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Find the average value of function f(x)= x^2-19 from x=0 to x=3.
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\[\frac{1}{3}\int_0^3(x^2-9)dx\]
rather \[\frac{1}{3}\int_0^3(x^2-19)dx\]
should be ok since \[\frac{1}{3}\int_0^3x^2dx-\frac{1}{3}\int_0^319dx\] is not too bad second integral is \(-19\) by your eyeballs first one is \(\frac{1}{3}\times \frac{3^3}{3}=3^3=27\)
wow that was wrong!!
\[\frac{3^3}{3^2}=3\] or the first one
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for a final answer of \(3-19=-16\) i think.
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