how to prove
"If F is an antiderivative of f on an interval I , then the most general
antiderivative of f on I is
F(x)+C"
it means there does not exist any functions other than F(x)+C group which can be the answer to the problem , right?
i can think of proving it by using fundemental theorem of calculus part 2 like :
suppose we have another group of function called g(x)+c where g(x) is not equal to F(x)
and g'(x) = f(x) then we get to contradiction when we try to calculate the definite integral of f(x) from a to b
because g(b)-g(a) is not necessarily equal to F(b)-F(a)
so we have two values for one area !
but are there any other ways to prove it ?
I don't believe your proof is correct.
Here's a proof that will work: suppose that $F$ and $G$ are two antiderivatives of $f$. It follows that $$ G'(x) - F'(x) = f(x) - f(x) = 0 \implies\\ [G - F]'(x) = 0 $$ That is, the function $G(x) - F(x)$ has a derivative of $0$. Since the only functions whose derivative are zero (on an interval) are constant functions, we have $$ G(x) - F(x) = C \implies\\ G(x) = F(x) + C $$ That is, any antiderivative $G(x)$ of $f(x)$ has the form $F(x) + C$.