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Dido525

Find the area of the plane region that is enclosed by the curves: \[y=\left| x^2-4 \right|\] and\[y=(\frac{ x^2 }{ 2 }) +4\]

  • one year ago
  • one year ago

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  1. Dido525
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    I just need to know what to do with the absolute value sign.

    • one year ago
  2. Dido525
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    @SmoothMath @Zarkon @UnkleRhaukus

    • one year ago
  3. Dido525
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    @Hero

    • one year ago
  4. Hero
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    You graph it. Remember y = |x^2 -4| = x^2 - 4 or 4 - x^2

    • one year ago
  5. Dido525
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    So it's really 3 curves I am dealing with right?

    • one year ago
  6. Hero
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    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.

    • one year ago
  7. Dido525
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    Okay now I am confused.

    • one year ago
  8. Dido525
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    I get that i consider the or part but then how do I find the area?

    • one year ago
  9. Hero
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    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

    • one year ago
  10. Dido525
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    Ohh okay! So I find the area between the curves for both cases and add them up right?

    • one year ago
  11. Hero
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    In theory

    • one year ago
  12. Dido525
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    Hmm let me do it...

    • one year ago
  13. Hero
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    Don't forget to find the enclosing points first.

    • one year ago
  14. Dido525
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    Yeah that's easy.

    • one year ago
  15. Dido525
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    yeah I know that @UnkleRhaukus but I jest need to know how to deal with the absolute value sign.

    • one year ago
  16. UnkleRhaukus
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    |dw:1355700012700:dw|

    • one year ago
  17. UnkleRhaukus
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    |dw:1355700073022:dw|

    • one year ago
  18. Dido525
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    Yep I got that... but you have to do this with a calculator don't you?

    • one year ago
  19. UnkleRhaukus
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    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

    • one year ago
  20. Dido525
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    I JUST noticed the symmetry :P .

    • one year ago
  21. UnkleRhaukus
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    |dw:1355700278738:dw|

    • one year ago
  22. Dido525
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    So I would go: |dw:1355700438412:dw|

    • one year ago
  23. Dido525
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    Right? To deal with:|dw:1355700518056:dw|

    • one year ago
  24. UnkleRhaukus
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    that looks right to me, (you draw your fours funny) but yes thats is the right integration

    • one year ago
  25. Dido525
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    My teachers said my 4's looked like 9 >.< .

    • one year ago
  26. UnkleRhaukus
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    looks more like a five to me

    • one year ago
  27. Dido525
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    |dw:1355700680945:dw|

    • one year ago
  28. UnkleRhaukus
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    so thats just where the first curve y=|x^2-4| is really y= 4-x^2

    • one year ago
  29. UnkleRhaukus
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    (like Hero was trying to saying )

    • one year ago
  30. Dido525
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    Ohh! So it would be: |dw:1355700839817:dw|

    • one year ago
  31. Dido525
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    Yeah I know hero said that :) .

    • one year ago
  32. UnkleRhaukus
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    yeah thats right ,

    • one year ago
  33. Dido525
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    Thanks a lot both of you :) @UnkleRhaukus @Hero .

    • one year ago
  34. Dido525
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    Could you give hero a medal too :) .

    • one year ago
  35. UnkleRhaukus
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    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

    • one year ago
  36. Dido525
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    Yep I know :) .

    • one year ago
  37. Hero
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    No @UnkleRhaukus. I wasn't trying to say it. I did say it.

    • one year ago
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