6.5 Average Value of a Function/1: Difference between revisions
Jump to navigation
Jump to search
No edit summary Tag: Manual revert |
No edit summary |
||
| Line 9: | Line 9: | ||
\begin{align} | \begin{align} | ||
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] | 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( 4x-\frac{x^3}{3}\bigg) \right] = \frac{1}{4} \bigg(8-\frac{8}{3} \bigg) \\[2ex] | ||
&= \frac{1}{4}[\frac{32}{4}] \\[2ex] | &= \frac{1}{4} \left[\bigg( \frac{32}{4}\bigg) \right] \\[2ex] | ||
&= 38 \frac{1}{3} | &= 38 \frac{1}{3} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 02:31, 29 November 2022
Find the average value of the function on the given interval.
![{\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]&=38{\frac {1}{3}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b8fd6730fb01e7a6bc1c595f4c32e136c94b192f)