Why does definite integral not depend on the variable? Why does definite integration not depend on the variable ?? If I can use any other variable which is defined in such a way that It lies outside of the given interval, then the expression for intergral may vary. 
Can you give me a mathematical proof to it(ie not just examples)
 A: This is what is called a bound variable.
We have $$\int_a^b f(x) \,\mathrm dx=\int_a^b f(t) \,\mathrm dt$$ just like we have $$\forall n\in\mathbb N\colon n^2\ge n\iff \forall k\in\mathbb N\colon k^2\ge k$$ or just like the correct answer to

What is $3x$ if $x=4$? 

is the same as to 

What is $3u$ if $u=4$?

As others have said, all this is more or less by definition. But you can give a proof for integrals using sustitution with the identity function if you desire, though that merely hide a direct proof from the definition that was used for showing the rule of substitution.
As a matter of fact, in measure theory there is a notation 
$$\int f\,\mathrm d\mu$$
for integrals that does not make use of a variable for the argument, and this makes it more obvious that we are using the function $f$ "as a whole" and not specific values $f(x)$ (or $f(t)$) when integrating.
A: why don't you give us an example of what you mean? No integral should depend on the variable you integrate against. $\int f(x) dx$  is a short hand $\int^x f(y) dy$ where the lower boundary is not given and can be chosen arbitarily, hence the +C
A: A definite integral works like this: you pick a function and you get a number. That's all. Of course you need to make things precise (the domain of the function must include the interval of integration, the function must have some properties etc.), but a definite integral is a machinery that produces a number (either real or complex, in some cases).
Therefore it is meaningless to ask for a proof: you cannot prove a definition!
