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Mean Value Theorem for integrals

  1. Apr 5, 2015 #1
    1. The problem statement, all variables and given/known data

    Prove the Mean Value Theorem for integrals by applying the Mean Value Theorem for derivatives to the function

    [tex]
    F(x) = \int_a^x \, f(t) \, dt
    [/tex]
    2. Relevant equations

    Mean Value Theorem for integrals: If f is continuous on [a, b], then there exists a number c in [a, b] such that

    [tex]
    \int_a^b \, f(x) \, dx = f(c) (b - a) \, .
    [/tex]

    Mean Value Theorem for derivatives, Fundamental Theorem of Calculus.


    3. The attempt at a solution

    Since the function f is continuous on [a, b], F(x) is continuous on [a, b] and differentiable on (a, b) by the FTC. The MVT for derivatives says

    [tex]
    F(b) - F(a) = F^\prime (c) (b - a) \quad \text{for} \, c \in (a, b) \, .
    [/tex]

    By FTC, F'(c) = f(c). Then f(c) (b - a) = F(b) - F(a). So

    [tex]
    f(c) = \frac{F(b) - F(a)}{b - a} = \frac{1}{b - a} \int_a^b \, f(x) \, dx
    [/tex]

    again by FTC.

    What I don't understand is why the book says c can be equal to a or b...F is differentiable on the open interval (a, b), not [a, b]. But the book says the MVT for integrals permits c to lie anywhere in the closed interval [a, b].
     
  2. jcsd
  3. Apr 5, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You derived a slightly stronger result - there has to be a "c" in (a,b). That implies the weaker result of the existence of a "c" in [a,b].
     
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