6.5 Average Value of a Function/1: Difference between revisions

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f_{avg} &= \frac{1}{4} \int_{0}^{4} (4x-x^2) dx  \\[2ex]
f_{avg} &= \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  = \frac{1}{4} \bigg(8-\frac{8}{3}\bigg)-\bigg(-8+\frac{8}{3}\bigg) \\[2ex]
&= \frac{1}{4} \left[\bigg( 4x-\frac{x^3}{3}\bigg)  \right]_0^4   
&= \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{1}{4} \left[\bigg( \frac{32}{4}\bigg)  \right]  \\[2ex]
&= \frac{8}{3}
&= \frac{8}{3}

Revision as of 02:46, 29 November 2022

Find the average value of the function on the given interval.


Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}f_{avg}&={\frac {1}{4-0}}\int _{0}^{4}(4x-x^{2})dx\\[2ex]f_{avg}&={\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}&={\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}}}