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Math Help - stuck on an integral

  1. #1
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    stuck on an integral

    49
    S dx/ (27+2x)^(2/3)
    0



    I'm pretty confused on what to do, help me figure it out please.
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  2. #2
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    u = 27+2x --> du = 2 dx
    change the value of upper/lower limits
    and it will be easy O_o
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  3. #3
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    I understand changing the limits but I can't figure out how to integrate this properly, I keep getting an incorrect answer
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  4. #4
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    \int_0^{49}\frac{1}{(27+2x)^{\frac{2}{3}}}dx

    So u = 27 + 2x

    du = 2dx
    dx = du/2

    change the limits, 27 to 125

    \int_{27}^{125}\frac{1}{2(u)^{\frac{2}{3}}}du


    \int_{27}^{125}\frac{1}{2}*u^{\frac{-2}{3}}du

    power rule:

    \frac{3}{2}*u^{\frac{1}{3}} from 27 to 125, can you take it from there?
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  5. #5
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    hmm that is exactly what I've been doing but when I input my answer through my online hw on webassign: (3/2)(277^(1/3)-3) it tells me I'm wrong. :/ anything wrong with that answer?
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  6. #6
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    \frac{3}{2}*u^{\frac{1}{3}}<br />

    \frac{3}{2}*(125^{\frac{1}{3}})-\frac{3}{2}*(27^{\frac{1}{3}})

    \frac{3}{2}*5-\frac{3}{2}*3

    I don't see where you get this 277 from.
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  7. #7
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    do you not replace u by (27+2x)?
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  8. #8
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    well you changed your limits to accomodate for u right?

    x goes from 0 to 49.

    u= 27 + 2x

    x = 0 => u=27 + 2*0 = 27

    x = 49 => u=27+2*49=27+98=125
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  9. #9
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    we covered the limits several replies ago, but 3 was correct thanks selb, I'm not sure where I got the idea i needed to throw 27+2x back in there
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  10. #10
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    I'm still not sure where you get this 277 from but the point of the last question was to check that you followed why I simply substituted 125 and 27 in for u instead of replacing it with the 27+2x, because i'm lazy and don't feel like substituting back in if I already know what it's going to be.

    But here is the solution with substituting the 27+2x back in:

    \frac{3}{2}*u^{\frac{1}{3}}<br />
u goes from 27 to 125


    \frac{3}{2}*(27+2x)^{\frac{1}{3}} x goes from 0 to 49

    =\frac{3}{2}*(27+2(49))^{\frac{1}{3}}-\frac{3}{2}*(27+2(0))^{\frac{1}{3}}


    =\frac{3}{2}*(27+98)^{\frac{1}{3}}-\frac{3}{2}*(27)^{\frac{1}{3}}

    =\frac{3}{2}*(125)^{\frac{1}{3}}-\frac{3}{2}*(27)^{\frac{1}{3}}

    =\frac{3}{2}*5-\frac{3}{2}*3

    does that help?
    Last edited by seld; September 14th 2009 at 08:34 PM. Reason: editing so it's easier to read
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