Tim_Lou Posted January 14, 2005 Report Posted January 14, 2005 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?
Bo Posted January 14, 2005 Report Posted January 14, 2005 a very fundamental proof is something like this: F: indefinite integral of fG(x)=int(a->x){f(t)dt) : indefinite integral of f F(x)=A(x)+C (always exist, C=constant)calculate F(a)-F(:hihi:, and realize that G(a)=0. Bo
Bo Posted January 16, 2005 Report Posted January 16, 2005 the antiderivative is the same as the indefinete integral... just 2 words for the same thing. Bo
Tim_Lou Posted January 17, 2005 Author Report Posted January 17, 2005 a very fundamental proof is something like this: F: indefinite integral of fG(x)=int(a->x){f(t)dt) : indefinite integral of f F(x)=A(x)+C (always exist, C=constant)calculate F(a)-F(:rant:, 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? :hihi:
Tim_Lou Posted January 17, 2005 Author Report Posted January 17, 2005 what i did is basically...mean of f'(x) = (f(:xx: - f(a)) / (b-a)"area" = mean height * baseso..... "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 :rant: :hihi:
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