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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how should i approach this problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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(640)=8 and sqrt(648) = sqrt(56) so then 8+sqrt(56)/8 im pretty sure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which would be square root of 56 whatever that number is..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that doesnt sound right...you can't just cancel the 8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it 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
 5 years ago
Best ResponseYou've already chosen the best response.0so the answer is \[\sqrt{56}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The average value of a function is given by: \[1/(ba) \int\limits_{a}^{b}f(x) dx\] In your problem, a=0, b=8.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah but that doesnt really work..i already tried it...can you walk me through it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nvrmind...im just gonna use Reiman Sum
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