Evaluate the definite integral...integrating by parts

**dokrbb** · Oct 5th 2013, 04:57 PM

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)?

**votan** · Oct 5th 2013, 05:25 PM

Originally Posted by dokrbb

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

**dokrbb** · Oct 5th 2013, 05:30 PM

Originally Posted by votan

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...),

**tom@ballooncalculus** · Oct 6th 2013, 03:45 AM

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

Consider it further

Edit: sorry, votan did say that!

**dokrbb** · Oct 6th 2013, 06:11 AM

[QUOTE=tom@ballooncalculus;799489]

Originally Posted by dokrbb

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...

**dokrbb** · Oct 6th 2013, 08:41 AM

never mind, I figured it out,

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