\}$$ $S_\tau$ is an upper bound approximation for our idea of $\int_a^b f(x) dx$ and $s_\tau$ is a lower bound approximation. Finally, iff $\lim_{\tau\rightarrow 0} S_\tau=\lim_{\tau\rightarrow 0} s_\tau$ you can define $f$ to be integrable with $$\int_a^b f(x) dx = \lim_{\tau\rightarrow 0} S_\tau=\lim_{\tau\rightarrow 0} s_\tau$$ Have a look at the Wikipedia article Fundamental theorem of calculus for an proof of the statement $$\int_a^b f(x) dx = F(b)-F(a)$$

Answer 2

This is a purely analytical approach.

Let $F$ be a function differentiable on $[a,b]$.

Suppose that $$\lim_{n\rightarrow\infty}\frac {b-a}n\sum_{i=1}^n F'(\xi_i^{(n)})$$exists and is independent of the tags $\,\xi_i^{(n)}$ .

(it is denoted by $\int_a^bF'$)

If you note , by the mean value theorem, that $$\forall\, n: \qquad F(b)-F(a)=\frac {b-a}n\sum_{i=1}^n F'(c_i^{(n)})$$you are done.

(I omitted some detail)

Answer 3

There are two issues involved here:

(a) Defining the area of more or less arbitrary shapes $S\subset{\mathbb R}^2$: Areas of rectangles are unproblematic, and so are areas of unions of finitely many rectangles having at most edges in common. The idea now is to approximate the given shape by such unions of ever smaller rectangles and "passing to the limit". The value so obtained is then defined to be ${\rm area}(S)$. This program is dealt with in detail in geometric measure theory, and it is shown that this area function has the desired properties.

(b) A particularly simple, but important, paradigm of shape is the "area under a curve" $y=f(x)$ for $a\leq x\leq b$. How does the area of this shape depend on the given $f$? The Fundamental Theorem of Calculus (FTC) gives an answer to this question which is absolutely miraculous. It says you can take any antiderivative (!) $F$ of $f$ and simply compute $F(b)-F(a)$ to get the area in question.

The essential idea to prove the FTC is to consider the function $x\mapsto A(x)$, where $A(x)$ denotes the area under the curve over the (variable) interval $[a,x]$. Comparing $A(x+h)$ with $A(x)$ for $h\to0+$ one recognizes that in fact $A'(x)=f(x)$; see the corresponding figure on the FTC-Wikipedia page.