integration the very concept so i was just taught about integration and one thing i do not understand is that say we integrate x dx from 1 to 2 ... there are an infinite number of numbers between them ... so how does the sum turn out to be finite ?
 A: Integration has got nothing to do with adding up numbers, let alone all numbers between two bounds which would be clearly impossible. It is rather about computing the area of the region bounded by the curve of the function, the $x$-axis, and the two vertical lines defined by $x=1$ and $x=2$.

This computation can be done by considering the limit of a sum of rectangle areas whose width gets smaller and smaller (and this point of view is called Riemann integration), but you should not fall in the trap of thinking like a physicist (with "infinitesimal rectangles of width $dx$") if your goal is to study maths.
A: The integral is a limit of finite sums. The sums do indeed consist of more and more terms, but the individual terms are also smaller and smaller, which makes it plausible that the limit should exist.
For your example, say $f(x)=x$. You approximate the integral by a finite sum of $n$ terms, then let $n\to\infty$ (i.e., consider the behavior as the number of terms increases without bound). Suppose you divide the interval $[1,2]$ into $n$ subintervals of equal length $\Delta x= \tfrac{2-1}{n}=\tfrac1n$. Choose a point $x_k$ in the $k^{\textrm{th}}$ subinterval--I'll pick the right endpoint $x_k = 1+\tfrac{k}{n}$, so $$x_1=1+\tfrac 1n\in[1,1+\tfrac 1n]$$
$$x_2=1+\tfrac 2n\in[1+\tfrac 1n,1+\tfrac 2n]$$
$$\cdots$$
$$x_n = 1+\tfrac nn =2 \in [1+\tfrac{n-1}n, 2]$$
Our approximation to the integral using $n$ terms is
$$S_n = \sum\limits_{k=1}^{n} f(x_k)\Delta x = \sum\limits_{k=1}^{n}(\underbrace{1+\tfrac kn}_{f(x_k)})(\underbrace{\tfrac 1n}_{\Delta x})$$
Let's fiddle with it to try to get a closed form:
$$S_n = \tfrac 1n\sum\limits_{k=1}^{n}(1+\tfrac kn) =
\tfrac 1n\left(\sum\limits_{k=1}^{n}1+\sum\limits_{k=1}^{n}\tfrac kn\right)$$
$$= \tfrac 1n\left(n+\tfrac 1n \sum\limits_{k=1}^{n}k\right)$$
$$=\tfrac 1n\left(n+\tfrac 1n \tfrac{n(n+1)}{2}\right)$$
$$=\tfrac 1n\left(n+\tfrac{n+1}{2}\right)$$
$$=\tfrac 1n\left(\tfrac{3n+1}{2}\right)$$
$$= \tfrac 32 + \tfrac 1{2n}$$
So our $S_n=\tfrac 32 + \tfrac 1{2n}$. The value of the integral is
$$\int\limits_1^2x\;dx = \lim\limits_{n\to\infty}S_n = \lim\limits_{n\to\infty}\left(\tfrac 32 + \tfrac 1{2n}\right)= \tfrac 32$$
