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Old 01-13-2005   #1 (permalink)
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fundamental theory of calculus

this is just a thread to see how many people we will have if we open a new calculus group.

has anyone ever tried to prove the integral of a function is the anti-derivative of it?
i did it using mean value theorm.... did you prove the theorem using someway else?


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Last edited by Tim_Lou; 01-13-2005 at 07:55 PM..
Old 01-14-2005   #2 (permalink)
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Re: fundamental theory of calculus

a very fundamental proof is something like this:

F: indefinite integral of f
G(x)=int(a->x){f(t)dt) : indefinite integral of f

F(x)=A(x)+C (always exist, C=constant)
calculate F(a)-F(b), and realize that G(a)=0.

Bo
Old 01-16-2005   #3 (permalink)
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Re: fundamental theory of calculus

What is the anti-derivative?


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Old 01-16-2005   #4 (permalink)
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Re: fundamental theory of calculus

the antiderivative is the same as the indefinete integral... just 2 words for the same thing.

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Old 01-16-2005   #5 (permalink)
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Re: fundamental theory of calculus

Quote:
Originally Posted by Bo
a very fundamental proof is something like this:

F: indefinite integral of f
G(x)=int(a->x){f(t)dt) : indefinite integral of f

F(x)=A(x)+C (always exist, C=constant)
calculate F(a)-F(b), and realize that G(a)=0.

Bo
Bo, how is it able to prove that the "area" under a curve is the antiderivative of that curve?
maybe i asked the wrong question?


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Old 01-16-2005   #6 (permalink)
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Re: fundamental theory of calculus

what i did is basically...
mean of f'(x) = (f(b) - f(a)) / (b-a)
"area" = mean height * base
so..... "area" = (f(b) - f(a)) / (b-a) * (b-a) which is integral of f'(x) from a to b

*"area" isnt really area, its actually the sum of it...
the prove is simple, and... not very supportive


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