Basic integral proof 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 $x$.

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.
