Is this expression of an integral invalid or misleading?

[humy]

I want to write an integration as;

∫[0, x] f(x) dx

where f(x) is some or any function of x the formula of which isn't relevant to my question.
But I notice I want to treat the "x" in the above "∫[0, x] " part of that integral expression as a fixed constant, not a variable, while treat the "x" in the above "f(x) dx" part of that integral expression as a variable, not a fixed constant. The x in the above "∫[0, x] " part and the "x" in the above "f(x) dx" are in all other respects the same kind of numerical variable because its a variable of the same kind of 'thing', so they only 'differ' in the narrow sense that the x in the above "∫[0, x] " part is given some specified constant numerical value to evaluate that integral while the "x" in the above "f(x) dx" isn't.

But can you argue that STILL means that we are talking about two DIFFERENT x things and thus they should NOT be given the SAME letter or symbol (of x in this case) but should be given DIFFERENT letters or symbols, such as in " ∫[0, x] f(X) dX " where one is capital X while the other is lower case x to make that valid or to prevent the two being confused with each other?

In other words, is writing it down as " ∫[0, x] f(x) dx " in some way invalid and/or misleading because I should instead write it down something
like " ∫[0, x] f(X) dX " or " ∫[0, x] f(y) dy " etc to make it clear what is being treated as a constant and what is being treated as a variable?

 

[HallsofIvy]

The form \(\displaystyle \int_0^x f(x)dx\) is not unheard of and I think people would know what you mean but a more "formal" and, in my opinion, better form would be \(\displaystyle \int_0^x f(y)dy\) (or \(\displaystyle \int_0^x f(t)dt\) or \(\displaystyle \int_0^x f(z)dz\), etc.).

The point is to distinguish the "dummy variable" within the integral from the "real" variable that would be part of the actual function. If, for example, \(\displaystyle f(x)= x^2\), \(\displaystyle f(y)= y^2\), \(\displaystyle f(t)= t^2\), and \(\displaystyle f(z)= z^2\), then we get the same result for all those above:
\(\displaystyle \int_0^x x^2dx= \frac{1}{3}x^3\), \(\displaystyle \int_0^x y^2 dy= \frac{1}{3}x^3\), \(\displaystyle \int_0^x t^2 dt= \frac{1}{3}x^3\), and \(\displaystyle \int_0^x z^2 dz= \frac{1}{3}x^3\).

The point does become important if the integrand itself also includes x: \(\displaystyle \int_0^x t^2x dt= \left[\frac{1}{3}t^3x\right]_0^x= \frac{1}{3}x^3(x)= \frac{1}{3}x^4\).

(Notice that there is no "+ C" on any of these. That only applies when you are finding the "indefinite integral" with no limits of integration.)

 

[humy]

The form \(\displaystyle \int_0^x f(x)dx\) is not unheard of and I think people would know what you mean but a more "formal" and, in my opinion, better form would be \(\displaystyle \int_0^x f(y)dy\) (or \(\displaystyle \int_0^x f(t)dt\) or \(\displaystyle \int_0^x f(z)dz\), etc.).

The point is to distinguish the "dummy variable" within the integral from the "real" variable that would be part of the actual function. If, for example, \(\displaystyle f(x)= x^2\), \(\displaystyle f(y)= y^2\), \(\displaystyle f(t)= t^2\), and \(\displaystyle f(z)= z^2\), then we get the same result for all those above:
\(\displaystyle \int_0^x x^2dx= \frac{1}{3}x^3\), \(\displaystyle \int_0^x y^2 dy= \frac{1}{3}x^3\), \(\displaystyle \int_0^x t^2 dt= \frac{1}{3}x^3\), and \(\displaystyle \int_0^x z^2 dz= \frac{1}{3}x^3\).

The point does become important if the integrand itself also includes x: \(\displaystyle \int_0^x t^2x dt= \left[\frac{1}{3}t^2x\right]_0^x= \frac{1}{3}x^2(x)= \frac{1}{3}x^3\).

(Notice that there is no "+ C" on any of these. That only applies when you are finding the "indefinite integral" with no limits of integration.)

Thanks for that. That's all I wanted to know