Tim_Lou Posted January 14, 2005 Report Share 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? Link to comment Share on other sites More sharing options...
Bo Posted January 14, 2005 Report Share 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 Link to comment Share on other sites More sharing options...
sanctus Posted January 16, 2005 Report Share Posted January 16, 2005 What is the anti-derivative? Link to comment Share on other sites More sharing options...
Bo Posted January 16, 2005 Report Share Posted January 16, 2005 the antiderivative is the same as the indefinete integral... just 2 words for the same thing. Bo Link to comment Share on other sites More sharing options...
Tim_Lou Posted January 17, 2005 Author Report Share 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: Link to comment Share on other sites More sharing options...
Tim_Lou Posted January 17, 2005 Author Report Share 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: Link to comment Share on other sites More sharing options...
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