Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

What is the average value of the function on the interval from x=-3 to x=-1

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__85.gif
The average value of the function would be \(\large \int_{-3}^{-1} g(x)dx \) Can you do the integration?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i exactly did that does not work
Sorry, the integral should be divided by (-1- (-3))=2 to get the average value.
?
what did you get?
Average value = \( \large \int_a^b f(x)dx / (b-a) \)
28/6
You have the correct answer, but you need to simplify it (14/3) and divide by 2 which comes from (-1-(-3)), the length of interval of integration.
In the end, you should get 7/3.
ya thx
@oldrin.bataku I think there was a mix-up with the substituted limits of integration.
i got it
That's good! Keep it up!

Not the answer you are looking for?

Search for more explanations.

Ask your own question