## swin2013 3 years ago Find the average integral?

1. swin2013

$y=x^{2} - 1, [0, \sqrt{3}]$

2. swin2013

|dw:1358903907628:dw|

3. swin2013

$\frac{ 1 }{ \sqrt{3} } ∫\left(\begin{matrix}\sqrt{3} \\ 0\end{matrix}\right)$

4. swin2013

and is it the area of a triangle - area of a triangle?

5. UnkleRhaukus

$\frac1{\sqrt3}\int\limits_0^{\sqrt3}(x^2-1)\mathrm dx$

6. swin2013

yea but i have to solve it with NINT :// the answer comes out to be 0 at x=1

7. swin2013

@UnkleRhaukus

8. UnkleRhaukus

are you sure?

9. UnkleRhaukus

i got √3-1 ~0.7~ 1

10. swin2013

it's in the back of my book lol

11. UnkleRhaukus

im not sure how x=1 has anything to do with this problem

12. swin2013

"At what point(s) in the interval does the function assume its average value?"

13. swin2013

@zepdrix ?? :((

14. 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.

15. UnkleRhaukus

ah the answer makes sense now you have stated the question @swin2013