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swin2013

  • 3 years ago

Integral with U sub!

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  1. swin2013
    • 3 years ago
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    integral y^2 (1+y)^2 dy

  2. swin2013
    • 3 years ago
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    My work: u = 1+y du/dy = 1 du=dy?

  3. swin2013
    • 3 years ago
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    then integral (1+y)^2 dy integral u^2 du

  4. swin2013
    • 3 years ago
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    (1_y)^3/3 +c

  5. swin2013
    • 3 years ago
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    *(1+y)^3

  6. swin2013
    • 3 years ago
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    do i keep the y^2?

  7. swin2013
    • 3 years ago
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    ohh ok.. but what about u sub?

  8. swin2013
    • 3 years ago
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    will u now be y^2+y^4?

  9. swin2013
    • 3 years ago
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    well my teacher made this worksheet specifically for u sub :(

  10. swin2013
    • 3 years ago
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    but i got 1 for u sub anyways. so i just keep y^2 ?

  11. swin2013
    • 3 years ago
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    no

  12. swin2013
    • 3 years ago
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    it's y^2 (1+y)^2

  13. swin2013
    • 3 years ago
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    the square is outside the parenthesis

  14. stamp
    • 3 years ago
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    I see my mistake. Let me reevaluate the problem and I will come at you with a response.

  15. zepdrix
    • 3 years ago
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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.

  16. stamp
    • 3 years ago
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    \[y^2(1+y)^2dy=y^2(y^2+2y+1)dy=(y^4+2y^3+y^2)dy\]

  17. swin2013
    • 3 years ago
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    i didn;t know i had to lol oops

  18. swin2013
    • 3 years ago
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    is there an indication to do so?

  19. stamp
    • 3 years ago
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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

  20. swin2013
    • 3 years ago
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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

  21. stamp
    • 3 years ago
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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.

  22. zepdrix
    • 3 years ago
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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.

  23. swin2013
    • 3 years ago
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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

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

  25. swin2013
    • 3 years ago
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    u sub can be used when the exponent is outside right?

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