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swin2013
 one year ago
Best ResponseYou've already chosen the best response.0\[y=x^{2}  1, [0, \sqrt{3}]\]

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0dw:1358903907628:dw

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ \sqrt{3} } ∫\left(\begin{matrix}\sqrt{3} \\ 0\end{matrix}\right)\]

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0and is it the area of a triangle  area of a triangle?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac1{\sqrt3}\int\limits_0^{\sqrt3}(x^21)\mathrm dx\]

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0yea but i have to solve it with NINT :// the answer comes out to be 0 at x=1

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i got √31 ~0.7~ 1

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0it's in the back of my book lol

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0im not sure how x=1 has anything to do with this problem

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0"At what point(s) in the interval does the function assume its average value?"

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1What's the ummm average of integral formula? I always forget that thing...\[\large f_{ave}=\frac{1}{ba}\int\limits_a^b f(x) dx\]I think it's that thing right? So if we apply that to our problem,\[\large f_{ave}=\frac{1}{\sqrt30}\int\limits_0^{\sqrt3} x^21 \;dx\]Which gives us,\[\large \frac{1}{\sqrt3}\left[\frac{1}{3}x^3x\right]_0^{\sqrt3} \qquad = \qquad 0\] 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.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0ah the answer makes sense now you have stated the question @swin2013
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