6.5 Average Value of a Function/1: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
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]
&= 38 \frac{1}{3}
&= \frac{32}{3}
\end{align}
\end{align}


</math>
</math>

Revision as of 02:33, 29 November 2022

Find the average value of the function on the given interval.