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Homework Help: Integration problem

  1. Mar 19, 2007 #1

    disregardthat

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    1. The problem statement, all variables and given/known data

    [tex]\int_{1}^{2}(\frac{2}{2x+1})^3 dx[/tex]

    2. Relevant equations

    Normal integral equations

    3. The attempt at a solution

    [tex]\int_{1}^{2}(\frac{2}{2x+1})^3 dx = 2^3\int_{1}^{2}\frac{1}{(2x+1)^3} dx[/tex]

    u=2x+1

    [tex]2^3\int_{1}^{2}\frac{1}{u^3} dx = 2^3\int_{1}^{2}u^{-3} dx[/tex]

    Antiderivate of [tex]u^{-3} = \frac{u^{-3+1}}{-3+1} = \frac{u^{-2}}{-2}= \frac{1}{-2u^2}[/tex]

    Plotting the real u in: [tex]\frac{1}{-2(2x+1)^2}[/tex]

    [tex]2^3\int_{1}^{2}\frac{1}{-2(2x+1)^2} = 2^3\left(\frac{1}{-2(2x+1)^2}\right)_1^2 =2^3\left(\frac{1}{-2(2 \cdot 2+1)^2}\right) - 2^3\left(\frac{1}{-2(2 \cdot 1+1)^2}\right) = \left(\frac{2^3}{-50}\right) - \left(\frac{2^3}{-18}\right)[/tex]

    [tex] = \left(\frac{72}{-450}\right) - \left(\frac{200}{-450}\right) = \left(\frac{72-200}{-450}\right) = \left(\frac{-128}{-450}\right) = \frac{128}{450} = \frac{64}{225}[/tex]

    This gives about: 0.284444...

    When I type this function on the calculator, the area between x=1 and x=2 is 0.1422222...

    That is half of my answer, what have I done wrong?
     
    Last edited: Mar 19, 2007
  2. jcsd
  3. Mar 19, 2007 #2
    u=2x+1
    du=2dx
    you need u du not u dx up there in your substitution.
    So you have to divide by 2 outside the integral.
     
    Last edited: Mar 19, 2007
  4. Mar 19, 2007 #3

    disregardthat

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    I see, I have to divide by two.

    But i did not understand why.

    EDIT: I mean, how do you find du?
     
    Last edited: Mar 19, 2007
  5. Mar 19, 2007 #4
    Hey,
    It's just the derivative of 2x+1 with respect to x, which is 2 dx. You always need to find du when you use the u substitution.
     
  6. Mar 19, 2007 #5

    disregardthat

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    All right.
    So how am I to set it up then?

    [tex]\int_{1}^{2}(\frac{2}{2x+1})^3 dx =[/tex]

    Any of these?

    [tex]\frac{d}{dx}2^3\int_{1}^{2}\frac{1}{(u)^3} dx[/tex]

    [tex]\frac{2^3}{du}\int_{1}^{2}\frac{1}{(u)^3} dx[/tex]

    Do you have an example of an solution to a task like this? I want to know how you should set up the solution so I see what is really going on.
     
    Last edited: Mar 19, 2007
  7. Mar 19, 2007 #6
    Hi,
    [tex]\int_{1}^{2}(\frac{2}{2x+1})^3 dx = 2^3\int_{1}^{2}\frac{1}{(2x+1)^3} dx[/tex]

    u=2x+1
    du=2dx

    [tex]\frac{2^3}{2}\int_{1}^{2}\frac{1}{u^3} du = \frac{2^3}{2}\int_{1}^{2}u^{-3} du[/tex]

    You just have to make sure that you've got [tex] u du[/tex] not [tex]u dx[/tex]. The derivative of the u=2x+1 has to be included. You don't want an extra 2 in there, so you just divide it out. That's all. You integrated this perfectly, just missed the du substitution.
    Hope this helps.
    CC
     
    Last edited: Mar 19, 2007
  8. Mar 19, 2007 #7
    or maybe more clear
    u=2x+1
    du=2dx
    [tex]\frac{1}{2}du=dx[/tex]
    then factor out the 1/2.
     
  9. Mar 19, 2007 #8

    disregardthat

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    Oh, right. So when I now have the du after the u^-3 then I just antiderivate just like before right?
    the du in the end doesn't have any meaning to how I am supposed to antiderivate the function, right? Without of course the division of 2^3.

    Uhm, I mean:

    When you have a function, and use substition, you have to find the derivative of the substibtution and then divide the function with it, (it is enough to divide the 2^3) and then just antiderivate as before, only with a du instead of dx. Is that correct?

    And should the du at the end stand there until i have antiderivated u or when i put 2x+1 back again?
     
    Last edited: Mar 19, 2007
  10. Mar 19, 2007 #9
    Correct. When you integrate, the du goes away. Then you put back in your u, just as you did.
     
  11. Mar 19, 2007 #10

    disregardthat

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    All right, thanks for your help
     
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