6.5 Average Value of a Function/1: Difference between revisions
No edit summary |
No edit summary |
||
| Line 16: | Line 16: | ||
&=\frac{1}{4} \left[\bigg( 4x-\frac{x^3}{3}\bigg) \right] = \frac{1}{4} \bigg(8-\frac{8}{3} \bigg) \\[2ex] | &=\frac{1}{4} \left[\bigg( 4x-\frac{x^3}{3}\bigg) \right] = \frac{1}{4} \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{ | &= \frac{8}{3} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 02:34, 29 November 2022
Find the average value of the function on the given interval.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle f(x) = 4x-x^2, (0, 4) }
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]={\frac {1}{4}}{\bigg (}8-{\frac {8}{3}}{\bigg )}\\[2ex]&={\frac {1}{4}}\left[{\bigg (}{\frac {32}{4}}{\bigg )}\right]\\[2ex]&={\frac {8}{3}}\end{aligned}}}