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In school, especscially, one is not taught "why" we can change variables, dummy variables in integration.

$$\int_{a}^{b} f(x) dx$$

We can change the $x$ variable to $y$ for example.

The idea is that the value of the expression does not depend on the given variable correct?

This is known as the bound variable idea.

But how do we know that: the value of the expression does not depend on the given variable

??

Is it an axiom, theorem?

Thanks.

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Another quick answer: get out of the 19th century! Functions are not $f$ of $x$. Functions are independent of the names of their variables. The following represent the same function: $$ \begin{align} x &\mapsto f(x) \\ t &\mapsto f(t) \\ u &\mapsto f(u) \\ \spadesuit &\mapsto f(\spadesuit) \\ \# &\mapsto f(\#) \end{align} $$ By contruction, the Riemann integral of a function depends only on the function (and on the interval $[a,b]$, but we do not take care of it now): therefore the notation $\int_a^b f(x)\, dx$ is not only heavy, but also misleading. We should write it $\int_a^b f$.

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  • $\begingroup$ Exactly, but math isnt a myth, (passed from generations) there has to be a reason why the value doesnt change. $\endgroup$ – anonymous Jan 9 '15 at 14:09
  • $\begingroup$ The value does not change because there is NO $x$ variable at all! $\endgroup$ – Siminore Jan 9 '15 at 14:10
  • $\begingroup$ Then what is the independent variable? $\endgroup$ – anonymous Jan 9 '15 at 14:11
  • $\begingroup$ "Independent variable" is a notion that applies to particular applications of functions. In the integral, if we use the $dx$ notation at all (although as shown in this answer, we don't have to!), it signifies that $f$ is (in effect) evaluated at each point in the interval $(a,b).$ $\endgroup$ – David K Jan 9 '15 at 14:22
  • $\begingroup$ @DavidK, but overall, is this from the bound variable definition? $\endgroup$ – anonymous Jan 9 '15 at 14:25
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This really harkens to computer science. When you do write a sum $$\sum_{k=1}^n a_k,$$ the variable $k$ is a "loop variable;" to wit, it is a placeholder. The sum sums the numbers in the list $\{a_1, a_2, \cdots a_n\}$ regardless of the name of the variable used to index the sum.

The variable $k$ has a scope confined to the summation. The same is true for variables of integration. In the integral $$\int_a^b f(x)\, dx,$$ the variable $x$ is merely a placeholder.

You do a similar thing when do make a definition like this one $$f(x) = \sqrt{x}, \qquad x \ge 0.$$ You can replace $x$ with cow as follows $$f(\text{cow}) = \sqrt{\text{cow}}, \qquad \text{cow} \ge 0.$$ and the definition is unaltered.

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  • $\begingroup$ But why doesnt this change the value $\endgroup$ – anonymous Jan 9 '15 at 14:14
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It's nothing like an axiom, it as the same meaning and you are just changing a name.

For example, let's say we have : $$a = b + c$$ Now, let $d = b$ and $e = c$. You have : $$a = d + e$$

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  • $\begingroup$ Not possible. Because then you cant solve the Gaussian Integral using $x = y$ $\endgroup$ – anonymous Jan 9 '15 at 14:08
  • $\begingroup$ Yet it's what happens. $\endgroup$ – servabat Jan 9 '15 at 14:09
  • $\begingroup$ Then try to solve the Gaussian Integral with the double integral method, you wont be able to. $\endgroup$ – anonymous Jan 9 '15 at 14:10
  • $\begingroup$ You are changing the problem, here we are talking of a one variable. If you want to ask something about multivariable calculus, you must specify your question, and this on another post. $\endgroup$ – servabat Jan 9 '15 at 14:11
  • $\begingroup$ $(+1)$ the part I dont get it, is this Because $x$ is a bound variable? $\endgroup$ – anonymous Jan 9 '15 at 14:26
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Put very simply, $x$ here in your example is a bound variable, before it was formed into an integrand it was a free variable. The integral sign is a variable-binding operator. We are free to change bound variables into other bound variables providing of course any labelling of the bound variable to the variable-binding operator is also suitabley changed.

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  • $\begingroup$ By definition?? $\endgroup$ – anonymous Jan 9 '15 at 14:10
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It does not follow from what is usually considered an "axiom".

Indeed, it is an application of a general rule of inference, called Instantiation of universals.

If you have time to spend on things like this, try follow and understand all the links in that Wikipedia page.

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