Uniqueness of an Anti Derivative up to a Constant.  TOPIC_SOLVED

Limits, differentiation, related rates, integration, trig integrals, etc.

Uniqueness of an Anti Derivative up to a Constant.

Postby sepoto on Mon Dec 16, 2013 6:49 am

http://sdrv.ms/1dgPhG8

If F'(x) = f(x), and G'(x) = f(x), then G(x)=F(x)+c for some constant c

Proof:
(G-F)'=f-f=0

Recall that we proved as a corollary of the Mean Value Theorem that if a function has a derivative zero the it is constant. Hence G(x)-F(x) = c ( for some constant c). That is, G(x)=F(x)+c.


I find that I don't really understand yet what is meant by the above theorem. I think perhaps I might understand it better if I could see an example.

Thank You...
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Re: Uniqueness of an Anti Derivative up to a Constant.  TOPIC_SOLVED

Postby buddy on Mon Dec 16, 2013 1:05 pm

sepoto wrote:If F'(x) = f(x), and G'(x) = f(x), then G(x)=F(x)+c for some constant c

Proof: (G-F)'=f-f=0

Recall that we proved as a corollary of the Mean Value Theorem that if a function has a derivative zero the it is constant. Hence G(x)-F(x) = c ( for some constant c). That is, G(x)=F(x)+c.

It's just saying that two functions are the same, other than for the constant terms, if they differentiate to the same thing. It's like f(x) = x^2 + 3 and g(x) = x^2 + 4. They both differentiate to 2x. They're only different in the number part: the 3 and the 4.
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