Definite Integral substitution rule

[white spade]

May someone please help with the last line as i'm unconvinced with their methods. If we set u=-x why do we then replace -u with -x?

[lyric verge]

We decompose our integral in two integrals
$$\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_{0}^a f(x) dx$$

we do the change the change u = -x
in the first integral
so you change u to -x and dx to -du
and lower bound -a to a
and upper bound 0 to 0

$$\int_{-a}^0 f(x) dx = \int_a^0 f(-u) (-du)$$

And if you exchange the bounds you have a - in front of the integral

so you have
$$\int_{-a}^a f(x) dx = - \int_{0}^a f(-u) (-du) + \int_{0}^a f(x) dx$$

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You can immediately cancels the two minus signs

$$\int_{-a}^a f(x) dx = \int_{0}^a f(-u) du + \int_{0}^a f(x) dx$$

And when you have any integral, the integration variable can have any name you want. The integration variable is called a dummy variable
That is to say that
$$ \int g(u) du = \int g(x) dx = \int g(s) ds$$

So you just $x$ name instead of $u$

$$\int_{-a}^a f(x) dx = \int_{0}^a f(-x) dx + \int_{0}^a f(x) dx$$

[proud flint]

un_decorateur

[white spade]

but the end feels like we've changed it for our own convenience, is it really mathematical?

also @lyric verge how do you always respond to my posts?

[lyric verge]

i didn't even notice i respond you again 🙂
i don't look at who i respond

the last argument i have the wikipedia page for you : https://en.wikipedia.org/wiki/Free_variables_and_bound_variables

[white spade]

i appreciate your help, you've saved my behind many times. Thank you!

[white spade]

.solved