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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

how should i approach this problem?
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
which would be square root of 56 whatever that number is..

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

that doesnt sound right...you can't just cancel the 8
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
so the answer is \[\sqrt{56}\]
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.
yeah but that doesnt really work..i already tried it...can you walk me through it
nvrmind...im just gonna use Reiman Sum

Not the answer you are looking for?

Search for more explanations.

Ask your own question