6.5 Average Value of a Function/1: Difference between revisions
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\begin{align} | \begin{align} | ||
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}[\frac{32}{4}] \\[2ex] | |||
&= 38 \frac{1}{3} | |||
\end{align} | |||
</math> | |||
Revision as of 23:12, 28 November 2022
Find the average value of the function on the given interval.
![{\displaystyle {\begin{aligned}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}}[{\frac {32}{4}}]\\[2ex]&=38{\frac {1}{3}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/c679c495f4dd44d501310bef09143a981048025b)