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Math Help - Evaluate the definite integral...integrating by parts

  1. #1
    Member dokrbb's Avatar
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    Evaluate the definite integral...integrating by parts

    So, I have got another problem and can't figure out where is my mistake,

    I have the following integral  \int_{2}^{5} t^{4}ln(2t)dt, I considered  u = ln(2t) => u' = \frac{1}{2t}; v' = t^{4} => v = \frac{t^{5}}{5}

    and proceeded as follows:  \left( \frac{t^{5}}{5}ln(2t) \right) - \int\frac{1}{2t}\frac{t^{5}}{5}dt = here, I considered that I can do some simplifications and

    obtained that  \left( \frac{t^{5}}{5}ln(2t) \right) - \int\frac{1}{10} t^{4}dt = \left( \frac{t^{5}}{5}ln(2t) \right) - \frac{t^{5}}{50} = ,

    but by evaluating it further and doing all the math I don't get the correct answer,

    I saw that and thought that I can't do the simplifications since they are still  u' and v,

    so, I considered them separately, even though it seems wrong for me  \left( \frac{t^{5}}{5}ln(2t) \right) - \frac{t^{6} ln|2t|}{30} , but this also, as I expected, wrong,

    please, can someone show me the mistake(s)?
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  2. #2
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    Re: Evaluate the definite integral...integrating by parts

    Quote Originally Posted by dokrbb View Post
    So, I have got another problem and can't figure out where is my mistake,

    I have the following integral  \int_{2}^{5} t^{4}ln(2t)dt, I considered  u = ln(2t) => u' = \frac{1}{2t}; v' = t^{4} => v = \frac{t^{5}}{5}

    and proceeded as follows:  \left( \frac{t^{5}}{5}ln(2t) \right) - \int\frac{1}{2t}\frac{t^{5}}{5}dt = here, I considered that I can do some simplifications and

    obtained that  \left( \frac{t^{5}}{5}ln(2t) \right) - \int\frac{1}{10} t^{4}dt = \left( \frac{t^{5}}{5}ln(2t) \right) - \frac{t^{5}}{50} = ,

    but by evaluating it further and doing all the math I don't get the correct answer,

    I saw that and thought that I can't do the simplifications since they are still  u' and v,

    so, I considered them separately, even though it seems wrong for me  \left( \frac{t^{5}}{5}ln(2t) \right) - \frac{t^{6} ln|2t|}{30} , but this also, as I expected, wrong,

    please, can someone show me the mistake(s)?
    did you apply the limits 2 and 5 correctly? also, the derivative u' is 1/t, not 1/2t
    Last edited by votan; October 5th 2013 at 06:32 PM.
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  3. #3
    Member dokrbb's Avatar
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    Re: Evaluate the definite integral...integrating by parts

    Quote Originally Posted by votan View Post
    did you apply the limits 2 and 5 correctly?
    with my first evaluation I get 1368,383397, with the second one I have 233,9377581 (but I suppose this one is a priori wrong, so...),
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    Re: Evaluate the definite integral...integrating by parts

    [QUOTE=dokrbb;799460]I considered  u = ln(2t) => u' = \frac{1}{2t}

    Consider it further

    Edit: sorry, votan did say that!
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    Member dokrbb's Avatar
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    Re: Evaluate the definite integral...integrating by parts

    [QUOTE=tom@ballooncalculus;799489]
    Quote Originally Posted by dokrbb View Post
    I considered  u = ln(2t) => u' = \frac{1}{2t}

    Consider it further

    Edit: sorry, votan did say that!
    and if you were answering without sarcasm...
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    Member dokrbb's Avatar
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    Re: Evaluate the definite integral...integrating by parts

    never mind, I figured it out,
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