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 ?

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
$$\int\limits_1^2x\;dx = \lim\limits_{n\to\infty}S_n = \lim\limits_{n\to\infty}\left(\tfrac 32 + \tfrac 1{2n}\right)= \tfrac 32$$
