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anonymous

  • 5 years ago

What is the average value of f(x)=sqrt 64-x^2 over the interval 0<=x<=8

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  1. anonymous
    • 5 years ago
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    how should i approach this problem?

  2. anonymous
    • 5 years ago
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    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

  3. anonymous
    • 5 years ago
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    which would be square root of 56 whatever that number is..

  4. anonymous
    • 5 years ago
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    that doesnt sound right...you can't just cancel the 8

  5. anonymous
    • 5 years ago
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    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

  6. anonymous
    • 5 years ago
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    so the answer is \[\sqrt{56}\]

  7. anonymous
    • 5 years ago
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    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.

  8. anonymous
    • 5 years ago
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    yeah but that doesnt really work..i already tried it...can you walk me through it

  9. anonymous
    • 5 years ago
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    nvrmind...im just gonna use Reiman Sum

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