# integrals differening by konstant

• Nov 2nd 2006, 08:01 AM
edgar davids
integrals differening by konstant
i think thsi is easy but cant think of the answer can some1 help me out
thanx
edgar

Suppose f is continuous on a domain D. Prove that any two primitives
of f (if they exist) differ by a constant.
• Nov 2nd 2006, 09:21 AM
ThePerfectHacker
Quote:

Originally Posted by edgar davids
i think thsi is easy but cant think of the answer can some1 help me out
thanx
edgar

Suppose f is continuous on a domain D. Prove that any two primitives
of f (if they exist) differ by a constant.

Primitive as in "anti-derivative"?
---
Theorem: If a function $\displaystyle f$ is differenciable with the property that $\displaystyle f'(x)=0$ for all points then, $\displaystyle f(x)=C$ for some real number $\displaystyle C$.

Proof: Let $\displaystyle a,b$ and be be points $\displaystyle a<b$. Then, the closed interval $\displaystyle [a,b]$ is continous (because differenciability implies continuity). And the open interval $\displaystyle (a,b)$ is differenciable because that is the property of the function. This satisfies Lagrange's Mean Value theorem. There exists a point $\displaystyle c\in (a,b)$ such as,
$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$
But, $\displaystyle f'(c)=0$
Thus,
$\displaystyle \frac{f(b)-f(a)}{b-a}=0$
Thus,
$\displaystyle f(b)-f(a)=0$ thus, $\displaystyle f(a)=f(b)$. Which means that the evaluation of the function is invariant, i.e. it is a constant function. Q.E.D.

Corrolarry
Given a function $\displaystyle f$ on some interval. If $\displaystyle F$ and $\displaystyle G$ are primitives, that is, $\displaystyle F'=G'=f$ for all points in the open interval. Then, $\displaystyle F'-G'=(F-G)'=0$, thus, $\displaystyle F-G$ is derivative zero throughtout the interval. By the theorem that means that, $\displaystyle F-G=C$ for some constanct function $\displaystyle C$. Thus, $\displaystyle F=G+C$. Q.E.D.