anonymous
  • anonymous
What is the average value of f(x)=sqrt 64-x^2 over the interval 0<=x<=8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
how should i approach this problem?
anonymous
  • anonymous
the way you do this is you plug in 0 into the function then you plug in 8 and you take the average so we have sqrt(64-0)=8 and sqrt(64-8) = sqrt(56) so then 8+sqrt(56)/8 im pretty sure
anonymous
  • anonymous
which would be square root of 56 whatever that number is..

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anonymous
  • anonymous
that doesnt sound right...you can't just cancel the 8
anonymous
  • anonymous
it isnt cancelling it... the 8 is due to the interval that just worked out that way because they chose the number 8 and the number 64
anonymous
  • anonymous
so the answer is \[\sqrt{56}\]
anonymous
  • anonymous
The average value of a function is given by: \[1/(b-a) \int\limits_{a}^{b}f(x) dx\] In your problem, a=0, b=8.
anonymous
  • anonymous
yeah but that doesnt really work..i already tried it...can you walk me through it
anonymous
  • anonymous
nvrmind...im just gonna use Reiman Sum

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