Answer
$$\overline f = \frac{1}{6}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x\left( {1 - x} \right){\text{ on the interval }}\left[ {0,1} \right] \cr
& {\text{Find the average value using }}\overline f = \frac{1}{{b - a}}\int_a^b {f\left( x \right)} dx \cr
& \overline f = \frac{1}{{1 - 0}}\int_0^1 {x\left( {1 - x} \right)} dx \cr
& \overline f = \int_0^1 {\left( {x - {x^2}} \right)} dx \cr
& {\text{Integrate}} \cr
& \overline f = \left[ {\frac{1}{2}{x^2} - \frac{1}{3}{x^3}} \right]_0^1 \cr
& \overline f = \left[ {\frac{1}{2}{{\left( 1 \right)}^2} - \frac{1}{3}{{\left( 1 \right)}^3}} \right] - \left[ {\frac{1}{2}{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr
& \overline f = \frac{1}{2} - \frac{1}{3} \cr
& \overline f = \frac{1}{6} \cr
& \cr
& {\text{Graph}} \cr} $$
