Interpreting the integral identity $\int c f(x) \,dx = c \int f(x) \,dx$ for $c = 0$

Question

Let me begin my question with these 2 facts:

1. The function $F(x)=3$ is an antiderivative of the function $f(x)=0$ since $F'(x)=f(x)$
2. Constant Multiple Rule: $\int cf(x)dx=c\int f(x)dx, \forall c \in R$.

So with the function $f(x)=0 $, we have: $\int f(x)dx=\int 0dx=0 \int dx=0 $ With the constant multiple rule I would never get $F(x)=3$ as an antiderivative of $f(x). What's wrong here ? I think I am wrong but I don't know where.

Answer 1

It's simply a matter of interpretation; if one allows $c = 0$ in the "Constant Multiple Rule" then for that value the r.h.s. is always zero and so one never gets the general antiderivative. We can rectify this by allowing $c = 0$ but defining $$0 \cdot \int f(x) \,dx := C,$$ where $C$ is an arbitrary constant. This isn't so unreasonable, as even to write down the "Constant Multiple Rule" we must define what it means to multiply a constant $c$ by a family of functions.

NB this rule works perfectly well without modification for definite integrals.

Answer 2

Note that $\int f(x)dx$ is not unique since you can always add a constant $C$ to it. To get back $3$ for $F(x)$ as you wish, you must have one more condition like $F(0) = 3$.