How and why do dummy variables work?

Question

When proving that $$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$$ the substitution $u = a+b-x$ is used, giving us the new integral $$\int_{a}^{b} f(u) du$$ After this, people will just swap $u$ with $x$ in order to satisfy the equality mentioned at the top of the post. Why is this allowed? Since $u$ is literally defined as being a different value than $x$, how can we make the two equal? When does this fail?

Answer 1

When substituting $u=a+b-x$, $du=-dx$, $x=a\rightarrow u=b,x=b\rightarrow u=a$ $$\int_a^bf\left(x\right)dx=\int_a^b -f\left(u\right)du$$ Because of $\int_a^b=-\int_b^a$, $$\int_a^b -f\left(u\right)du=\int_a^bf\left(u\right)du$$

Answer 2

The value of the integral $\int_a^b f(x)dx$ depends only on the function $f$ and on the constants $a,b$; if you want, it's the area under $f$ between $a$ and $b$. It doesn't depend on any specific value of $x$; in fact we're using $x$ as a placeholder variable taking (in a sense) all the values between $a$ and $b$.

When you make the substitution for $u$, you make it at each specific value of $x$, and at a specific value, $u$ and $x$ are different, as you noted. But you end up with the integral $\int_a^b f(u)du$, which again just means we're taking the area under $f$ between $a$ and $b$. We got there using a different variable, but the value of the integral itself doesn't care which variable is used. That's what a dummy variable is.