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Improper integral

  1. Oct 20, 2007 #1

    tony873004

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    evaluate [tex]\int_0^{33} {\left( {x - 1} \right)^{ - 1/5} } \,dx[/tex]
    The back of the book gives 75/4 as the answer. I get 80/4. It has an asymptote at 1 since 0^(-1/5) DNE, and 0.00000001^(1/5) equals a number that grows larger as I add more 0's.

    But any time I try to raise a negative number ^(1/5) I get an error. It just doesn't seem like 0-1 is in the Domain of this function. Even when I graph it, I see it shoot down from the x=1 asymptote, and then make a hard turn right as it approaches the x-axis. I don't see any Range in this function below x=1.

    But somehow, the integral from 0 to 1 must equal -5/4 for my answer to match the back of the book's.

    Here's my effort:
    [tex]
    \begin{array}{l}
    \int_0^{33} {\left( {x - 1} \right)^{ - 1/5} } \,dx \\
    \\
    \,\,\,\,\,\,\,\,\,\,u = x - 1,\,\,\frac{{du}}{{dx}} = 1,\,\,dx = du \\
    \\
    \int_*^* {u^{ - 1/5} } \,du = \mathop {\lim }\limits_{t \to 0^ + } \int_*^* {u^{ - 1/5} } \,du \\
    \\
    \mathop {\lim }\limits_{t \to 0^ + } \left. {\frac{{u^{4/5} }}{{4/5}}} \right|_*^* \,\, = \,\,\mathop {\lim }\limits_{t \to 0^ + } \left. {\frac{{u^{4/5} }}{{4/5}}} \right|_*^* \left. { = \mathop {\lim }\limits_{t \to 0^ + } \frac{{5\left( {x - 1} \right)^{4/5} }}{4}} \right|_t^{33} = \mathop {\lim }\limits_{t \to 0^ + } \left[ {\left( {\frac{{5\left( {33 - 1} \right)^{4/5} }}{4}} \right) - \left( {\frac{{5\left( {t - 1} \right)^{4/5} }}{4}} \right)} \right] = \\
    \\
    \left( {\frac{{5\left( {33 - 1} \right)^{4/5} }}{4}} \right) - \left( {\frac{{5\left( {0^ + - 1} \right)^{4/5} }}{4}} \right) = \frac{{5 \cdot 16}}{4} - \frac{{5\left( { - 1} \right)^{4/5} }}{4} = \frac{{80}}{4} - \frac{{5\left( { - 1} \right)^{4/5} }}{4} \\
    \end{array}
    [/tex]
    But here's my problem.
    If [tex]
    \frac{{5\left( { - 1} \right)^{4/5} }}{4}=\frac{5}{4}[/tex] then I'd have the right answer.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Oct 20, 2007
  2. jcsd
  3. Oct 20, 2007 #2

    Curious3141

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    If you restrict the range of the function to reals, then f(x) = x^(4/5) will give f(-1) = 1. Negative numbers can have real odd roots.

    But I still can't follow your working. In the original integral, your asymptote occurs at x = 1. You should be splitting up your integral like so :

    [tex]\int_{0}^{33}(x-1)^{-\frac{1}{5}}dx = \lim_{t \rightarrow 1}\int_{0}^{t}(x-1)^{-\frac{1}{5}}dx + \lim_{t \rightarrow 1}\int_{t}^{33}(x-1)^{-\frac{1}{5}}dx[/tex]

    then working out the individual integrals.

    You decided to make the substitution, which changes the bounds and the value of the asymptote. In this case, the bounds go from u = -1 to u = 32, and the value of t becomes 0. You still get the same answer.
     
  4. Oct 20, 2007 #3

    Gib Z

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    Try [tex]\lim_{a\to 0} \left( \int_0^{1-a} (x-1)^{-1/5} dx + \int_{1+a}^{33} (x-1)^{-1/5} dx[/tex].

    EDIT: was too slow :(
     
  5. Oct 20, 2007 #4

    tony873004

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    Thanks for the replies :)

    I know that substitution changes the bounds. In my examples, I just put astericks on the bounds after converting to du, so I don't have to compute them. I switch back to dx before solving which is why I don't bother to compute the new bounds.

