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Improper Integral- Check Answer

  1. May 5, 2007 #1
    Hie, I have attempted the following inproper intergral, and I just wanted to check that I have answered it the correctly, if not, any tips would be great. I attached them as pics.

    I attached my steps as pics, as im not sure how to writ ein text on these forums.

    My answer is thus = 0



    PS if you cant view the pics for some reason?, um the Q is

    Evaluate the improper intergral:

    Intergal from 0 to 4 of

    dx /[(4-x)^(1/3)]

    Attached Files:

  2. jcsd
  3. May 5, 2007 #2


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    The attachement is still pending, but here's a tip:

    Take the derivative of your answer (not the constant for your final answer, but the function that the integral evaluated to, which you then plugged the endpoints into), and see if it matches up with what you're integrating
  4. May 6, 2007 #3


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    If you graph y= (4-x)^(-1/3) does it have some portion of its graph (between 0 and 4) below the x-axis? If not, the integral can't be 0!

    Did you forget to evaluate the anti-derivative at x= 0?
  5. May 6, 2007 #4

    Gib Z

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    I get [tex]-3\cdot \sqrt[3]{2}[/tex]
  6. May 6, 2007 #5


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    can't be negative, the function is above the x-axis from 0 to 4

    edit: unless of course you meant anti-derivative evaluated at 0 only
    Last edited: May 6, 2007
  7. May 6, 2007 #6

    Gib Z

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    Sorry I meant [itex]3\cdot \sqrt[3]{2}[/itex]
  8. May 6, 2007 #7
    I forgot to change my limits of integration (ie seen as I used a substitution method to evaluate the integral)

    hence this is what I did

    lower limit now = 4 and upper limit now = 4-a

    hence the definite integral at the end becomes

    lim a-> 4 from the left of -3(a)^(2/3) / 2

    which gives me an answer of


    which doesnt match your answer??? so what am I doing wrong here??

  9. May 6, 2007 #8
    found it.....was doin sum crazy stuff....really fundamentally stupid

    now end up with POSITIVE


    which is equivalent to your answer I think

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