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Take an indefinite integral on $f(x)$ is $x$ bound in this expression, as on the other side we get an expression that depends on $x$ for example $$\int f(x) dx = e^x+c$$

We get an expression that depends on $x$ but subsituting for $x$ in left -hand expression is entirely meaningless.

user37577
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  • Substituting for $x$ on the right-hand expression is also pretty much meaningless. The way I view an indefinite integral $\int f(x)dx$ is the set of functions $F$ such that $\frac{d}{dx}F(x)=f(x)$. – Varun Vejalla Aug 18 '22 at 17:00
  • As I mentioned in your other question (though that's about derivatives) it's again a matter of quickly getting the point across, not formal correctness. See for example the first half of my answer here for an alternative way of writing it (obviously clunky and inconvenient for practical computations) – peek-a-boo Aug 18 '22 at 17:02
  • @peek-a-boo so we don't call it 'bound' even though to substitute into it is meaningless, it's sort of just a clunky expression/name for a value that is the anti-derivative of the function in the integrand, but we can't substitute into it? On the last question, it's not bound in the differential operator either? – user37577 Aug 19 '22 at 07:55
  • No. the introduction of the letter $x$ everywhere is just a historic relic. If you want, you can define $\int f$ to denote the set (equivalence class) of all differentiable functions whose derivative is $f$, but this is highly non-standard. – peek-a-boo Aug 19 '22 at 08:19
  • Ok, but I see, but however we still have it equal to a number, also we have it's relation to the idea of a reimann sum, so can I view it as a way or referring to the general form expression for a function $g(x)$ where f$(x)=g'(x)$ – user37577 Aug 19 '22 at 14:36

1 Answers1

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Note that, given the anti-derivative $g(x)$ of $f(x)$, an alternative expression reads

$$\int_b^x f(t) dt =g(x)+C$$ where $C=-g(b)$. So, compared to

$$\int f(x) dx = g(x)+C$$ $x$ may be viewed as a variable (indefinite) bound of the integral.

Quanto
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    The OP is talking about a variable to be bounded (as opposed to a free variable), not about a numeric bound of the integral. – jjagmath Aug 18 '22 at 20:36