# Limit of integration can't be the same as variable of integration?

I am told that an expression like $$\int_a^x f(x)dx$$ is not well formed, i.e. it should be $$\int_a^xf(t)dt$$ or similar.

Why is it that the limits of integration can't depend on the variable of integration?

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It's a bit like writing $\sum_{i = 1}^i f(i)$. What is it supposed to mean? – Dylan Moreland Feb 14 '12 at 0:17
There is no real reason why it shouldn't (in the end the integration variable is a dummy variable) -- however, choosing the integration variable to be the same as one of the limits opens many doors for error... – Fabian Feb 14 '12 at 0:18
Try $\frac{d}{dx}$ on the first one. No confusion? – Raymond Manzoni Feb 14 '12 at 0:19
You can do as you have in the first integral, but you shouldn't. The x's play entirely different roles when you use x to denote both a limit and a dummy variable – The Chaz 2.0 Feb 14 '12 at 0:20
Poor style, but in principle OK. – André Nicolas Feb 14 '12 at 0:29

It confuses a free variable and a bound variable. In effect you are saying "let $x$ range from $a$ to $x$ when taking the integral of $f(x)$".
It is also ambiguous. There is a risk some people might expect $\int_a^x f(x)dx = (x-a)f(x)$ in the same way as $\int_a^x f(x)dt = (x-a)f(x)$.
It is easier to show the problem as a sum. The sum of the first $n$ positive integers can be written $\sum_{i=1}^{i=n} i = \frac{n(n+1)}{2}$ but if you wrote it as $\sum_1^n n$, some people might expect the answer to be $n^2$. Meanwhile the following looks very strange $$1+2+3+\cdots+n+\cdots+(n-1)+n$$