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Prove some more properties of the averages of functions on an interval

Define A_a^b (f) to be the average of a function f on the interval [a,b],

    \[ A_a^b (f) := \frac{1}{b-a} \int_a^b f(x) \, dx. \]

  1. If c \in \mathbb{R} with a < c < b, prove there exists t \in \mathbb{R} with 0 < t < 1 such that

        \[ A_a^b (f) = t A_a^c (f) + (1-t)A_c^b (f). \]

  2. Prove part (a) holds for weighted averages of functions where for a nonnegative weight function w we define the weighted average of f on [a,b] by

        \[ \frac{\int_a^b w(x) f(x) \, dx}{\int_a^b w(x) \, dx}. \]


  1. Proof. Let

        \[ t = \frac{c-a}{b-a}. \]

    Then, since a < c < b, we have 0 < t < 1. Furthermore,

        \[ 1-t = 1 - \frac{a-c}{a - b} = \frac{a-b-a+c}{a-b} = \frac{c-b}{a-b}. \]

    So,

        \begin{align*}  tA_a^c (f) + (1-t)A_c^b (f) &= \frac{c-a}{b-a} \cdot \frac{1}{c-a} \int_a^c f(x) \, dx + \frac{c-b}{a-b} \cdot \frac{1}{b-c} \int_c^b f(x) \, dx \\  &= \frac{1}{b-a} \int_a^c f(x) \, dx - \frac{1}{a-b} \int_c^b f(x) \, dx \\  &= \frac{1}{b-a} \left( \int_a^c f(x) \, dx + \int_c^b f(x) \, dx \right)\\  &= \frac{1}{b-a} \int_a^b f(x) \, dx \\  &= A_a^b (f). \qquad \blacksquare \end{align*}

  2. Proof. Let

        \[ t = \frac{\int_a^c w(x) \, dx }{\int_a^b w(x) \, dx}. \]

    Then,

        \[ 1-t = \frac{\int_a^b w(x) \, dx - \int_a^c w(x) \, dx}{\int_a^b w(x) \, dx} = \frac{\int_c^b w(x) \, dx}{\int_a^b w(x) \, dx}. \]

    Thus, 0 < t < 1 (since \int_c^b w(x) \, dx < \int_a^b w(x) \, dx since a < c < b and since w is nonnegative both of these are nonnegative).
    Then,

        \begin{align*}  t \cdot A_a^c (f) + (1-t) \cdot A_c^b (f) &= \frac{\int_a^c w(x) \, dx}{\int_a^b w(x) \, dx} \cdot \frac{\int_a^c w(x) f(x) \, dx}{\int_a^c w(x) \, dx} + \frac{\int_c^b w(x) \, dx}{\int_a^b w(x) \, dx} \cdot \frac{\int_c^b w(x) f(x) \, dx}{\int_c^b w(x) \, dx} \\  &= \frac{\int_a^c w(x) f(x) \, dx + \int_c^b w(x) f(x) \, dx}{\int_a^b w(x) \, dx} \\  &= \frac{\int_a^b w(x) f(x) \, dx}{\int_a^b w(x) \, dx} \\  &= A_a^b (f). \qquad \blacksquare \end{align*}

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