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virtus

  • 4 years ago

find the area bounded by the graph y = 4x^2 -1, the x axis and the ordinates x = -1/2 and x=1

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  1. dumbcow
    • 4 years ago
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    \[A = \int\limits_{1/2}^{1}(4x^{2} -1) dx\]

  2. virtus
    • 4 years ago
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    smartcow, i thought the ordinates were from -1/2 where did you get the 1/2 from?

  3. dumbcow
    • 4 years ago
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    well when i graphed it, the parabola from -1/2 to 1/2 is below the x-axis ?

  4. virtus
    • 4 years ago
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    so ....?

  5. dumbcow
    • 4 years ago
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    is it bounded by the x-axis above or below ?

  6. virtus
    • 4 years ago
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    |dw:1328176928988:dw|

  7. virtus
    • 4 years ago
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    |dw:1328176997411:dw|

  8. virtus
    • 4 years ago
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    so is the shaded the area? |dw:1328177068433:dw|

  9. dumbcow
    • 4 years ago
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    hmm i guess that makes more sense...why do they say it is bounded by the x-axis then

  10. virtus
    • 4 years ago
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    but when i find the area i get 0 or maybe i have done some stupid error lemme check

  11. dumbcow
    • 4 years ago
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    no i get 0 as well, if the limits are -1/2 to 1

  12. dumbcow
    • 4 years ago
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    thats because the area under the x-axis is negative

  13. virtus
    • 4 years ago
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    the answer in my solution book says the answer is 4/3 square units

  14. dumbcow
    • 4 years ago
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    hmm well the area from 1/2 to 1 is 2/3 are you sure its not x=-1 an x=1

  15. virtus
    • 4 years ago
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    i am sure it is x=1

  16. virtus
    • 4 years ago
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    perhaps the answer in the solution book is wrong then or the question is wrong

  17. dumbcow
    • 4 years ago
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    |dw:1328177655800:dw| \[\int\limits_{-1}^{-1/2}(4x^{2} -1)dx + \int\limits_{1/2}^{1}(4x^{2} -1)dx = 2/3 + 2/3 = 4/3\] thats my best guess :)

  18. virtus
    • 4 years ago
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    WOAH!!!!!!!!!!!!!!!! that is brilliant thinking goodjob smartcow, thanks for the help greatly appreciated

  19. dumbcow
    • 4 years ago
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    your welcome

  20. FoolForMath
    • 4 years ago
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    But the only solution for \( 4x^2-1\) is \(\pm\frac 12 \)

  21. virtus
    • 4 years ago
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    genius what are you trying to say?

  22. FoolForMath
    • 4 years ago
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    The parabola intersects the x axis at \(\pm \frac 12 \)

  23. FoolForMath
    • 4 years ago
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    This question is not correct in my opinion.

  24. virtus
    • 4 years ago
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    oh i see, i think i support your opinion

  25. FoolForMath
    • 4 years ago
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    See here, the area is not well defined : http://www.wolframalpha.com/input/?i=+y+%3D+4x%5E2+-1%3Bx%3D-1%2F2%3Bx%3D1%3Bx%3D0%3By%3D0

  26. virtus
    • 4 years ago
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    ya

  27. FoolForMath
    • 4 years ago
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    ya ya :D

  28. FoolForMath
    • 4 years ago
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    So I think dumcow's interpretation is not correct either.

  29. virtus
    • 4 years ago
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    ya ya :D

  30. dumbcow
    • 4 years ago
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    haha sorry i should have clarified , my drawing assumes the question is wrong at to get area of 4/3 the bounds must be x=-1 and x=1 and x-axis

  31. virtus
    • 4 years ago
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    no need to apologise! you have a case indeed hehehe

  32. FoolForMath
    • 4 years ago
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    But the bound x=-1 and x=1 makes no sense right?

  33. virtus
    • 4 years ago
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    ya

  34. FoolForMath
    • 4 years ago
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    ya ya :D

  35. FoolForMath
    • 4 years ago
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    I am leaving now, take care :)

  36. virtus
    • 4 years ago
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    farewell genius, and thanks for the help once again !

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