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Integral and series

  1. Dec 10, 2006 #1
    1. The problem statement, all variables and given/known data Evaluate the indefinite integral as an infinite series.
    [tex]\int{\sqrt{x^3 +1}} [/tex]

    2. Relevant EquationsTaylor Series: f(x) = [tex]\Sigma_{n=0}^{inf} \frac{\f^{n}(a)}{n!} (x-a)^n[/tex]

    Maclaurin Series --> taylor series with a = 0

    3. The attempt at a solution
    I have no idea how to evaluate this integral. I can't use u substitution. From the chapter, I have seen no general formula for such an integral.
    Please give a hint.
  2. jcsd
  3. Dec 10, 2006 #2


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    Firstly, I'm going to assume you meant to write [itex]\int \sqrt{x^3 + 1} \, dx[/itex].

    Read the problem!

    Evaluate the indefinite integral as an infinite series.​

    This suggests infinite series are going to be involved. Is there anything here you can write as an infinite series?
  4. Dec 10, 2006 #3
    Yeah, I know that I'm supposed to use an infinite series...but my book only gives examples that are similar to e^x, sin x, cos x, [tex]tan^{-1}x[/tex], and 1/(1-x)...so I can't really think of what to do.

    [tex]\int{\sqrt{x^3 +1}} [/tex] is not similar to any of those, except maybe the integral of [tex]tan^{-1}x[/tex].
  5. Dec 10, 2006 #4


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    Well, you've learned how to compute infinite series, haven't you? You could try that.

    P.S. Your book doesn't have infinite series expressions for things like [itex]\sqrt{x+1}[/itex]?
  6. Dec 10, 2006 #5
    Could you solve by first expanding the integrand using binomial expansion and then integrate each individual part?
  7. Dec 10, 2006 #6
    Umm...I don't think so. My professor did not cover this section in the detail I would have liked.
    I didn't quite understand everything in the section.

    Please tell me how to compute infinite series! :)
  8. Dec 10, 2006 #7
    I haven't gotten to binomial expansions yet.
  9. Dec 10, 2006 #8
    Oh I see, well, just in case you want to know, the binomial expansion series says you can expand a binomial according to the following formula:

    [tex]1+kx+\frac{k(k-1)}{2!}x^2+\frac{k(k-1)(k-2)}{3!}x^3[/tex] and so on, where k is your exponent.

    In this integral, [tex]\int{\sqrt{x^3 +1}} [/tex] your function is raised to the (1/2) power so k = (1/2). Then you just simply plug in for x and k in the formula for a couple of terms and integrate each term individually.

    I worked out your integral on my TI-89 and evaluated from 0 to 1 and got the answer 1.111448. When I worked it with the binomial expansion method (3 terms) I got 1.149. So you can see the values are pretty close. If I worked out more terms, my answer would have been any closer.

    I know you want to use a different technique, but just in case you can't find it through your method, I just wanted to give you an alternative method to use. Hope it helps and I wish I could help you with the other way, but I am not too sure about that way.
  10. Dec 10, 2006 #9


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    You already know how!

    As you've said (correcting the typos):

    [tex]f(x) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n[/tex]

    at least... you know how to compute the first several terms of the infinite series. (And hopefuly, you can then figure out what the general term is)
  11. Dec 11, 2006 #10


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    Incidentally, the binomial formula you wrote is valid only when k is an integer. So advising him to plug k=1/2 is nonsensical.

  12. Dec 11, 2006 #11


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    Actually, the name "binomial formula" is often applied to the Taylor series for (1+x)^k, even when k is not an integer. You can even write it with binomial coefficients, if you take the generalized definition.
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