6.5 Average Value of a Function/1: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| (9 intermediate revisions by the same user not shown) | |||
| Line 2: | Line 2: | ||
<math> | <math> | ||
f(x) = 4x-x^2, (0, 4) | f(x) = 4x-x^2, \quad (0, 4) | ||
</math> | </math> | ||
| Line 11: | Line 11: | ||
f_{avg} &= \frac{1}{4-0} \int_{0}^{4} (4x-x^2) dx \\[2ex] | 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} | |||
&= \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} | ||
Latest revision as of 02:50, 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]&={\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)