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I have the following statement:

Prove that if $f$ is continuous then $\int_a^b{f(-x)dx}= \int_{-b}^{-a}f(x)dx$

My attempt was:

Starting from LHS, Let $u=-x$ therefore i have $-\int_{-a}^{-b}f(u)du=\int_{-b}^{-a}f(u)du$. But from here i don't know how to back to $x$.

ESCM
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    $x$ and $u$ are "dummy" integration variables. Doesn't matter what you call the integration variable...you can call it $x$ or $u$ or $z$ or anything else you like. – lulu Sep 28 '20 at 15:02
  • well, but $x = -u$ therefore $-f(u)du$ will be equal to $f(x)dx$ only if $f$ is odd because i will have $f(u)du = -f(-x)dx = -(-f(x))dx = f(x)dx$ But $f$ is not necessary odd – ESCM Sep 28 '20 at 15:05
  • That was one substitution, but now forget that and make the substitution $u=x$. Suppose you were just handed the integral $\int_a^b f(x),dx$. Would you not agree that this is the same as $\int_a^b f(u),du$? The integral doesn't "remember" the prior substitutions. – lulu Sep 28 '20 at 15:07
  • @lulu Well, i understood that is a dummy variable like $j$ in $\sum_{j=0}^n j$ in this case the role of $j$ is be a counter, but what is the role of $x$ in this case of the definite integral? – ESCM Sep 29 '20 at 00:21
  • It's the same. It's just an index that tells you haw to run the integral. If you prefer you can write the integral as the limit of a sum...in that case the dummy variable will be exactly of the form you understand. – lulu Sep 29 '20 at 10:08
  • @lulu so, why don't write $\int_{a}^b f dx$? – ESCM Sep 29 '20 at 16:01
  • What's the difference? You still have the dummy variable $x$. – lulu Sep 29 '20 at 16:04
  • Well, is shorter. By the other hand, you know or have some links to investigate why and from where $dx$ appears? what is its equivalent in the limit notation of the definite integral? – ESCM Sep 29 '20 at 16:42
  • In the limit definition, whichever one you had in mind, you consider a number of values in the range and consider $f$ evaluated at each. $x$ then just represents the values you choose. – lulu Sep 29 '20 at 16:44
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    Should stress: I am not trying to defend the notation for the integral. There are lots of reasons to be unhappy with it (not least because "$dx$" is unreasonably mysterious). I'm just describing how it works. – lulu Sep 29 '20 at 16:45
  • Thanks for your answer, you have some text that speak about the $dx$ in definite integration? I couldn't find any associated question in MSE – ESCM Sep 29 '20 at 16:47
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    I don't, no. I would suggest not getting too hung up on the notation. We use a lot of that notation for historical reasons and would probably not choose it today. – lulu Sep 29 '20 at 16:59
  • Oh, thanks lulu. – ESCM Sep 29 '20 at 21:09

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You are handling definite integrals. Therefore, as lulu mentions in a comment, the dummy variable does not affect the value of the integral: $$ \int_a^bg(x)dx=\int_a^bg(t)dt=\int_a^bg(y)dy=\int_a^bg(z)dz=\cdots $$

\begin{align} \int_a^b f(-x)dx&=\int_{-a}^{-b}f(u)(-1)du\\ &=\int_{-b}^{-a}f(u)du\\ &=\int_{-b}^{-a}f(x)dx \end{align}

According to your comment above, you seem to have confusion with indefinite and definite integrals.

If you set $u=-x$, then formally, $$ f(-x)dx = -f(u)du\tag{1} $$ In this case, $u$ is a function of $x$, and (1）is an identity of differentials.

One the other hand, when you write $$ \int_a^b f(-x)dx=\int_a^b f(-u)du\tag{2} $$ both $u$ and $x$ are dummy variables for the definite integral. There is no relation between them.

  • Well, i understood that is a dummy variable like $j$ in $\sum_{j=0}^n j$ in this case the role of $j$ is be a counter, but what is the role of $x$ in this case of the definite integral? – ESCM Sep 29 '20 at 00:20
  • @EduardoSebastian: it is easier to see this by looking at examples than talking about it abstractly. Consider $\int_0^1x^2dx$ and $\int_0^1y^2dy$. In general, you can look at the fundamental theorem of calculus: $\int_a^bf(x)dx=F(b)-F(a)$. There is no variable on the RHS. –  Sep 29 '20 at 22:31