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anonymous

  • 5 years ago

how do you find the area enclosed by a line, a parabola, and the x-axis?

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  1. anonymous
    • 5 years ago
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    integrate :)

  2. anonymous
    • 5 years ago
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    i know but integrate what

  3. myininaya
    • 5 years ago
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    we need to know what parabola and what line

  4. anonymous
    • 5 years ago
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    if you have x^2 from 0 to 1 its int (0 to 1) of x^2 dx

  5. anonymous
    • 5 years ago
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    as an example

  6. anonymous
    • 5 years ago
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    Find the area enclosed by \[y=\sqrt{2x}, y=6-2x, and the xaxis\]

  7. myininaya
    • 5 years ago
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    first find where they intersect

  8. anonymous
    • 5 years ago
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    but they intersect at three different places

  9. amistre64
    • 5 years ago
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    those equation only intersect in one place :)

  10. anonymous
    • 5 years ago
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    rearrange for x in the equations, your parabola will then be the lower limit, the straight line will be the upper limit and your y limits will be from 0 to the y component of the intersection point, then it is a trivial double integral

  11. myininaya
    • 5 years ago
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    i will scan and show you my drawing

  12. myininaya
    • 5 years ago
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  13. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{y component of intersection}\int\limits_{(y^2)/2}^{1/2y+3}..and so on\]

  14. myininaya
    • 5 years ago
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    i always have to draw picture first

  15. anonymous
    • 5 years ago
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    me too by visual inspection you can simplify the problem a lot

  16. myininaya
    • 5 years ago
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    dana, have you looked at the pic?

  17. myininaya
    • 5 years ago
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    i also put what i integrated

  18. myininaya
    • 5 years ago
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    oops that second integral should e from 2 to 3 i messed up my line a bit

  19. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{2}\int\limits_{.5y^2}^{-0.5y-3}dxdy\]

  20. anonymous
    • 5 years ago
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    yeaa that makes sence now, thank you!

  21. myininaya
    • 5 years ago
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    there was not suppose to be an e in that sentence lol

  22. myininaya
    • 5 years ago
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    because the x intercept of that y=6-2x is 3 not 4

  23. anonymous
    • 5 years ago
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    yea his way if fine too, the double integral is just another way you can check your answer

  24. myininaya
    • 5 years ago
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    she*

  25. myininaya
    • 5 years ago
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    her*

  26. anonymous
    • 5 years ago
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    oh sorry

  27. anonymous
    • 5 years ago
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    wasn't paying attention to the name there sorry

  28. anonymous
    • 5 years ago
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    but on the other hand that double integral is pretty epic? can someone be my fan i think i deserve it for going all out on this question

  29. myininaya
    • 5 years ago
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    i will be ur fan

  30. anonymous
    • 5 years ago
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    thanks! i'm your fan too.

  31. anonymous
    • 5 years ago
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    helped people a few times but it seems a lot of people just make an account to ask a question and then leave without saying thank you

  32. myininaya
    • 5 years ago
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    your welcome dana lol

  33. myininaya
    • 5 years ago
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    he didnt become our fan :(

  34. myininaya
    • 5 years ago
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    dana, if you come back you could have done int(3-y/2-y^2/2,y=0..2) and got the same thing you could have look at the functions as if y-axis was x-axis

  35. anonymous
    • 5 years ago
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    ohh thats true, to switch the problem around. and i did say thank you, for the record. you all were a lot of help! :)

  36. myininaya
    • 5 years ago
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    lol im sorry i didn't notice im an idiot

  37. anonymous
    • 5 years ago
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    are aunder a curve, use integration & subtract lower curve from upper

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