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

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

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

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

swin2013
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0i got √31 ~0.7~ 1

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

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

swin2013
 2 years 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
 2 years 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
 2 years 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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