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kirbykirby

  • one year ago

Integral 2b(1+t) dt

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  1. kirbykirby
    • one year ago
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    \[\int\limits2b(1+t)dt\] How does my prof get b(1+t)^2 + C but I got 2bt +bt^2 +C

  2. Mimi_x3
    • one year ago
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    \[\int\limits2b(1+t)dt \] \[\int\limits2b(1+t)dt \] \[2b\int\limits1dt +2b\int\limits td t\] bt^2+2bt+c

  3. kirbykirby
    • one year ago
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    It seems like my prof might have done lik a usubstituion on (1+t) but how come it doesn't give the same answer?

  4. Mimi_x3
    • one year ago
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    \[2b*t+2b*\frac{t^{2}}{2} +c\] \[2bt+bt^2+c\] it is the same..

  5. Azteck
    • one year ago
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    \[\int\limits2b(1+t)dt= (1+t)^2 + k\] Leave the 2b alone and put it back later since it's a constant. \[=\frac{1}{2}(1+t)^2 +k\] Put back the 2b. \[=\frac{2b}{2}(1+t)^2 +k\] \[b(1+t)^2 + k\]

  6. sami-21
    • one year ago
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    your professor is right . because \[\Large \int\limits_{}^{} (ax+b)=\frac{(ax+b)^2}{a}+C\]

  7. sami-21
    • one year ago
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    its 2a !

  8. Azteck
    • one year ago
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    k is a constant. Just replace that k with a big C.

  9. kirbykirby
    • one year ago
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    b(1+t)^2 = b(1+2t+t^2)=b+2bt+bt^2 but this is not the same as 2bt+bt^2 o_O

  10. Mimi_x3
    • one year ago
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    but im still right!

  11. kirbykirby
    • one year ago
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    am I missing something?

  12. sami-21
    • one year ago
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    yes you are !

  13. Azteck
    • one year ago
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    @kirbykirby You meant to differentiate it.

  14. Azteck
    • one year ago
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    You expand then differentiate.

  15. Azteck
    • one year ago
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    Wait sorry you differentiate by not expanding.

  16. Mimi_x3
    • one year ago
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    differentiate?? this is integration!

  17. Azteck
    • one year ago
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    No, he wants to get the answer back to the original expression.

  18. Azteck
    • one year ago
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    When you integrate it, you can differentiate it back to the expression that you originally integrated.

  19. Azteck
    • one year ago
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    @kirbykirby You differentiate that expanding expression you had, to get back your original expression. Integration is the reverse of differentiation.

  20. kirbykirby
    • one year ago
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    Ok. I'm trying to understand though how the 2 methods though give a different answer though...

  21. Azteck
    • one year ago
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    differentiate b+2bt+bt^2 with respect to t and you get: 2b+2bt=2b(1+t)

  22. sami-21
    • one year ago
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    \[\Large \int\limits_{}^{}2b(1+t)dt=2b \frac{(1+t)^2}{2}+C=b(1+t)^2+C\] differentiate \[\Large \frac{d}{dt}(b(1+t)^2)+C=2b(1+t)\] you get original back . means it is correct .

  23. Azteck
    • one year ago
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    Which is the original expression you had in the question...

  24. kirbykirby
    • one year ago
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    oh is it because the b + C gets absorbed into another constant "C" using the other method o_O?

  25. kirbykirby
    • one year ago
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    isn't it a constant here with respect to t?

  26. Azteck
    • one year ago
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    Yes.

  27. Azteck
    • one year ago
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    If it was respect to b, then it's a different story.

  28. kirbykirby
    • one year ago
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    b(1+t)^2 = b(1+2t+t^2)=b+2bt+bt^2 but this is not the same as 2bt+bt^2 So is this difference because of the b being absorbed??

