Mean Value Theorem: Suppose 1.) f continuous on [a, b] 2.) f differentiable on (a, b) Then there exists c ∈ (a, b) such that f′(c) = \frac{f(b)−f(a)}{b−a} Ex 3: If f′(x) = 0 for all x ∈ (a, b), then f(x) = c for some constant c. Proof: Take x_{0} ∈ (a, b) Show f(x) = f(x_{0}) for all x ∈ (a, b) [i.e., c = f(x_{0})]. Without loss of generality, assume x > x_{0}. (proof is similar when x < x_{0}). f is continuous on [x_{0}, x]. f is differentiable on (x_{0}, x). By MVT, there exists c ∈ (x_{0}, x) such that \frac{f(x)−f(x_{0})}{x−x_{0}} = f′(c) = 0. Thus f(x) − f(x_{0}) = 0. Hence f(x) = f(x_{0}). Ex 4: If f′(x) = g′(x) for all x ∈ (a, b), then f(x) = g(x) + c for some constant c. Proof: f′(x) = g′(x) implies (f − g)′(x) = f′(x) − g′(x) = 0. Thus (f − g)(x) = f(x) − g(x) = c for some constant c. Thus f(x) = g(x) + c.

The Fundamental Theorem of Calculus: Suppose f continuous on [a, b].

1.) If g(x) = ∫_{a}^{x} f(t)dt, then g′(x) = f(x).

2.) ∫_{a}^{b} f(t)dt = F (b) − F (a) where F is any antiderivative of f, that is F′ = f.