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How would you integrate this?

  1. May 29, 2014 #1

    Nathanael

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    ∫x(x[itex]^{2}[/itex]+1.44)[itex]^{-1/2}[/itex](dx)


    The only technique of integration I know of is to recognize the antiderivative (which I can do for this one but) I'm tired of being limited by antiderivatives.


    I was wondering how would you go about solving this?
     
    Last edited: May 29, 2014
  2. jcsd
  3. May 29, 2014 #2

    jbunniii

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    Have you tried a substitution?
     
  4. May 29, 2014 #3

    Ray Vickson

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    If you are tired of antiderivatives, you are tired of integration. The basic way we do integration is by finding antiderivatives or else---when we cannot find them---using infinite series or approximations of various kinds.
     
  5. May 29, 2014 #4

    Nathanael

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    I don't know what that is. What's the general idea behind substitutions?

    Edit: Sorry if this is basic and I'm wasting your time, it's just that I've never seen a single method other than anti-derivatives.
     
  6. May 29, 2014 #5

    jbunniii

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    You haven't seen a method of solution that starts with "let ##u = \ldots##"?
     
  7. May 29, 2014 #6

    micromass

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    Please don't get me wrong, I really am trying to be helpful. But I think you should read a basic calculus text. Or watch some Khan Academy videos on integration. None of the explanations we give will be helpful to you if you don't read read up on integration first.
     
  8. May 29, 2014 #7

    Nathanael

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    Oh, yes, but isn't that just a (more "careful") way of finding the antiderivative?

    I understand the concept of integration pretty well (this integral comes from a physics question). I can find correct integrals easily, it's just that I don't know any computational techniques (except for one). I was under the impression that there are various ways to compute an integral, and was just curious about an alternate way.

    No explanation has even been given so how do you know that?...
    Besides, I didn't post this with the intention of being taught a whole new idea step by step, I simply wanted to be introduced to a new concept that I could explore by myself.

    But perhaps my impression (of there being multiple ways to compute integrals) was wrong.
     
  9. May 29, 2014 #8

    jbunniii

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    Sure, but it's an important enough technique that it's usually the first thing to try unless the antiderivative is obvious.

    So looking at your problem, what is a good candidate for "let ##u = \ldots##"?
     
  10. May 29, 2014 #9
    There are. One can for example use the Residue theorem, the parity properties of functions, numerical methods and various other methods to calculate (some) definite integrals. However, calculating the indefinite integral of a function means finding the function(s) whose derivative the integrand is -- its antiderivative. So, one way or another, that's what you're doing.
     
    Last edited: May 29, 2014
  11. May 29, 2014 #10

    Nathanael

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    Ahh, I should have said, in the physical problem I only need to integrate it from x=1 to x=3.
    So this limitation could open up more ways of integrating it?

    I guess it would probably be unneccessarily difficult to do it an alternate way, wouldn't it?

    I would suspect u=(x[itex]^{2}+1.44)[/itex] but I don't understand how that does any good.

    edit: I suppose the proper way to solve this would be? to recognize that x is the derivative of (x[itex]^{2}+1.44)[/itex] and so [itex](x^{2}+1.44)^{1/2}[/itex] should be proportional to the answer (and the constant of proportionality just happens to be 1) ?
     
    Last edited: May 29, 2014
  12. May 29, 2014 #11
    It would most likely make it more difficult indeed, especially considering the integrand has a simple and elementary antiderivative. One of the advantages of the "tricks" you can do with definite integrals is that they might allow calculating a definite integral even if the integrand has no elementary antiderivative: The canonical example is [itex]\exp(-x^2)[/itex] whose antiderivative is non-elementary, even though [itex]\intop_{-\infty}^{\infty} \exp(-x^2)\mathrm{d}x=\sqrt{\pi}[/itex] is something that's easy to calculate. But typically if the answer is as simple as it is in this case, simply finding the antiderivative with the "usual" methods is the most effective way of doing things.

    EDIT: Also, check the first post. I think there might be a - that's supposed to be a + in your integrand.
     
  13. May 29, 2014 #12

    Nathanael

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    Thank you, and yes, you're right, I meant to put +1.44
     
  14. May 29, 2014 #13

    jbunniii

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    Here is the usual way of expressing it:

    Let ##u = x^2 + 1.44##. Then ##du = 2x dx##. Therefore, ##x dx = \frac{1}{2}du##. Substituting into
    $$\int \frac{x}{(x^2 + 1.44)^{1/2}}dx$$
    we obtain
    $$\frac{1}{2}\int \frac{du}{u^{1/2}}$$
    In this way, we have transformed the problem into one whose antiderivative is known at a glance. We calculate that antiderivative (a function of ##u##), and then substitute ##u = x^2 + 1.44## into the result.

    If this had been a definite integral, we would also have to compute the new endpoints for the integral with respect to ##u##, or wait until we transform the answer back into a function of ##x## before applying the endpoints.
     
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