anonymous
  • anonymous
Find the area of the plane region that is enclosed by the curves: \[y=\left| x^2-4 \right|\] and\[y=(\frac{ x^2 }{ 2 }) +4\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I just need to know what to do with the absolute value sign.
anonymous
  • anonymous
@SmoothMath @Zarkon @UnkleRhaukus
anonymous
  • anonymous
@Hero

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More answers

Hero
  • Hero
You graph it. Remember y = |x^2 -4| = x^2 - 4 or 4 - x^2
anonymous
  • anonymous
So it's really 3 curves I am dealing with right?
Hero
  • Hero
You have to pay attention to the OR part. You will have two separate graphs. However, my theory is that the areas of both will be the same.
anonymous
  • anonymous
Okay now I am confused.
anonymous
  • anonymous
I get that i consider the or part but then how do I find the area?
Hero
  • Hero
On one graph, you will find the area enclosing y = x^2 - 4 and y = x^2/2 + 4 On the other graph you will find the are enclosing y = 4 - x^2 and y = x^2/2 + 4
anonymous
  • anonymous
Ohh okay! So I find the area between the curves for both cases and add them up right?
Hero
  • Hero
In theory
anonymous
  • anonymous
Hmm let me do it...
Hero
  • Hero
Don't forget to find the enclosing points first.
anonymous
  • anonymous
Yeah that's easy.
anonymous
  • anonymous
yeah I know that @UnkleRhaukus but I jest need to know how to deal with the absolute value sign.
UnkleRhaukus
  • UnkleRhaukus
|dw:1355700012700:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1355700073022:dw|
anonymous
  • anonymous
Yep I got that... but you have to do this with a calculator don't you?
UnkleRhaukus
  • UnkleRhaukus
i would break the graph in half first , because it is symmetric about, the y axis then i would calculate the area when x is from 0 to 2, and add this to the area from x=2 to x=4
anonymous
  • anonymous
I JUST noticed the symmetry :P .
UnkleRhaukus
  • UnkleRhaukus
|dw:1355700278738:dw|
anonymous
  • anonymous
So I would go: |dw:1355700438412:dw|
anonymous
  • anonymous
Right? To deal with:|dw:1355700518056:dw|
UnkleRhaukus
  • UnkleRhaukus
that looks right to me, (you draw your fours funny) but yes thats is the right integration
anonymous
  • anonymous
My teachers said my 4's looked like 9 >.< .
UnkleRhaukus
  • UnkleRhaukus
looks more like a five to me
anonymous
  • anonymous
|dw:1355700680945:dw|
UnkleRhaukus
  • UnkleRhaukus
so thats just where the first curve y=|x^2-4| is really y= 4-x^2
UnkleRhaukus
  • UnkleRhaukus
(like Hero was trying to saying )
anonymous
  • anonymous
Ohh! So it would be: |dw:1355700839817:dw|
anonymous
  • anonymous
Yeah I know hero said that :) .
UnkleRhaukus
  • UnkleRhaukus
yeah thats right ,
anonymous
  • anonymous
Thanks a lot both of you :) @UnkleRhaukus @Hero .
anonymous
  • anonymous
Could you give hero a medal too :) .
UnkleRhaukus
  • UnkleRhaukus
so take those integrals, add them together, then dont forget to Double your answer because we were only looking at the RHS of the graph
anonymous
  • anonymous
Yep I know :) .
Hero
  • Hero
No @UnkleRhaukus. I wasn't trying to say it. I did say it.

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