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.

Find the average integral?

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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

\[y=x^{2} - 1, [0, \sqrt{3}]\]
|dw:1358903907628:dw|
\[\frac{ 1 }{ \sqrt{3} } ∫\left(\begin{matrix}\sqrt{3} \\ 0\end{matrix}\right)\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

and is it the area of a triangle - area of a triangle?
\[\frac1{\sqrt3}\int\limits_0^{\sqrt3}(x^2-1)\mathrm dx\]
yea but i have to solve it with NINT :// the answer comes out to be 0 at x=1
are you sure?
i got √3-1 ~0.7~ 1
it's in the back of my book lol
im not sure how x=1 has anything to do with this problem
"At what point(s) in the interval does the function assume its average value?"
@zepdrix ?? :((
What's the ummm average of integral formula? I always forget that thing...\[\large f_{ave}=\frac{1}{b-a}\int\limits_a^b f(x) dx\]I think it's that thing right? So if we apply that to our problem,\[\large f_{ave}=\frac{1}{\sqrt3-0}\int\limits_0^{\sqrt3} x^2-1 \;dx\]Which gives us,\[\large \frac{1}{\sqrt3}\left[\frac{1}{3}x^3-x\right]_0^{\sqrt3} \qquad = \qquad 0\] Does it have something to do with that maybe? So the average of our integral occurs when f(x)=0... Am I interpreting that correctly?? So we have \(f(x)=0 \quad \text{when} \quad x=1\). I dunno... This is a bit of a confusing problem. That's my guess at least.
ah the answer makes sense now you have stated the question @swin2013

Not the answer you are looking for?

Search for more explanations.

Ask your own question