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swin2013 Group Title

Integral with U sub!

  • one year ago
  • one year ago

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  1. swin2013 Group Title
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    integral y^2 (1+y)^2 dy

    • one year ago
  2. swin2013 Group Title
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    My work: u = 1+y du/dy = 1 du=dy?

    • one year ago
  3. swin2013 Group Title
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    then integral (1+y)^2 dy integral u^2 du

    • one year ago
  4. swin2013 Group Title
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    (1_y)^3/3 +c

    • one year ago
  5. swin2013 Group Title
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    *(1+y)^3

    • one year ago
  6. swin2013 Group Title
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    do i keep the y^2?

    • one year ago
  7. swin2013 Group Title
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    ohh ok.. but what about u sub?

    • one year ago
  8. swin2013 Group Title
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    will u now be y^2+y^4?

    • one year ago
  9. swin2013 Group Title
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    well my teacher made this worksheet specifically for u sub :(

    • one year ago
  10. swin2013 Group Title
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    but i got 1 for u sub anyways. so i just keep y^2 ?

    • one year ago
  11. swin2013 Group Title
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    no

    • one year ago
  12. swin2013 Group Title
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    it's y^2 (1+y)^2

    • one year ago
  13. swin2013 Group Title
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    the square is outside the parenthesis

    • one year ago
  14. stamp Group Title
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    I see my mistake. Let me reevaluate the problem and I will come at you with a response.

    • one year ago
  15. zepdrix Group Title
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    Yah you wouldn't use a `U sub` for this one I'm afraid swin. You would just expand out the brackets and integrate each term individually.

    • one year ago
  16. stamp Group Title
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    \[y^2(1+y)^2dy=y^2(y^2+2y+1)dy=(y^4+2y^3+y^2)dy\]

    • one year ago
  17. swin2013 Group Title
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    i didn;t know i had to lol oops

    • one year ago
  18. swin2013 Group Title
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    is there an indication to do so?

    • one year ago
  19. stamp Group Title
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    No, you are simplifying your y terms and happen to end up with \[ (y^4+2y^3+y^2)dy \]Integrate each term dy

    • one year ago
  20. swin2013 Group Title
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    oh ok. that makes sense. usually i don't have to expand a problem or distribute so i didn't know if using u sub for 1+y is correct

    • one year ago
  21. stamp Group Title
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    Had you done some usub, you would had \[(y^2u)du\] and would not really have gotten anywhere as far as the integral goes. Be sure to have + C to your integral term since the integral is undefined.

    • one year ago
  22. zepdrix Group Title
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    To apply a `u sub`, you want to look for a suitable `u` and `u'` somewhere in your problem. \(2y(1+y^2)\) See how if \(u=1+y^2\) then \(u'=2y\) ? In the problem you were given, letting your inner function be `u` did NOT result in the outer function being u'. So that is usually your indicator.

    • one year ago
  23. swin2013 Group Title
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    OHH ok! I'll make sure to tell my teacher to clarify this to the class. I did not know that! hahaha

    • one year ago
  24. swin2013 Group Title
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    oh but in the case you're presenting it's (1+y^2) not (1+y)^2

    • one year ago
  25. swin2013 Group Title
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    u sub can be used when the exponent is outside right?

    • one year ago
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