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mlddmlnog

  • 2 years ago

evaluate the following: ( calculus )

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  1. mlddmlnog
    • 2 years ago
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    |dw:1358278438909:dw|

  2. mlddmlnog
    • 2 years ago
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    @zepdrix

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

  4. mlddmlnog
    • 2 years ago
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    ohh thank you so much! @exraven !

  5. exraven
    • 2 years ago
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    you are welcome :)

  6. mlddmlnog
    • 2 years ago
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    |dw:1358278984085:dw|

  7. mlddmlnog
    • 2 years ago
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    |dw:1358279110805:dw|

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

  9. zepdrix
    • 2 years ago
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    |dw:1358279244554:dw|Let's pretend this is what P(x) looks like.

  10. mlddmlnog
    • 2 years ago
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    oh, so when you switch the two, you mulply the integration by -1?

  11. mlddmlnog
    • 2 years ago
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    ok :) @zepdrix

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

  13. zepdrix
    • 2 years ago
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    So what do you think that area will be? c:

  14. zepdrix
    • 2 years ago
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    Integral represents the area under the curve. This part was 4.|dw:1358279422383:dw|

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

  16. mlddmlnog
    • 2 years ago
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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 !

  17. exraven
    • 2 years ago
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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\[-\int\limits_{3}^{5} Q(u) du\]

  18. zepdrix
    • 2 years ago
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    ya c: good times.

  19. mlddmlnog
    • 2 years ago
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    but thank you @exraven . and yup thanks as always @zepdrix

  20. zepdrix
    • 2 years ago
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    Hmm rave came up with -4, maybe I made a silly mistake somewhere.. I better check again.

  21. mlddmlnog
    • 2 years ago
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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

  22. zepdrix
    • 2 years ago
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    Oh i see :o

  23. mlddmlnog
    • 2 years ago
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    |dw:1358279740263:dw|

  24. zepdrix
    • 2 years ago
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    Oh that's a fun one c:

  25. exraven
    • 2 years ago
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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 \[\int\limits_{3}^{1}P(x) dx\]

  26. mlddmlnog
    • 2 years ago
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    could you please explain how you did that..? i am completely lost...

  27. exraven
    • 2 years ago
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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\[\int\limits_{1}^{5} P(x) dx = A_1 + A_2 = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]

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

  29. zepdrix
    • 2 years ago
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    I'm confused why you're swapping the limits D: hmmm

  30. zepdrix
    • 2 years ago
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    Oh cause that's what we need to solve for :D I read it incorrectly haha

  31. mlddmlnog
    • 2 years ago
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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 \[\int\limits_{3}^{5}p(x)dx\]

  32. mlddmlnog
    • 2 years ago
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    are

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

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

  35. mlddmlnog
    • 2 years ago
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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?

  36. exraven
    • 2 years ago
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    yes

  37. mlddmlnog
    • 2 years ago
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    oh yay thank you guys! :)

  38. exraven
    • 2 years ago
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    you are welcome :)

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