Let f be a continuous function on [a,b] and let F(x) and G(x) be two antiderivatives of f(x) such that F′(x)=f(x) and G′(x)=f(x) for all x in [a,b].
(a) Prove that F(x) and G(x) differ only by a constant; that is, G(x)=F(x)+k for some constant k.
(b) Using part (a), establish that ∫abf(x)dx=F(b)−F(a)=G(b)−G(a), demonstrating that the definite integral is independent of the choice of antiderivative and of the constant k.
(c) Verify the result in part (b) for the function f(x)=3x2−4x+1 on [0,2] by evaluating the definite integral once with the antiderivative F(x)=x3−2x2+x+7 and again with G(x)=x3−2x2+x−4.