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

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<math>
<math>
f(x) = 4x-x^2, (0, 4)
f(x) = 4x-x^2, \quad (0, 4)
</math>
</math>


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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]
 
f_{avg} &= \frac{1}{4} \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} \left[\bigg( 4x-\frac{x^3}{3}\bigg)  \right]_0^4 \\[2ex]  

Latest revision as of 02:50, 29 November 2022

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