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Find a value at which functions are equal to their mean on an interval

  1. Let f(x) = x^2 on the interval [0,a]. Find c \in \mathbb{R} with 0 < c < a such that f(c) equals the average of f on the interval.
  2. Do part (a) with f(x) = x^n for n \in \mathbb{Z}_{>0}.

  1. We want f(c) to equal the average of f on [0,a], so we set them equal and solve:

        \begin{align*}  f(c) = A(f) && \implies && c^2 &= \frac{1}{a} \int_0^a x^2 \, dx \\  && \implies && c^2 &= \frac{a^2}{3} \\  && \implies && c &= \frac{a}{\sqrt{3}}.  \end{align*}

  2. Again, we set f(c) equal to the average of f on [0,a]:

        \begin{align*}  f(c) = A(f) && \implies && c^n &= \frac{1}{a} \int_0^a x^n \, dx \\  && \implies && c^n &= \frac{a^n}{n+1} \\  && \implies && c &= \frac{a}{(n+1)^{1/n}}. \end{align*}

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