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swin2013

  • 3 years ago

Find the average integral?

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  1. swin2013
    • 3 years ago
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    \[y=x^{2} - 1, [0, \sqrt{3}]\]

  2. swin2013
    • 3 years ago
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    |dw:1358903907628:dw|

  3. swin2013
    • 3 years ago
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    \[\frac{ 1 }{ \sqrt{3} } ∫\left(\begin{matrix}\sqrt{3} \\ 0\end{matrix}\right)\]

  4. swin2013
    • 3 years ago
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    and is it the area of a triangle - area of a triangle?

  5. UnkleRhaukus
    • 3 years ago
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    \[\frac1{\sqrt3}\int\limits_0^{\sqrt3}(x^2-1)\mathrm dx\]

  6. swin2013
    • 3 years ago
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    yea but i have to solve it with NINT :// the answer comes out to be 0 at x=1

  7. swin2013
    • 3 years ago
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    @UnkleRhaukus

  8. UnkleRhaukus
    • 3 years ago
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    are you sure?

  9. UnkleRhaukus
    • 3 years ago
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    i got √3-1 ~0.7~ 1

  10. swin2013
    • 3 years ago
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    it's in the back of my book lol

  11. UnkleRhaukus
    • 3 years ago
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    im not sure how x=1 has anything to do with this problem

  12. swin2013
    • 3 years ago
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    "At what point(s) in the interval does the function assume its average value?"

  13. swin2013
    • 3 years ago
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    @zepdrix ?? :((

  14. zepdrix
    • 3 years ago
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    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
    • 3 years ago
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    ah the answer makes sense now you have stated the question @swin2013

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