110 1. Measure theory Proof. Apply Rolle’s theorem to the function x → F (x) − F (b) − F (a) b − a (x − a). Remark 1.6.6. As Rolle’s theorem is only applicable to real scalar-valued functions, the more general mean value theorem is also only applicable to such functions. Exercise 1.6.4 (Uniqueness of antiderivatives up to constants). Let [a, b] be a compact interval of positive length, and let F : [a, b] → R and G: [a, b] → R be differentiable functions. Show that F (x) = G (x) for every x ∈ [a, b] if and only if F (x) = G(x) + C for some constant C ∈ R and all x ∈ [a, b]. We can use the mean value theorem to deduce one of the fundamental theorems of calculus: Theorem 1.6.7 (Second fundamental theorem of calculus). Let F : [a, b] → R be a differentiable function, such that F is Riemann integrable. Then the Riemann integral b a F (x) dx of F is equal to F (b)−F (a). In particular, we have b a F (x) dx = F (b) − F (a) whenever F is continuously differentiable. Proof. Let ε 0. By the definition of Riemann integrability, there exists a finite partition a = t0 t1 . . . tk = b such that | k j=1 F (tj ∗)(tj − tj−1) − b a F (x)| ≤ ε for every choice of tj ∗ ∈ [tj−1,tj]. Fix this partition. From the mean value theorem, for each 1 ≤ j ≤ k one can find tj ∗ ∈ [tj−1,tj] such that F (tj ∗)(tj − tj−1) = F (tj) − F (tj−1) and thus by the telescoping series |(F (b) − F (a)) − b a F (x)| ≤ ε. Since ε 0 was arbitrary, the claim follows. Remark 1.6.8. Even though the mean value theorem only holds for real scalar functions, the fundamental theorem of calculus holds for complex or vector-valued functions, as one can simply apply that theorem to each component of that function separately. Of course, we also have the other half of the fundamental theorem of calculus:

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