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] = \frac{1}{4} \bigg(8-\frac{8}{3} | &=\frac{1}{4} \left[\bigg( 4x-\frac{x^3}{3}\bigg) \right] = \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 02:37, 29 November 2022
Find the average value of the function on the given interval.