Find the average value of the function on the given interval.
f ( x ) = 4 x − x 2 , ( 0 , 4 ) {\displaystyle f(x)=4x-x^{2},_{(}0,4)}
f a v g = 1 4 − 0 ∫ 0 4 ( 4 x − x 2 ) d x = 1 4 ∫ 0 4 ( 4 x − x 2 ) d x = 1 4 [ ( 4 x − x 3 3 ) ] 0 4 = 1 4 ( 8 − 8 3 ) − ( − 8 + 8 3 ) = 1 4 [ ( 32 4 ) ] = 8 3 {\displaystyle {\begin{aligned}f_{avg}&={\frac {1}{4-0}}\int _{0}^{4}(4x-x^{2})dx\\[2ex]&={\frac {1}{4}}\int _{0}^{4}(4x-x^{2})dx\\[2ex]&={\frac {1}{4}}\left[{\bigg (}4x-{\frac {x^{3}}{3}}{\bigg )}\right]_{0}^{4}\\[2ex]&={\frac {1}{4}}{\bigg (}8-{\frac {8}{3}}{\bigg )}-{\bigg (}-8+{\frac {8}{3}}{\bigg )}\\[2ex]&={\frac {1}{4}}\left[{\bigg (}{\frac {32}{4}}{\bigg )}\right]\\[2ex]&={\frac {8}{3}}\end{aligned}}}
![{\displaystyle {\begin{aligned}f_{avg}&={\frac {1}{4-0}}\int _{0}^{4}(4x-x^{2})dx\\[2ex]&={\frac {1}{4}}\int _{0}^{4}(4x-x^{2})dx\\[2ex]&={\frac {1}{4}}\left[{\bigg (}4x-{\frac {x^{3}}{3}}{\bigg )}\right]_{0}^{4}\\[2ex]&={\frac {1}{4}}{\bigg (}8-{\frac {8}{3}}{\bigg )}-{\bigg (}-8+{\frac {8}{3}}{\bigg )}\\[2ex]&={\frac {1}{4}}\left[{\bigg (}{\frac {32}{4}}{\bigg )}\right]\\[2ex]&={\frac {8}{3}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/fb7dcaac02e10017401ccc151aeda5d1b2d9e6f5)