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

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

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

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

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

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

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
@UnkleRhaukus
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
are you sure?
 one year ago

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

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

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

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

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix ?? :((
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
What'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.
 one year ago

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