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Homework Help: Extra Credit Definite Integral - No Solution In Book- Did I do This right?

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \int^{1}_{-1} 2x\sqrt{(1-x^{2})^{3} +1} dx [/tex]


    3. The attempt at a solution

    This is one of those questions in a math book that is almost never assigned and can be seen as "going above and beyond" what the chapter teaches you. I kind of improvised on the solution by substituting variables twice throughout the problem, so I am not sure if the solution is correct. I know that this is a definite integral but I only solved the indefinite portion of it, at least I think that I did. If I can get someone to give me the thumbs up on the indefinite solution I'll plug in the numbers for fun, but this is the part that counts right? ha, anyway:

    Let:

    [tex] u = 1-x^{2} [/tex]

    [tex] du = -2x dx [/tex]

    [tex] dx = -\frac{1}{2}x^{-1} du [/tex]

    So now the problem is going to look like so:

    [tex] \int^{1}_{-1} -\frac{1}{2} \frac{2x}{x} \sqrt{(u)^{3}+1}du [/tex]

    and then:

    [tex] \frac {2x}{x} = 2 \rightarrow -\frac{1}{2}(2) = -1 [/tex]

    so now its:

    [tex] \int^{1}_{-1} - \sqrt{(u)^{3}+1} du [/tex]

    or better yet:

    [tex] \int^{1}_{-1} -((u)^{3}+1)^{1/2} du [/tex]

    and now it gets to the part of where I am not sure if this is legal to do:

    Let:

    [tex] s = (u)^{3}+1 [/tex]

    [tex] ds =3u^{2} du[/tex]

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

    and since:

    [tex] s = (u)^{3} +1 \rightarrow u = (s-1)^{1/3} = (s^{1/3}-1)[/tex]

    so now the problem looks like:

    [tex] \int^{1}_{-1} -\frac{1}{3}(s)^{1/3}(u)^{-2} ds [/tex]

    which is the same as:

    [tex] \int^{1}_{-1} -\frac{1}{3}(s)^{1/3}(s^{1/3}-1)^{-2} ds [/tex]

    and then to get rid of the -2 exponent:

    [tex] \int^{1}_{-1} -\frac{1}{3}(s)^{1/3}(s^{-2/3}-1) ds [/tex]

    then distribute out the binomial:

    [tex] \int^{1}_{-1} -\frac{1}{3}(s^{-1/3} - s^{1/2}) ds [/tex]

    finally distribute the -1/3:

    [tex] \int^{1}_{-1} -\frac{1}{3}s^{-1/3} + \frac{1}{3}s^{1/2} ds [/tex]

    now lets take the anti-derivative of what we have:

    [tex] -\frac{1}{2}(s)^{2/3}+\frac{2}{9}(s)^{3/2} [/tex]

    and then substitute:

    [tex] -\frac{1}{2}(u^{3}+1)^{2/3}+\frac{2}{9}(u^{3}+1)^{3/2} [/tex]

    and once more:

    [tex] -\frac{1}{2}((1-x^{2})^{3}+1)^{2/3}+\frac{2}{9}((1-x^{2})^{3}+1)^{3/2} [/tex]

    then finally to complete the indefinite integral of the problem:

    [tex] -\frac{1}{2}((1-x^{2})^{3}+1)^{2/3}+\frac{2}{9}((1-x^{2})^{3}+1)^{3/2} + C[/tex]
     
  2. jcsd
  3. Apr 29, 2010 #2

    Dick

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    No, it's not right. (s-1)^(1/3) is not the same as (s^(1/3)-1), and then the algebra problems continue. You aren't going to able to find an indefinite integral for that without using elliptic functions. It's much simpler than that. You have a definite integral from x=(-1) to x=1. Any other thoughts?
     
  4. Apr 29, 2010 #3
    I don't understand why you are saying that I have a definite integral that I don't have to find the indefinite integral for, isn't that the whole point? To solve a definite integral you need to find the indefinite integral (which is to say the anti-derivative) of the definite integral and then it is F(b)-F(a), right?
     
  5. Apr 29, 2010 #4

    Dick

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    Yes, but there are some definite integrals you can evaluate without having an indefinite integral. Here's a big hint. If f(x) is odd, i.e. f(-x)=(-f(x)), what's the integral from -1 to 1 of f(x)?
     
  6. Apr 29, 2010 #5

    Dick

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    Yes, but there are some definite integrals you can evaluate without having an indefinite integral. Here's a big hint. If f(x) is odd, i.e. -f(x)=f(-x), what's the integral from -1 to 1 of f(x).
     
    Last edited: Apr 29, 2010
  7. Apr 29, 2010 #6
    Well, I am not sure of exactly what you mean, if you mean this:

    [tex] \int^{1}_{-1} -x dx [/tex]

    then the answer would be 0... is that what you getting at?
     
  8. Apr 29, 2010 #7

    Dick

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    Yes. The integral of sin(x), x*exp(x^2), cos(x)*x, cos(x)*sin(x) etc from -1 to 1 is also 0. You don't have to find the indefinite integral to know that. Could your function be one of this group?
     
  9. Apr 29, 2010 #8
    Right, well I typed the function into a graphing calculator to look at the graph and I know that the answer is 0 just from the way the graph looks, I was just interested in doing the math to find the indefinite integral, that's all. Let me ask you this, could the problem have been solved the way that I tried to solve it? I know that the algebra is bad in there, but saying that it is right, would it have worked out or is it just not possible to find the indefinite integral of this function using this method?
     
  10. Apr 29, 2010 #9

    Dick

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    No, the indefinite integral is not an elementary function. Like I said, you can find one by using exotic functions like elliptic integrals, but that's out of your league. You did the right thing by graphing it. Now can you tell me how to figure out that it's zero without graphing it? What property of the function makes the integral zero?
     
  11. Apr 30, 2010 #10
    Well I think that the integral going from -1 to 1 makes it an odd function, and definite integrals of odd functions makes it = 0.
     
  12. Apr 30, 2010 #11

    Mark44

    Staff: Mentor

    You're approaching this from the wrong end. The limits of integration aren't what make a function odd or not: it's the function you're integrating that is either odd or it isn't. If the integrand is an odd function, then you can take advantage of the limits of integration to say something about the value of the definite integral.
     
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