Why do we use dummy variables in integrals?

Bear with me.

Let's say we want to model a function that represents the number of boxes a factory outputs per hour. The input, of course would be number of hours, and the output would be number of boxes. Now, we need to name our function, because we don't want to keep referring to the function by its definition -- that would take up a lot of hour time. Let's call it Efficiency.

So, $\text{Efficiency}(\text{number of hours)} = \text{number of boxes}. \tag1$

We could do stuff with this function. We can differentiate it for optimization, or integrate it to find the average values, or anything in between.

Of course though, this is a time killer, since the function in $(1)$ is tedious to write out. We can introduce other variables, like $f$ for the name of our function, $t$ for our time (in hours), and $y$ for the output (in boxes).

Now, $(1)$ is equivalent to $$f(t) = y.$$

But look, we didn't really change anything. Our function still models the original problem in the same exact way. The variables are just abbreviations. We abbreviate them for convenience.

Now suppose the factory makes different types of boxes. We can call these new types of boxes $y$ also, but that would be awfully confusing. The same reasoning is behind the use of dummy variables in definite integrals.

When we introduce dummy variables into integrals, we're just doing it for convenience. The meaning behind the algebra does not change the original problem; it just makes it nicer to analyze.

I'll interpret the question as

Why are we not using the orignal one instead of a dummy variable?
Why use $\int_0^xf(x')dx'$ and not $\int_0^xf(x)dx$?

(If this is the case, please clarify you question with an marked(!) edit, others interpreted in differently.)

Because it has a different meaning and there is a problem with how to define it.
To show the difference define

$$g(x):= \int_0^xf(x,x')dx'$$ for some $f:\mathbb{R}^2\to \mathbb{R},(x,y)\mapsto f(x,y)$ continuously differentiable with respect to both arguments (being only continuous instead of differentiable would be enough).

Now compute

$$\frac{dg(x)}{dx}= \frac{d}{dx}\int_0^xf(x,x')dx' =f(x,x) +\int_0^x\frac{df(x,x')}{dx}dx' =f(x,x) + \int_0^x f_1(x,x')dx',$$ where $f_1$ means the first partial derivative of $f$.
The right part happens, because the inner part of the integral was dependend of the outher variable $x$.
So $\int_0^xf(x,x')dx'$ means something different from $\int_0^xf(x',x')dx'$

This is the reason why $$\int_0^xf(x,x)dx$$ is somewhat not well defined.

Because the definite integral depends only on the function and the limits of integration. So writing $\int_{a}^{b} f(x)dt$, $\int_{a}^{b} f(t)dt$, $\int_{a}^{b} f(s)ds$ makes no difference as long as $x$, $t$ and $s$ are "integrated out" in your final answer.

For more information look here and here (section 7.2).