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Another method of solving integrals?

  1. Aug 14, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\text {Find m such that }\displaystyle\int^m_4 \frac{1}{x\sqrt{x}}\,dx = .9[/tex]

    2. Relevant equations




    3. The attempt at a solution

    [tex]\displaystyle\int^m_4 \frac{1}{x\sqrt{x}}\,dx = .9 \Rightarrow \displaystyle\int^m_4 x^{-3/2}\,dx = .9 \Rightarrow -2m^{-1/2} +2(4)^{-1/2} = .9 \Rightarrow m = \frac {4}{1.9^{2}} = 1.108[/tex]


    If I plug this value back into the original integral, it comes out as [tex]-.9[/tex]. Should I solve this integral another way? Also, an extention of the problem is

    [tex] \text {Explain why there is no number m such that} \displaystyle\int^m_4 \frac{1}{x\sqrt{x}}\,dx = 1.1[/tex]

    It seems like that has to deal with convergence issues, something I'm just beginning to learn. Any hints for the first step?
     
  2. jcsd
  3. Aug 14, 2008 #2

    nicksauce

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    Check your numbers again, I get m = 400.

    To do the second part, show that the integral is maximal if m=infinity, and show that the value of the integral in that case is less than 1.1.
     
  4. Aug 14, 2008 #3

    I don't know latex but that bold part is wrong (Addition mistake) I also get 20^2

    You added when you should subtract
     
  5. Aug 14, 2008 #4
    Thanks, in order to show that it is maximal at infinity, does it suffice to say that since x must always be greater than 0, then the integral is maximal at infinity? That doesn't sound very rigorous...how would you phrase it?
     
  6. Aug 14, 2008 #5

    nicksauce

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    I think it should be enough to show that the function
    [tex]g(t)=\int_4^{t}\frac{dx}{x\sqrt{x}}[/tex] is always increasing (you could use FTC, with g'(t)>0), and so it follows that g(t) is maximal as t->infinity.
     
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