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Math Help - Integration by Part

  1. #1
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    Integration by Part

    So my professor was going over integration by part and using the tabular method. He use the problem:

    (the integral sign, big swoosh like thing) (3t+9)^3(2t+e^2t)dt

    Now, to find the derivative, I believe he used the chain rule, perhaps? Only thing is, I am not certain how I would use the chain rule on this one? Somehow, I do recall him showing on the outside he obtained

    3x3(something) for the first derivative
    and
    3x2x9(something) for the second derivative

    But I do not know how.

    Any help would be appreciated!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mike5055 View Post
    So my professor was going over integration by part and using the tabular method. He use the problem:

    (the integral sign, big swoosh like thing) (3t+9)^3(2t+e^2t)dt

    Now, to find the derivative, I believe he used the chain rule, perhaps? Only thing is, I am not certain how I would use the chain rule on this one? Somehow, I do recall him showing on the outside he obtained

    3x3(something) for the first derivative
    and
    3x2x9(something) for the second derivative

    But I do not know how.

    Any help would be appreciated!
    The derivative of which? Which did you select to be the "u" term?
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  3. #3
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    I'm almost positive he set (3t+9)^3 = u
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mike5055 View Post
    I'm almost positive he set (3t+9)^3 = u
    Ok then if we let f(x)=(3x+9)^3 then

    f'(x)=9(3t+9)^2

    f''(x)=54(3x+9)

    f'''(x)=162
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  5. #5
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    That was it!

    Alright, my question is, how did you get the 9, the 54, and eventually, the 162?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mike5055 View Post
    That was it!

    Alright, my question is, how did you get the 9, the 54, and eventually, the 162?
    \left(3x+9\right)^3=\left(3(x+3)\right)^3=27(x+3)^  3

    And then do you know the chain rule?

    \frac{d}{dx}\bigg[f\left(g(x)\right)\bigg]=f'\left(g(x)\right)\cdot{g'(x)}
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    Ok then if we let f(x)=(3x+9)^3 then

    f'(x)=9(3t+9)^2

    f''(x)=54(3x+9)

    f'''(x)=162
    This is all correct. You want :
    f(x)=(3x+9)^3
    f'(x)=3(3x+9)^2*3
    f'(x)=9(3x+9)^2

    Do this in a similar manner until you get down to 162. Do you understand about using the tabular method and integrating 2t+e^(2t)?
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  8. #8
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    Oh! Okay! I get it now. I was doing it wrong at first and wasn't getting anything near those. Thank you!
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  9. #9
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    Quote Originally Posted by kathrynmath View Post
    Do this in a similar manner until you get down to 162. Do you understand about using the tabular method and integrating 2t+e^(2t)?
    Somewhat. My professor never really explains things very clearly so I have to rely on the textbook for most of it.
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  10. #10
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    Quote Originally Posted by Mike5055 View Post
    Somewhat. My professor never really explains things very clearly so I have to rely on the textbook for most of it.
    Ok, what helped me learn it was by making a chart with one colum labeled D(for taking the derivative) and the other I(for integrating).

    So, I'll start you out:

    D
    (3t+9)^3
    9(3t+9)^2

    I
    2t+e^(2t)
    1/3t^3+1/2e^(2t)

    You understand that you once you get all the terms, you start by multiplying (3t+9)^3 ( 1/3t^3+1/2e^(2t))?
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  11. #11
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    Yeah, the 2 column thing is what a friend taught me to do and it definitely helps a lot.
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