Answer
Average value = $0$
x = 0, x = $\displaystyle \frac{3}{4}$
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 = $\displaystyle \frac{1}{1-0}\int_{0}^{1}(4x^{3}-3x^{2})dx$
$=[4\displaystyle \cdot\frac{x^{4}}{4}-3\cdot\frac{x^{3}}{3}]_{0}^{1}$
$=[x^{4}-x^{3}]_{0}^{1}$
$=1-1-(0-0)$
$=0$
To find x where f(x)=average value, solve (on [0,1] )
$4x^{3}-3x^{2} = 0$
$x^{2}(4x-3)=0$
x = 0 or x =$\displaystyle \frac{3}{4}$
