Critique the following "proofs" that an arbitrary constant must be $0$.

Question

1) $C=\int 0\, dx$=$\int 0\times0\, dx$=$0\times\int 0\,dx$=$0$. Basically, why can't you factor a zero out of the integral.

2) $0=(\int x\,dx)-(\int x\,dx)=\int(x-x)dx=\int 0\, dx=C$

Answer 1

The problem is that the notation $\int f$ does not mean any specific function. In fact, it means a family of functions. More precisely, the set of all the solutions to the differential equation $ F'=f. $ The "identity" $$ \int f=F+C\quad \text{for arbitrary constant }C, $$ is not really an identity at all. A more appropriate way to understand it might be $$ \int f=\{F+C\mid F'=f\text{ and }C\in\mathbb{R}\}. $$

In the first argument (1), $C$ is a fixed number and the expression $C=\int 0\ dx$ can not be thought as an "identity". It means the derivative of $f=C$ is $0$. Otherwise, by $$ 0=\int 0\ dx $$ and $$ 2016=\int 0\ dx $$ one could conclude that $0=2016$.