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The property: $$ \int_a^b f(x) \, dx=\int_a^b f(a+b-x) \, dx $$

Derivation given in my textbook:

Let $t = a+b-x$. Then $dt = -d x$. When $x=a, t=b$ and when $x=b, t=a$. Therefore, $$ \begin{aligned} \int_a^b f(x) \, dx &= -\int_b^a f(a+b-t) \, dt \\ &= \int_a^b f(a+b-t) \, dt \\ &= \int_a^b f(a+b-x) \, dx \ \end{aligned} $$

I don't understand how in step 3 of the derivation $t$ changes to $x$ when $t = a+b-x$ was assumed in the first step. Is there some implicit conversion being done here?

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    $\begingroup$ Recycling the symbol $x$. $\endgroup$
    – A rural reader
    Commented Mar 6 at 1:24
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    $\begingroup$ In a definite integral the "variable of integration" is a dummy variable. You can change its name to whatever you like. mathworld.wolfram.com/DummyVariable.html $\endgroup$
    – Ethan Bolker
    Commented Mar 6 at 1:26
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    $\begingroup$ It is not important to remember what substitutions you make with definite integrals because when evaluated, they are simply numbers, and the last line says these numbers are equal. None of the written integrals are functions of $x$ or $t$. It would be important to remember what substitutions you made if the integrals were indefinite, because they evaluate to (the family of) primitive functions, which are functions of $x$ or $t$. $\endgroup$
    – Ennar
    Commented Mar 6 at 1:46

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