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kirbykirby

  • 2 years ago

Integral 2b(1+t) dt

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  1. kirbykirby
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    \[2b*t+2b*\frac{t^{2}}{2} +c\] \[2bt+bt^2+c\] it is the same..

  5. Azteck
    • 2 years 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
    • 2 years ago
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    your professor is right . because \[\Large \int\limits_{}^{} (ax+b)=\frac{(ax+b)^2}{a}+C\]

  7. sami-21
    • 2 years ago
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    its 2a !

  8. Azteck
    • 2 years ago
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    k is a constant. Just replace that k with a big C.

  9. kirbykirby
    • 2 years 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
    • 2 years ago
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    but im still right!

  11. kirbykirby
    • 2 years ago
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    am I missing something?

  12. sami-21
    • 2 years ago
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    yes you are !

  13. Azteck
    • 2 years ago
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    @kirbykirby You meant to differentiate it.

  14. Azteck
    • 2 years ago
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    You expand then differentiate.

  15. Azteck
    • 2 years ago
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    Wait sorry you differentiate by not expanding.

  16. Mimi_x3
    • 2 years ago
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    differentiate?? this is integration!

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

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

  19. Azteck
    • 2 years 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
    • 2 years ago
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    Ok. I'm trying to understand though how the 2 methods though give a different answer though...

  21. Azteck
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Which is the original expression you had in the question...

  24. kirbykirby
    • 2 years 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
    • 2 years ago
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    isn't it a constant here with respect to t?

  26. Azteck
    • 2 years ago
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    Yes.

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

  28. kirbykirby
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    integrate*

  32. kirbykirby
    • 2 years 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
    • 2 years ago
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    \[2bt+bt^2 \neq=2b(t+1)\]

  34. Azteck
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    oops forget the extra "2" in the last expression

  39. Azteck
    • 2 years 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
    • 2 years ago
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    I am not learning how to integrate :( I know how to integrate

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

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

  43. Azteck
    • 2 years 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
    • 2 years ago
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    first put*

  45. Azteck
    • 2 years ago
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    x^2*

  46. Azteck
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Then use one method. Simple as that.

  51. kirbykirby
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    If you really want to know, then go to your professor and ask him why.

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

  55. Azteck
    • 2 years ago
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    *ask him/her

  56. kirbykirby
    • 2 years 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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