  29. Azteck
    • one year ago
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    If you want to know why there's a C at the end everytime you integrate, it's because there are many curves that can still produce the same derivative. For example. The derivative of this curve y=x^2+1 is the same as y=x^2

  30. Azteck
    • one year ago
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    That's for you to understand why there's a constant attached everytime you itnegrate. If you don't attach a C or k whatever, you get it wrong.

  31. Azteck
    • one year ago
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    integrate*

  32. kirbykirby
    • one year ago
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    No I am not asking why there is a C constant... I am trying to understand why: Integrating it gives either "2bt+bt^2 + C" or "b(t+1)^2 + C"... which gives "b+2bt+bt^2+C" which is not the same as above

  33. Azteck
    • one year ago
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    \[2bt+bt^2 \neq=2b(t+1)\]

  34. Azteck
    • one year ago
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    You're meant to differentiate to get the original result. The word differentiate is the key word here.

  35. Azteck
    • one year ago
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    Once you integrate, you must differentiate to get back the original, otherwise, it's going to be certain that it's not the same as the original.

  36. kirbykirby
    • one year ago
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    Because One method is using \[2b( \int\limits(1)dt+\int\limits(t)dt)\] = 2bt+bt^2 +C OR \[2b \int\limits(1+t)dt=2b \frac{(1+t)^2}{2}=2b(1+t)^2+C\]

  37. Azteck
    • one year ago
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    Integration is just getting the original equation of a curve or function. When you're integrating something, that "something" is the derivative.

  38. kirbykirby
    • one year ago
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    oops forget the extra "2" in the last expression

  39. Azteck
    • one year ago
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    To be frank with you kirby, you shouldn't be using formulas when you're beginning to learn how to integrate.

  40. kirbykirby
    • one year ago
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    I am not learning how to integrate :( I know how to integrate

  41. kirbykirby
    • one year ago
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    I just don't see how these 2 methods are giving different answers

  42. Azteck
    • one year ago
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    Still, these easy indefinite integrals require you to just think about differentiating, but working backwards.

  43. Azteck
    • one year ago
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    If you were to differentiate 3x, I would probably first but an indice of 2 beside x giving me x^2 and then when you integrate s^2, you get 2x. but the original equation just has an x without the two. SO then you put a half in front of the x^2. That gives you (1/2)x^2. and then once you put back the 3, you get (3/2)x^2.

  44. Azteck
    • one year ago
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    first put*

  45. Azteck
    • one year ago
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    x^2*

  46. Azteck
    • one year ago
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    That's how you should do it. If you have that mindset, nothing can stop you from getting good marks in integration.

  47. kirbykirby
    • one year ago
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    I just want to be clear... I know how to differentiate... Factoring 2b and integrating "1" and "t" individually gives "2bt+bt^2+C" But if you factor 2b and integrate (1+t) using a substitution, you get "integral 2bu" = 2bu^2/2 +C= bu^2+C => b(1+t)^2+C

  48. Azteck
    • one year ago
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    Why would you do it that way. You rather keep integration as simple as possible. Don't go into that mucky stuff. You will end up confusing yourself in the end.

  49. kirbykirby
    • one year ago
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    I know it is stupid to do that, but shouldn't the answers AGREE? I mean what is the point if the two methods don't agree???

  50. Azteck
    • one year ago
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    Then use one method. Simple as that.

  51. kirbykirby
    • one year ago
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    But why does one not give the same answer as the other :( THIS is what I wanna understand

  52. Azteck
    • one year ago
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    No point in pondering on two methods. I'm not using any of the two but integrating through differentiation.

  53. Azteck
    • one year ago
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    If you really want to know, then go to your professor and ask him why.

  54. Azteck
    • one year ago
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    Your professor's there to help you understand, not bore you to death.

  55. Azteck
    • one year ago
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    *ask him/her

  56. kirbykirby
    • one year ago
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    is it because the extra "b" in the second method gets absorbed into the constant C?? (which is why they give the same derivative)... but I never saw an answer like "2x+2+C" being absorbed as 2x+C

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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