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anonymous

  • 5 years ago

Find the exact area between y=ln(x) and y= ln(x^2) when 1=/<x </=2

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  1. anonymous
    • 5 years ago
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    How would you set up this equation?

  2. anonymous
    • 5 years ago
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    Hi handy dandy helpers

  3. anonymous
    • 5 years ago
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    |dw:1326752403115:dw|

  4. myininaya
    • 5 years ago
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    \[\int\limits_{1}^{2}(\ln(x^2)-\ln(x)) dx\]

  5. anonymous
    • 5 years ago
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    Could you explain why mininaya?

  6. myininaya
    • 5 years ago
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    because of your graph lol

  7. anonymous
    • 5 years ago
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    lol

  8. anonymous
    • 5 years ago
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    lol you're right any questions rl?

  9. anonymous
    • 5 years ago
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    Think i got it

  10. anonymous
    • 5 years ago
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    Thanks myin and no data

  11. myininaya
    • 5 years ago
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    it always helps to draw a picture first or to graph using your calculator

  12. anonymous
    • 5 years ago
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    You're welcome rl

  13. myininaya
    • 5 years ago
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    then do \[\int\limits_{lowerlimit}^{upperlimit} (upperfunction -lowerfunction) dx\]

  14. anonymous
    • 5 years ago
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    that is great mininaya.

  15. anonymous
    • 5 years ago
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    and the upper function wld be the one with x^2 obviously?

  16. anonymous
    • 5 years ago
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    But what if he don't have a graphing calculator at hand?

  17. myininaya
    • 5 years ago
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    right

  18. myininaya
    • 5 years ago
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    he will have to draw it with out one then

  19. anonymous
    • 5 years ago
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    No data i am not sam now LOL

  20. anonymous
    • 5 years ago
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    who is sam?

  21. anonymous
    • 5 years ago
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    I am a she at the prsesnt moment

  22. myininaya
    • 5 years ago
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    also you may come of course some these when the functions with change from upper to lower and then you will have to break the integral up

  23. anonymous
    • 5 years ago
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    what do u mean?

  24. myininaya
    • 5 years ago
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    for example |dw:1326753034373:dw|

  25. myininaya
    • 5 years ago
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    if i wanted the area btw a and b bounded by g and f

  26. myininaya
    • 5 years ago
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    \[\int\limits_{a}^{c}(f-g) dx+\int\limits_{c}^{b}(g-f) dx\]

  27. myininaya
    • 5 years ago
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    then i would set up like that since f is the upper from a to c and g is the upper from c to b

  28. anonymous
    • 5 years ago
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    oh ok i c so basically u will have to draw the function and then decide which goes first

  29. myininaya
    • 5 years ago
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    yep drawing or seeing the graph is totally important

  30. myininaya
    • 5 years ago
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    the upper goes first

  31. anonymous
    • 5 years ago
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    Ok thanks for the tip myin u r awesome :DDDD

  32. myininaya
    • 5 years ago
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    :)

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