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I came across the following:

$$ F(x) = \int x^3 \cos(x)dx $$

where $F$ is understood to be a primitive of $x^3 \cos(x)$. I find this confusing, because of the "same" $x$ appearing on both sides of the equality. To me, $x$ is "integrated out" on the right side, and I prefer the notation:

$$ F(x) = \int_{0}^{x} u^3 \cos(u) du $$

or possibly:

$$ F = \int x^3 \cos(x) dx $$

without mentioning the variable for F.

Is the first notation widely used?

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    it's plain wrong but often seen.2012-06-05
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    I don't think it's wrong. The indefinite integral notation is just poor because in a definite integral the variable is integrated out.2012-06-05
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    @Neal Using $x$ on both sides of the equation, as variable on the lhs and bound to the integral on the rhs is the item the post was asking for, if I got it right. Indefinite integral I don't mind (a constant of integration would be a good thing (TM)).2012-06-05
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    Do you have problems with $(F(x))' = x^3 \cos(x)$. It's the same thing. Also, indefinite integral and a definite integral in interval [0, x] are somewhat different things.2012-06-05
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    @KarolisJuodelė No, it's not the same thing.2012-06-05
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    @Neal - yes, using $x$ on both sides is to me very confusing, although I can understand the shortcut. I like the equality to mean equality.2012-06-05
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    @Karolis - agree with Thomas, it's not the same. I believe the correct definition of the antiderivative is precisely the second form above $F(x) = \int_{0}^{x} x^3 cos(x) dx$, which is not apparent in the first notation.2012-06-05
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    @Thomas It looks to me like he is confused by the statement $F(x) = \int f(x)\ dx$, which is an indefinite integral with no bounds. The variable is just $x$, there's no integration to make it go away.2012-06-06

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The answer depends on what $\int f(x) \,dx $ means, about which there is no universal agreement. One interpretation is: the set of all functions whose derivative is $f$. If this definition is accepted, then $=$ should really be read as $\in$. This is the same convenient abuse of notation as in $\sqrt{x^2+1}=O(x)$. The other abuse is in writing $F(x)$ when you mean $F$, and this is also convenient at times.

So, this is how $F\in \int x^3\,dx$ becomes $F(x)=\int x^3\,dx$.

Notice that there is no integration involved in the above interpretation.

2nd interpretation: Someone may say that $\int f(x)\,dx$ is actually an integral, namely $\int_a^x f(t)\,dt$ with unspecified $a$. If you subscribe to this point of view, then $\sin x=\int \cos x\,dx$ is a true statement while $\sin x+5=\int \cos x\,dx$ is false.

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    It's a bit scary when you have to wonder what $\int f(x) dx$ actually means :-) Better that it means only one thing that everybody agrees on, IMHO.2012-06-05
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    @Frank which is what, exactly?2012-06-05
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    I believe your suggested first interpretation should be the only one accepted, or better, never use the shortcut, and be precise each time. I don't like maths to be open to "interpretation" = waste of time.2012-06-05
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    @Frank I'm all for being precise, which is why I stated both definitions that people use, either consciously or not. When someone speaks of "variable of integration" etc, their words suggest there is *integration* involved in $\int\cos x\,dx$ (as in the 2nd interpretation). Personally, I stick with the first version: **there is no integral whatsoever in $\int \cos x\,dx$.**2012-06-05
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    Beyond the superficial notation issue, is the set of functions whose derivative is $f(x)$ different from the set ${\int_{a}^{x} f(t) dt, a \in \mathbb{R}}$ (supposing we are just talking about integrating on (subsets of) $\mathbb{R}$)?2012-06-05
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    @Frank Yes, see my example with $\int \cos x\,dx$: it includes $\sin x+5$ under 1st interpretation, but not under second.2012-06-05
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    Got it. Thanks.2012-06-05
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It is indeed widely used, as is the uglier notation $ F(x) = \int^x f(w)dw $. $ F(x) = \int f(x)dx $ means the family of primitives or antiderivatives: all $F$ such that $F'(x)=f(x)$. Nothing to worry about.

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    It is readily understandable, IMHO, although I see it as a shortcut with some confusion on $x$ on both sides of $=$.2012-06-05