Answer
Average value = $\displaystyle \frac{1}{4}$
$ x=2^{-2/3}\approx$0.62996
Work Step by Step
If $f$ is integrable on the closed interval $[a, b]$, then the average value of $f$ on the interval is
$\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx$.
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Average value = $\displaystyle \frac{1}{1-0}\int_{0}^{1}x^{3}dx$
$=[\displaystyle \frac{x^{4}}{4}]_{0}^{1}$
$=\displaystyle \frac{1}{4}$
To find x where f(x)=average value,
solve $x^{3}=\displaystyle \frac{1}{4}$ for x (on [$0,1$])
$ x=(\displaystyle \frac{1}{4})^{1/3}=2^{-2/3}\approx$0.62996
