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In old mathematics books, I see a lot of notations like $\int_{0}^{x} f(x) dx$. For example, Courant-Hilbert: Methods of mathematical physics. However, when I wrote it in this site, it was sometimes edited like $\int_{0}^{x} f(t) dt$.

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It shouldn’t just be frowned on: $\int_0^x f(x)dx$ is simply wrong. – Brian M. Scott Sep 2 '12 at 21:58
Although the notation $\int_0^x f(x)\,dx$ is fine from a logical point of view, I have always avoided it in teaching. Just as I would never write $\forall x(F(x)\land\forall x G(x))$. – André Nicolas Sep 2 '12 at 21:58
In most cases it's kind of a nit-picky thing. "Everyone knows" what is meant by $\int_0^x f(x)\,dx$, but there is something unsatisfactory about $x$ pulling double duty. If a student uses the same variable as a dummy variable and as a limit, I tend to overlook it. But in a textbook, it seems kind of shoddy. Why not use a different symbol for total clarity? It's not like you're going to pay extra for using a different letter. – Bill Cook Sep 2 '12 at 21:59
Sometimes I wonder why we don't just write it as $\int_0^x f$. – Rahul Sep 2 '12 at 22:12
Because it violates the very strong notational convention against using one variable name to refer to two different variables in the same expression. It also subverts one of the main functions of mathematical notation, which is to facilitate understanding. – Brian M. Scott Sep 2 '12 at 22:36

Over the summer I came up with an exercise for the kind of people who like to write $\int_0^x f(x) \, dx$: evaluate

$$\int_1^x \int_x^{x^2} \int_{x^2}^{x^3} x^4 x^5 x^6 \, dx \, dx \, dx.$$

I hope that my point is clear.

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Ouch! :-) I love it. – Brian M. Scott Sep 3 '12 at 9:46
Once upon a time I was a student in a course in which the professor wrote a rather complicated proof on the board, and in one section of the proof used the letter $j$ to refer to one thing, and in another to another. When I mentioned this, she said "Pretend one of them is a Greek $j$." – Michael Hardy Sep 3 '12 at 12:58
Here is a case where multiplication is not commutative: compare this with $\int_1^x \int_x^{x^2} \int_{x^2}^{x^3} x^4 \, dx\, x^5 \, dx\, x^6 \, dx.$ – Ross Millikan Sep 3 '12 at 21:04
This doesn't really generalize the notation $\int_{0}^xf(x)dx$. A better generalization would be $$\int_1^x \int_x^{x^2} \int_{x^2}^{x^3} x^4 y^5 z^6 \, dx \, dy \, dz$$ or some such. As far as I can tell, this isn't ambiguous. – goblin Oct 11 '13 at 17:23

Note that there are two different $x$'s in $\int_{0}^{x} f(x) dx$, which is made explicit when one is changed to $t$. One is the upper limit of integration, which is still free, and the other is the dummy variable bound inside the integral. On careful reading one can tell them apart, but it is easier on the reader and less mistake-prone to distinguish them.

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This is wrong because it would not allow (without confusion) something like this:

$$\int_0^x f(t, x)\ dt$$

The variable of integration (or summation when doing sums) should always differ from all other variables because, to use expressions I recall (possibly incorrectly) from my youth, the other variables are "bound" (the $x$ above) and the variable of integration is "free", so that the expression is unchanged if the variable is replaced by another.

For example, what would you make of this:

$$\int_0^x \int_0^x f(x,x)\ dx\ dx$$

instead of this:

$$\int_0^u \int_0^y f(x,y)\ dx\ dy$$

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If you write $\int_0^x f(x)\,dx$, you have two different $x$s. One has scope inside of the integral, the other outside. The term "scope" is somewhat strange in the mathematical world, but it means "where the variable has meaning."

In this case it is far better to write $\int_0^x f(t)\,dt$; the variable $t$ is a "loop variable" or place-holder.

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$\int_0^x f(t) \,dt$

is an expression that, by its limit definition, is basically the sum of an infinite number of areas of rectangular pieces of infinitesimally small width and height $f(t)$ for each value of $t$ between $0$ and $x$. So, if we wrote

$\int_0^x f(x) \,dx$

instead, it would mean to add up the areas of these rectangles as the value of $x$ ranges from $0$ to $x$. Hopefully, it is clear that this makes no sense. $x$ can not simultaneously vary and stay constant.

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