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Integratiion by parts

  1. Sep 26, 2007 #1
    (2x+7)squareroot(7-x)dx from x=6 to x=7

    i've been trying to solve this integral with the integration by parts formula but it wont work.
    would appreciate some help.

    would like to see the whole solution and not just the answer. (the answer should bee in fractions)
  2. jcsd
  3. Sep 26, 2007 #2


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    I would like to see some work...

    You have to show some work in order to get help here. Let us see what you have tried so far.
  4. Sep 26, 2007 #3
    (2x+7)squareroot(7-x)dx from x=6 to x=7



    dx=-2 du


    (2(7-u^2)+7)u dx

    (14-2u^2+7)u dx

    (21-2u^2)u * -2u


    dont know how to continue from here.
  5. Sep 26, 2007 #4
    You aren't performing integration by parts to solve this problem, from what I can see.


    Read that article.
  6. Sep 26, 2007 #5


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    As l46kok said, what you are doing is not integration by parts. You are trying to integration by substitution from what I can tell.

    Here is an article on integration by parts from Wolfram MathWorld:

  7. Sep 26, 2007 #6


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    I've put in some parentheses to make the expression unambiguous.

    Now multiply out and integrate each term.
  8. Sep 27, 2007 #7
    i think i need to see a complete solution to this problem, i've been working on this for three days now and stiill dont understand it completly.

    [tex]\int_{1}^{2}(9x+6)\sqrt{7-3x} dx\ [/tex]
  9. Sep 27, 2007 #8


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    Well, (21- 2u^2)(-2udu)

    Well, that's NOT integration by parts, it's a straight forward substitution. But it looks to me like you are doing fine. I don't see why you only multiplied one of the "u"s and not both. : (21- 2u^2)(-2u du)= (4u^3- 42u)du. You might want to change the limits of integeration also. When x= 6, what is u? When x= 7 what is u?
  10. Sep 28, 2007 #9
    i know now how to get it right, thanks for your help.
  11. Sep 28, 2007 #10
    i have a new problem, i've tried to solve it but i think i need some advice.

    [tex]\int_{0}^{\pi /2}sin7xcos6x\hspace{6} dx\\ \\ f=cos6x\hspace{6} F=\frac{sin6x}{6}\\ g'=7cos7x\hspace{6} g=sin7x\\[/tex]

    and this is what i have done so far

    [tex]\int_{0}^{\pi /2}sin7xcos6x\hspace{6} dx\\ \\ (\frac{1}{6}sin6x)(sin7x)-\int_{0}^{\pi /2}(\frac{1}{6}sin6x)(7cos7x)=\\ \\ (\frac{1}{6}sin6x)(sin7x)-\frac{7}{6}\int_{0}^{\pi /2}(sin6x)(cos7x)\hspace{6} [/tex]
    Last edited: Sep 28, 2007
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