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

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&=\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]
&= \frac{32}{3}
&= \frac{32}{4}
\end{align}
\end{align}


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Revision as of 02:34, 29 November 2022

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle f(x) = 4x-x^2, (0, 4) }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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] &=\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{32}{4} \end{align} }