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Approximation methods that can be applied

  1. Sep 11, 2004 #1
    How do I find integrals like
    [tex] \int_{a}^{b} \left( x^2 + 1 \right)^2 \ dx [/tex]. This one is easy, since I can just turn it into [tex] \int_{a}^{b} \left( x^4 + 2x^2 + 1 \right) \ dx [/tex]. But what if it would say [tex] \int_{a}^{b} \left( x^2 + 1 \right)^{40} \ dx [/tex]? What technique should I use?
     
  2. jcsd
  3. Sep 11, 2004 #2
    There are certain approximation methods that can be applied.
     
  4. Sep 11, 2004 #3

    matt grime

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    Or substituion methods to obtain a (possibly hyperbolic) trig integral. Often you'll end up with trying to integrate something that we can write as:

    I(n) = int f(x)dx

    where f( x) is some function of x, and the n indicates some exponent or coefficient (the 40 in your case), then doing something like integration by parts you'll end up with

    I(n-1) + something= I(n)

    and you get an iteration going that lets you figure out how to solve the integral.
     
  5. Sep 11, 2004 #4

    matt grime

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    I think you need to fix that latex.
     
  6. Sep 11, 2004 #5
    Yeah, I tried to (and then I deleted my post before I noticed that you had replied to it. Doh). LaTeX doesn't seem to like the combination of an integral sign, sum sign and a binomial coefficient... Let's try again:

    [tex]\int (x^2 + 1)^{40} dx = \int \sum^{40}_{k=0} C^{40}_{k} x^{2k} dx = \left( \sum^{40}_{k=0} \frac{C^{40}_{k}}{2k + 1} x^{2k + 1} \right) + D.[/tex]

    Actually calculating those binomial coefficients is going to suck though :P
     
    Last edited: Sep 11, 2004
  7. Sep 11, 2004 #6
    Leaving it as a series solution is absolutely fine--- consider

    [tex]\int e^{x^2} dx[/tex]

    There is no substitution you can use, but you can use the general series form of e^x and plug in x^2 and then integrate the sum as a whole, and leave you answer as that.
     
  8. Sep 11, 2004 #7
    this maybe silly, but wouldn't a trig sub then a integral table work for something like that?
     
  9. Sep 11, 2004 #8

    Dr Transport

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    try the trig substitution for the original problem, let
    [tex] x = \tan(y) [/tex], then the portion in the parentheses becomes [tex] \sec(y)^2 [/tex] from there you can make the substitutions and complete the integral. As for the [tex] e^{x^{2}} [/tex] over a finite interval, you have to use an integral table because it can only be done numerically, if the limits are infinite in extent, then the integral is [tex] \frac{\pi}{2} [/tex],
     
  10. Sep 11, 2004 #9
    [tex]\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}[/tex]
     
  11. Sep 12, 2004 #10

    HallsofIvy

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    Yes, but the theelectricchild's point was that there is no "elementary" anti-derivative for \[e^{x^2}\]. The sad fact is that the great majority (in a very specific sense, "almost all") of nice "elementary" functions, even though they have an anti-derivative, the anti-derivative is not an "elementary" function.
     
  12. Sep 12, 2004 #11

    Dr Transport

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    I forgot the square root.........
     
  13. Sep 12, 2004 #12
    Hehe its ok, thats one of the first places I have seen a "squareroot" of pi... its an integral you can actually solve using a double integration step.

    [tex]\int_{0}^{\infty} e^{-x^2} dx[/tex]

    Convert to DI problem:

    [tex]{(\int_{0}^{\infty} e^{-x^2} dx)}^2[/tex]

    [tex]=(\int_{0}^{\infty} e^{-x^2} dx) (\int_{0}^{\infty} e^{-y^2} dy)[/tex]

    [tex]=\int_{0}^{\infty}\int_{0}^{\infty} e^{-x^2}e^{-y^2} dxdy[/tex]

    [tex]=\int_{0}^{\infty}\int_{0}^{\infty} e^{-{(x^2+y^2)}} dxdy[/tex]

    From here using a Change of Variables step or noticing the region we are integrating is simply a semicircular arc, we can use polar coordinates:

    [tex]=\int_{0}^{\frac{\pi}{2}}\int_{0}^{R} e^{-r^2}r drd\theta[/tex]

    Where we take the limit of this as R --> infinity and the extra r in the integrand is from the Jacobian when switching to polar coordinates (proven in the change of variables theorem)

    And NOW we can use a U-substitution to easily solve this iterated double integral:

    [tex]{(\int_{0}^{\infty} e^{-x^2} dx)}^2 = \frac{\pi}{4}[/tex]

    So [tex]\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}[/tex]

    It seems like we can solve quite a few "definite" or "improper" integrals of non elementary functions using multivar. calculus. Does anyone know any other methods? Thanks.
     
    Last edited: Sep 12, 2004
  14. Sep 12, 2004 #13
    I like integrating things.
     
  15. Sep 13, 2004 #14
    Where did the R come from again?
     
  16. Sep 15, 2004 #15
    Well taking an improper integral requires you to take the limit of the evaluated integral as some letter in place of infinity goes to it--- I didnt learn to do limits in TeX yet Ill show it later.
     
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