How to find integrals using limits? How to find integrals using limits?
The question arise when I see that to find the derivative of a function $f(x)$ we need to find: $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ and it works fine for finding derivatives of every function you can give. But is there a similar approach to find integrals using limits?
Thanks!
 A: That is the definition of the derivative of a function, not simply a way to find it (though it can be directly used).
The Riemann integral is sort of defined in terms of limits (see the Wiki article for example), though in a slightly subtle way. For many nice functions but not in general, it is equivalent to the following naive definition:
$$\int_a^b f(x) \;\mathrm d x = \lim_{N\to \infty} \sum_{n=1}^{N} f\left(a+\frac{(b-a)n}N\right) \times \frac{b-a}N$$
That is, calculate the area under $N$ rectangles placed along the curve, then let $N\to\infty$.
A: For an integrable function $f(x)$, the integral in limit form is as follows:
$$\int_a^bf(x)\,dx=\lim_{\Delta{x}\to0}\sum_{k=0}^{N}f(a+k{\Delta}x)\,\Delta{x}$$
where $N=\frac{b-a}{\Delta{x}}$.
In other words, an approximation of the integral is the summation of a set of rectangles of width $\Delta{x}$ and the height equals the function value at each discrete point in the interval of integration. The smaller $\Delta{x}$ is (the closer it is to $0$), the more precise the approximation. As $\Delta{x}$ approaches $0$, the summation approaches the precise value of the integral.
A: Yes.
The simplest way to write
$\int_a^b f(x)\, dx$
as a limit
is to divide the range of integration
into $n$ equal intervals,
estimate the integral over each interval
as the value of $f$
at either the start, end, or middle
of the interval,
and sum those.
Mathematically,
$$\int_a^b f(x)\, dx
= \lim_{n \to \infty} \frac1{n} \sum_{k=0}^{n-1} f(a+(k+c)\frac{b-a}{n})
$$
where $c = 0, 1/2, 1$
for the function to be evaluated at,
respectively,
the start, middle, or end
of the subinterval.
For a reasonable class of functions,
this limit converges to
the integral.
Generalizations of this
have provided many careers
in mathematics.
A: I would say that Sharkos or Marty is probably fine for all of the usual functions one learns in a first course (I am writing this up for completeness, and also for intellectual exercise). It is certainly sufficient for continuous functions, and even functions with a finite number of jump discontinuities (I would have to think carefully about $f(x)=\sin(\frac{1}{x}))$. In fact this may be overkill. But here a formal proper definition. So here we go!
Let us start with some definitions. 



*

*A patrician of $[a,b]$ is a sequence of numbers, 
$$0=x_0<x_1<x_2<x_3\cdots x_{n-1}<x_n=1.$$ We will denote our partisans by $\mathcal{P}$ 
with a possible subscript.


*The mesh size of a patrician, $\mathcal{P}$, denoted by $\mu(\mathcal{P})$ will be $\max(x_{i+1}-x_i)$.


*A Patrician $\mathcal{P}'$ is finer than $\mathcal{P}$, denoted $\mathcal{P}<\mathcal{P}'$ if each element of the sequence of $\mathcal{P}$ is an element of the patrician $\mathcal{P}'$.


*A tagged partician, $\mathcal{P}$, is a partician, $$0=x_0<x_1<x_2<x_3\cdots x_{n-1}<x_n=1.$$, and a sequence of numbers, $$y_1<y_2<\cdots y_{n-1}<y_n$$ such that 
$$x_{n-1}\leq y_i\leq x_n$$


*The underlying partisan of a tagged patrician is simply the sequence, $$0=x_0<x_1<x_2<x_3\cdots x_{n-1}<x_n=1.$$ A tagged patrician $\mathcal{P}$ is finer another tagged patrician $\mathcal{P}'$, if the underlying patrician of $\mathcal{P}$ is finer than the underlying patrician of $\mathcal{P}'$. The mesh size of a tagged patrician is the same as the mesh size of the underlying tagged patrician.


*We will call a sequence of tagged patricians, $$\mathcal{P}_0, \mathcal{P}_1, \mathcal{P}_2, \cdots, \mathcal{P}_n, \mathcal{P}_{n+1}, \cdots$$ is good if 
a. $$\mathcal{P}_i<\mathcal{P}_{i+1}$$, and 
b. $$\lim_{i\to \infty}\mu{\mathcal{P}_i}=0$$

We will now set up some notation: Let $\mathcal{P}$ be a tagged partician, 
$$0=x_0<x_1<x_2<x_3\cdots x_{n-1}<x_n=1.$$, and a sequence of numbers, $$y_1<y_2<\cdots y_{n-1}<y_n.$$
 Then we write define $$\Sigma_{\mathcal{P_i}}f(y_i)\mu(\mathcal{P}_i)=\Sigma_{k=1}^n f(y_1)\mu(\mathcal{P}).$$

We say that a function $f(x)$ is Riemann integrable if 
a. for every good sequence of tagged particians, $\{\mathcal{P}_i\}_{i=0}^{\infty}$ if the limit, $$\lim_{n\to \infty}\Sigma_{\mathcal{P_i}}f(y_i)\mu(\mathcal{P}_i) $$ , exists
b. If for any two tagged particians,$\{\mathcal{P}_i\}_{i=0}^{\infty}$, and $\{\mathcal{P'}_i\}_{i=0}^{\infty}$ , the limits, $$\lim_{n\to \infty}\Sigma_{\mathcal{P_i}}f(y_i)\mu(\mathcal{P}_i)=\lim_{n\to \infty}\Sigma_{\mathcal{P'_i}}f(y_i)\mu(\mathcal{P'}_i)$$ 
We then define $$\int_a^b f(x)dx=\lim_{n\to \infty}\Sigma_{\mathcal{P_i}}f(y_i)\mu(\mathcal{P}_i) $$, for any good partician (it does not matter which one since by b the limit will be the same).
As has been mentioned above, for continuous functions, the definition by Sharkos  or Marty is fine.
