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mlddmlnog Group Title

evaluate the following: ( calculus )

  • one year ago
  • one year ago

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  1. mlddmlnog Group Title
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    |dw:1358278438909:dw|

    • one year ago
  2. mlddmlnog Group Title
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    @zepdrix

    • one year ago
  3. exraven Group Title
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    u can evaluate this easily by using the properties of definite integral \[\int\limits_{3}^{5} \left( 2P(x)+Q(x) \right)dx=\int\limits_{3}^{5} 2P(x) dx+\int\limits_{3}^{5}Q(x)dx\]\[=2\int\limits_{3}^{5}P(x) dx + \int\limits_{3}^{5}Q(x) dx\]

    • one year ago
  4. mlddmlnog Group Title
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    ohh thank you so much! @exraven !

    • one year ago
  5. exraven Group Title
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    you are welcome :)

    • one year ago
  6. mlddmlnog Group Title
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    |dw:1358278984085:dw|

    • one year ago
  7. mlddmlnog Group Title
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    |dw:1358279110805:dw|

    • one year ago
  8. exraven Group Title
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    for the last one, you can swap the limits of integration and multiply the whole thing by -1\[\int\limits_{5}^{1} P(x) dx = -\int\limits_{1}^{5} P(x) dx\]

    • one year ago
  9. zepdrix Group Title
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    |dw:1358279244554:dw|Let's pretend this is what P(x) looks like.

    • one year ago
  10. mlddmlnog Group Title
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    oh, so when you switch the two, you mulply the integration by -1?

    • one year ago
  11. mlddmlnog Group Title
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    ok :) @zepdrix

    • one year ago
  12. zepdrix Group Title
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    P(-x) will be the same function but reflected over the Y-axis.|dw:1358279336129:dw|

    • one year ago
  13. zepdrix Group Title
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    So what do you think that area will be? c:

    • one year ago
  14. zepdrix Group Title
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    Integral represents the area under the curve. This part was 4.|dw:1358279422383:dw|

    • one year ago
  15. zepdrix Group Title
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    Darn they're suppose to be the same, I didn't draw that left curve very well though lol

    • one year ago
  16. mlddmlnog Group Title
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    @zepdrix - ohh the area stays the same since it's only changed to negative but the mount of space under curve? stays the same !

    • one year ago
  17. exraven Group Title
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    for this one\[\int\limits_{-3}^{-5} Q(-x) dx\] use the substitution rule, let\[u = -x\]\[du = -dx\]\[x = -3, u = 3\]\[x = -5, u = 5\]the definite integral becomes\[-\int\limits_{3}^{5} Q(u) du\]

    • one year ago
  18. zepdrix Group Title
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    ya c: good times.

    • one year ago
  19. mlddmlnog Group Title
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    but thank you @exraven . and yup thanks as always @zepdrix

    • one year ago
  20. zepdrix Group Title
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    Hmm rave came up with -4, maybe I made a silly mistake somewhere.. I better check again.

    • one year ago
  21. mlddmlnog Group Title
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    no the right answer is 4 :) she gave us the answer on the study guide. but just the answers, and no process. :p

    • one year ago
  22. zepdrix Group Title
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    Oh i see :o

    • one year ago
  23. mlddmlnog Group Title
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    |dw:1358279740263:dw|

    • one year ago
  24. zepdrix Group Title
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    Oh that's a fun one c:

    • one year ago
  25. exraven Group Title
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    \[\int\limits_{1}^{5} P(x) dx = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]\[\int\limits_{1}^{5} P(x) dx = -\int\limits_{3}^{1} P(x) dx + \int\limits_{3}^{5} P(x) dx\]solve for \[\int\limits_{3}^{1}P(x) dx\]

    • one year ago
  26. mlddmlnog Group Title
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    could you please explain how you did that..? i am completely lost...

    • one year ago
  27. exraven Group Title
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    okay, first of all suppose P(x) is like this|dw:1358280041569:dw|notice that I divide the area from x = 1 to x = 5 into two parts, and that's why I can write it like this\[\int\limits_{1}^{5} P(x) dx = A_1 + A_2 = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]

    • one year ago
  28. exraven Group Title
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    and then, just swap the limit of integration for\[\int\limits_{1}^{3} P(x) dx\]it becomes\[-\int\limits_{3}^{1} P(x) dx\]

    • one year ago
  29. zepdrix Group Title
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    I'm confused why you're swapping the limits D: hmmm

    • one year ago
  30. zepdrix Group Title
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    Oh cause that's what we need to solve for :D I read it incorrectly haha

    • one year ago
  31. mlddmlnog Group Title
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    oh i understand that part now, but i still dont understand how to solve for the value when you do not know what \[\int\limits_{1}^{3}p(x)dx\] and \[\int\limits_{3}^{5}p(x)dx\]

    • one year ago
  32. mlddmlnog Group Title
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    are

    • one year ago
  33. exraven Group Title
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    actually we know\[\int\limits_{3}^{5} P(x) dx\] and \[\int\limits_{1}^{5} P(x) dx\]

    • one year ago
  34. mlddmlnog Group Title
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    oh waiat! could you just subtract \[\int\limits_{3}^{5}p(x)dx\] from \[\int\limits_{1}^{5}p(x)dx\] ?

    • one year ago
  35. mlddmlnog Group Title
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    so like.. \[\int\limits_{1}^{5}p(x)dx - \int\limits_{3}^{5}p(x)dx = -1-3 = -4\] and then, since you have to switch the limits to 3 to 1, you multiply the -4 by -1 to get 4 as our answer?

    • one year ago
  36. exraven Group Title
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    yes

    • one year ago
  37. mlddmlnog Group Title
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    oh yay thank you guys! :)

    • one year ago
  38. exraven Group Title
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    you are welcome :)

    • one year ago
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