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

1. anonymous

how should i approach this problem?

2. 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

3. anonymous

which would be square root of 56 whatever that number is..

4. anonymous

that doesnt sound right...you can't just cancel the 8

5. 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

6. anonymous

so the answer is $\sqrt{56}$

7. 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.

8. anonymous

yeah but that doesnt really work..i already tried it...can you walk me through it

9. anonymous

nvrmind...im just gonna use Reiman Sum