If I define $F(x)=\int f(x)\:dx$, is right to say that $F(u)=\int f(u)\:du$

Question

I thought that the $dx$ sign can't be changed to $du$. For example, instead of $x=u$, If I had $x=2$, then I would have to evaluate $F(2)=\int f(2) \: d2 $, but this can't be right. It makes sense to define a function $F(x)=\int f(x)\:dx$ then say that $F(u)=\int f(u)\:du$?

Answer 1

Well, let us make some clarifications over the symbol $\int f(x) dx$. In general, whith this, we are used to note a whole family of functions, not just one function - you can see the specific construction of this family in the first part of this answer. So, writing: $$F(x)=\int f(x)dx$$ has no meaning, since the left hand side of this equation is a function and the right one is a set of functions. Even writing: $$F(t)=\int f(x)dx$$ has no meaning, for the same reasons.

So, keeping in mind that the indefinite integral is a symbol for a set of functions - with a common property - what one could use is the symbol of the definite integral, as mentioned in the comments section by @Mark Viola. So, for instance, one can write: $$F(x)=\int_a^xf(t)dt$$ which is, if $f(x)\geq0$ over its doamin, the area under $f$ from $a$ to $x$ - or the opposite of this area, if $x<a$. Note that "inside" the integral we use a different variable than that we use at its limits. So, one can also write: $$F(x)=\int_a^xf(u)du$$ Using the same variable "inside" and "outside" the integral would cause a large confusion as of with respect to what are we integrating.

You can also make both limits of integration be variable: $$F(x)=\int_x^{x+1}f(t)dt$$ or, more generally, if $g,h$ are two "nice" functions - for instance, continuous - we can consider: $$F(x)=\int_{g(x)}^{h(x)}f(t)dt$$

Note, however, that the following has a meaning: $$F(x)=\int_a^xf(x)dt$$ To interpret this, one can think as follows: $dt$ tells us with respect to which variable are we integrating. So, in our case, this variable is $t$ and, hence, everything that does not contain $t$ is considered to be a fixed number. So, for this case $f(x)$ is a plain number, and, we have: $$F(x)=\int_a^xf(x)dt=f(x)\int_a^x1dt=f(x)(x-a)$$ So, this was a more "tricky" way to write the function $f(x)(x-a)$.

Finally, we can have even more complex cases, as the following: $$F(x)=\int_a^xg(x-t)f(t)dt$$ but the discussion has already gone too far...