How can you prove using the definition of anti-differentiation, that if two functions are equal, then their anti-derivatives must differ by a constant?
Also can this be used to prove the constant rule for integration??
How can you prove using the definition of anti-differentiation, that if two functions are equal, then their anti-derivatives must differ by a constant?
Also can this be used to prove the constant rule for integration??
First, if H'(x) = 0, then by definition of the derivative, it must be that H(x) = constant.
If you have F and G s.t. F'=G' = f, then (F - G)' = F' - G' = f - f = 0.
Since (F - G)'(x) = 0, then it must be that (F - G)(x) = constant, or F(x) - G(x) = constant, or F(x) = G(x) + constant.
That is if both F and G belong to the set of anti-derivatives of f, they may differ by a constant.
The statement is not true. Choose for example:
0,1)\cup (2,3)\rightarrow{\mathbb{R}}" alt="F,G0,1)\cup (2,3)\rightarrow{\mathbb{R}}" />
Then, , however and don't differ by a constant.
The statement is true if the domain of definition for and is an interval of .
Hint : Use the Mean Value Theorem.
Fernando Revilla
P.S Edited: Sorry, I didnīt see the previous posts. Anyway, perhaps the counter-example could be useful.
I suppose you meant the first part. The second one has already been proved in this thread.
For every we verify . That is, and are primitives of on . However:
i.e. is a non constant function.
Fernando Revilla