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sha0403

help me to solve this question... Find the total area between the curve y=1-x^2, and the x-axis over the interval (0,2)...

  • one year ago
  • one year ago

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  1. AbhimanyuPudi
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    you have to integrate the given curve's equation in the interval (0,2) \[\int\limits_{0}^{2}1-x^2 = (2-0) + (-8/3 + 0) = 2 - 8/3 = -2/3\] Since are cannot be negative, area = 2/3 sq.units

    • one year ago
  2. AbhimanyuPudi
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    sorry I'm little confused while typing equations on OpenStudy..so I've been a little late in answering

    • one year ago
  3. sha0403
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    oo its ok..the working just like that AbhimanyuPudi ?

    • one year ago
  4. AbhimanyuPudi
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    may be u have to put another step in between..showing the integration.. \[\int\limits_{0}^{2}1-x^2 = x(from 0 \to 2) - x^3/3 (from 0 \to 2)\]

    • one year ago
  5. AbhimanyuPudi
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    the negative sign in the answer indicates that the area is formed below the x-axis

    • one year ago
  6. RadEn
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    for this case, first u have to figure out of the function y=1-x^2 |dw:1355817094537:dw| so, the total area = A1 + A2

    • one year ago
  7. shubhamsrg
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    i dont think you'll have to consider case of negative area here separately , so integrating directly from 0 to 2 should be correct..i may be wrong though..

    • one year ago
  8. shubhamsrg
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    forexample, integral(cosx) from x=0 to pi = 0 ..-ve and +ve areas cancel..

    • one year ago
  9. RadEn
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    the result will be different if u use integration (0,2) with i said above :)

    • one year ago
  10. shubhamsrg
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    i know, i am saying if we integrate directly from 0 to 2, we will get correct ans.

    • one year ago
  11. RadEn
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    no, @shubhamsrg for A1 = [x-1/3*x^3] [0,1] = 2/3 for A2 = [1/3*x^3-1] [1,2] = 4/3 so, total = 2/3 + 4/3 = 6/3 = 2

    • one year ago
  12. shubhamsrg
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    why will you treat negative area separately ? area can be -ve during integration!

    • one year ago
  13. RadEn
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    because the rule, if u want calculate area in under x-axist, u have to give -ve in front integral (because the integration will be -ve also), so we get positive area

    • one year ago
  14. sha0403
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    ok2 which one i should follow?

    • one year ago
  15. RadEn
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    btw, for A1 i take int (f(x)) dx [0,1] and A2 i take int(-f(x)) dx [1,2]

    • one year ago
  16. shubhamsrg
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    so if ask you integral(cos x) from 0 to pi? ans according to you should be 2 right ?

    • one year ago
  17. Chlorophyll
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    I absolutely agree with @RadEn as we have to check the position of the area :)

    • one year ago
  18. Chlorophyll
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    Yes, A = 2 unit square!

    • one year ago
  19. RadEn
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    but Abhimanyu's job not 2, but 2/3 :)

    • one year ago
  20. shubhamsrg
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    i just wanted to confirm : http://www.wolframalpha.com/input/?i=integral+%28cosx%29+from+0+to+pi

    • one year ago
  21. Chlorophyll
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    @sha0403 I leave the computing part for you, questions?

    • one year ago
  22. shubhamsrg
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    yes i have a question// why is wolfram giving value of integral =0 ? -ve is also real no.,, -ve area we do study, -ve indicates direction..please someone clarify..

    • one year ago
  23. RadEn
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    @shubhamsrg , is it possible the area be 0 :)

    • one year ago
  24. shubhamsrg
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    http://www.wolframalpha.com/input/?i=integral+%28cosx%29+from+x%3D+pi%2F2+to+3pi%2F2

    • one year ago
  25. shubhamsrg
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    and also, it'll be 0 when you add together the +ve and -ve areas!

    • one year ago
  26. RadEn
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    if just processing integration, yeah that's right but this case to find the area, so impossible be 0 or negative

    • one year ago
  27. shubhamsrg
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    i see what you mean,,hmm.. now i am also thinking if we go by my methodology , to find area of a circle, suppose x^2 + y^2 =1 , area turns out to be 0 i see..hmm you were right,,we got the take the absolute values..my apologies..

    • one year ago
  28. Chlorophyll
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    @shubhamsrg I'm not a fan of worf, honestly it's just a simple calculation!

    • one year ago
  29. shubhamsrg
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    i have maybe understood the concept sir,,sorry for the confusion..

    • one year ago
  30. sha0403
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    ok thanks a lot for you guys for help me... =)

    • one year ago
  31. RadEn
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    very welcome ... ::)

    • one year ago
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