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.
It's by a simple fact that if $\displaystyle f:\mathbb{R}\to\mathbb{R}$ (or any connected subspace of $\displaystyle \mathbb{R}$ in fact) is such that $\displaystyle f'\equiv 0$ then $\displaystyle f$ is constant. To see this fact let $\displaystyle x,y\in\mathbb{R}$ be arbitrary. By the MVT there exists some $\displaystyle \xi\in(x,y)$ such that $\displaystyle f(x)-f(y)=(x-y)f'(\xi)=0$ and thus $\displaystyle f(x)=f(y)$. Since $\displaystyle x,y$ were arbitrary the ocnclusion follows.
The statement is not true. Choose for example:
$\displaystyle F,G0,1)\cup (2,3)\rightarrow{\mathbb{R}}$
$\displaystyle F(x)=\begin{Bmatrix} 0& \mbox { if }& x\in (0,1)\\1 & \mbox{if}& x\in (2,3)\end{matrix}\quad G(x)=\begin{Bmatrix} 1 & \mbox{ if }& x\in (0,1)\\0 & \mbox{if}& x\in (2,3)\end{matrix}$
Then, $\displaystyle F'=G'$ , however $\displaystyle F$ and $\displaystyle G$ don't differ by a constant.
The statement is true if the domain of definition for $\displaystyle F$ and $\displaystyle G$ is an interval of $\displaystyle \mathbb{R}$ .
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 $\displaystyle x\inS=(0,1)\cup (2,3)$ we verify $\displaystyle F'(x)=G'(x)=0$ . That is, $\displaystyle F$ and $\displaystyle G$ are primitives of $\displaystyle f=0$ on $\displaystyle S$. However:
$\displaystyle (F-G)(x)=\begin{Bmatrix} -1& \mbox { if }& x\in (0,1)\\\;\;1 & \mbox{if}& x\in (2,3)\end{matrix}$
i.e. $\displaystyle F-G$ is a non constant function.
Fernando Revilla