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

by RoRi

August 17, 2015

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.
Do part (a) with $f(x) = x^n$ for $n \in \mathbb{Z}_{>0}$ .

We want $f(c)$ to equal the average of $f$ on $[0,a]$ , so we set them equal and solve:
$\begin{align*} f(c) = A(f) && \implies && c^2 &= \frac{1}{a} \int_0^a x^2 \, dx \\ && \implies && c^2 &= \frac{a^2}{3} \\ && \implies && c &= \frac{a}{\sqrt{3}}. \end{align*}$
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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