    I tried splitting it up from 0 to 1 and 1 to 33, but I ran into the exact same problem. Here's my effort. This is only the 0-1 part. The 1 to 33 part I get 80/4:
    [tex]\begin{array}{l}
    \int_0^1 {\left( {x - 1} \right)^{ - 1/5} \,dx} = \mathop {\lim }\limits_{t \to 1^ - } \int_0^1 {\left( {x - 1} \right)^{ - 1/5} \,dx} = \mathop {\lim }\limits_{t \to 1^ - } \left( {\frac{5}{4}\left( {\left( {t - 1} \right)^{4/5} - \left( { - 1} \right)^{4/5} } \right)} \right) = \\
    \\
    \frac{5}{4}\left( {0 - \left( { - 1} \right)^{4/5} } \right) = \frac{5}{4}\left( { - \left( { - 1} \right)^{4/5} } \right) = \frac{{ - 5\left( { - 1} \right)^{4/5} }}{4} \\
    \end{array}[/tex]
     
  6. Oct 20, 2007 #5

    Gib Z

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    In the Reals [itex](-1)^{4/5} = 1^{1/5} = 1[/itex], giving you the correct answer.
     
  7. Oct 21, 2007 #6

    tony873004

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    Thanks. I'll take your word for it, but how do you get that? I just looked in a pre-calc textbook about imaginary numbers and didn't find anything that allowed me to draw that conclusion.
     
  8. Oct 21, 2007 #7

    Gib Z

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    haha well you did look at a pre calc textbook about imaginary numbers and this integral is talking completely about the reals. Your integral is over the real interval [0,33], no complex numbers.
     
  9. Oct 21, 2007 #8

    tony873004

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    ok, but if I type (-1)^(4/5) into Google, it tells me (-1)^(4 / 5) = -0.809016994 + 0.587785252 i . I guess that's where I got confused. My TI-84 tells me (-1)^(4/5)=1 like you say. My computer eval calculator gives me an error. As long as my TI-84 says it's so, I guess I don't have to justify that step. But I'm still curious: how do you get 1^(1/5) from (-1)^(4/5)? And why does Google calculator want to give me the answer in imaginary form?
     
  10. Oct 21, 2007 #9

    tony873004

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    and also the online graphing calculator refuses to plot anything less than x=1. How can there be an area when there's no curve?
     
  11. Oct 21, 2007 #10

    Gib Z

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    Are you checking all quadrants? Because graphamatica has no problem graphing it...and the inbuilt numerical integrator confirms the 75/4 answer to 3 sig figs.

    As a side exercise, note that the (x-1) arguement rather than x just shifts the graph 1 unit to the right. So graph y=(x-1)^(-1/5). It is the same curve as xy^5 =1.

    What that tells me is that its a similar shape to a rectangular hyperbola, but the y axis becomes as asmytote much faster than the x axis because of the higher power of y. To see what i mean, graph it on some program.
     
  12. Oct 21, 2007 #11
    Is it correct to say that you used the following steps:

    (-1)^(4/5)=[(-1)^4]^(1/5)=[1]^(1/5)=1

    This is tricky. I would start with (-1)^(1/5). This presents 5 different numbers. These are approximately (0.809+i0.588); (0.809-i0.588); (-0.309+i0.951); (-0.309-i0.951) and -1.
    So, only one real number. Raising these to the power 4 gives approx. (-0.809+i0.588); (-0.809-i0.588); (0.309-i0.951); (0.309+i0.951) and 1. The outcome is the same, only
    one real number. If this is used in the integral you get the correct answer 75/4. This way of working gives a detailed view on what is happening, albeit a way bit more work.
    Also the reason for the different results with the different methods.

    Integral solved, case closed.
     
  13. Oct 22, 2007 #12

    tony873004

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    I asked in office hours. The teacher pointed out that if I treat -0.809016994 + 0.587785252 i, with the 1st number on the x-axis, and the 2nd number on the y-axis, and then use pothagorean theorum, I'd get 1. She was right:
    sqrt(0.809016994^2 + 0.587785252^2)=1

    Thanks everyone for your helpful replies.
     
  14. Oct 23, 2007 #13

    Gib Z

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    That is true but you shouldn't have needed to go there at all! It is a completely real integral, and the only solution in the reals for [itex](-1)^{4/5}[/itex] is 1 !! This kind of thinking might harm you when doing functions that are one to one in the reals, but not so in the complex numbers such as the exponential function.
     
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