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Integral substitution

  1. Aug 4, 2009 #1
    How do I evaluate the integral [tex]\int_a^b[/tex] f(x4) dx = [tex]\int_y^z[/tex] f(u) dx/du du, where u=x4
    y=a4
    z=b4
     
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  3. Aug 4, 2009 #2

    HallsofIvy

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    How you integrate the function f depends strongly on what f is, don't you think?
     
  4. Aug 4, 2009 #3
    f is [tex]\int_-1^1[/tex] 1+x4 dx
     
  5. Aug 4, 2009 #4
    f is [tex]\int_1^1[/tex] 1+x4 dx
    meant to be a -1 at the bottom of the integral.
    How could I do this then.
     
  6. Aug 4, 2009 #5

    jgens

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    Well, if your integral is simply

    [tex]\int_{-1}^1 1 + x^4\, dx[/tex]

    Then it's pretty simple. Just seperate the integrand so that,

    [tex]\int_{-1}^1 \,dx + \int_{-1}^1 x^4 \, dx[/tex]

    But, based on the context of the question I'm not sure that's what you're looking for. Perhaps you could repost exactly what you mean clearly?
     
  7. Aug 5, 2009 #6
    Sure. Consider the integral [tex]\int_{-1}^1[/tex] 1+x4 dx
    How do I evaluate this integral using the substitution u=x4 and the formula:
    [tex]\int_a^b[/tex] f(x4) dx = [tex]\int_{a^4}^{b^4}[/tex] f(u) dx/du du, where u=x4
     
  8. Aug 5, 2009 #7

    jgens

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    Why would you use that method to evaluate the integral? The integral you have posted can easily be evaluated using the power rule for integration.
     
  9. Aug 5, 2009 #8
    I thought this was a more interesting way but just don't know how to get started.
     
  10. Aug 5, 2009 #9
    Also I someone could help me with this problem.
    How could I evaluate the integral above using u=x4, but seperating it into 2 integrals.
     
  11. Aug 5, 2009 #10

    Cyosis

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    Taking detours isn't a more interesting way, but here goes.
    [itex]u=x^4 \Rightarrow du=4x^3dx[/itex]

    [tex]
    \int 1+x^4 dx=\int \frac{1+u}{4(u^{3/4})} du
    [/tex]

    Making life a lot harder!
     
    Last edited: Aug 5, 2009
  12. Aug 5, 2009 #11

    Gib Z

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    Personally I wouldn't discourage him, I find it a good way to learn how to compute integrals and what methods to use can be supplemented by trying other methods that come to mind, and seeing why they may not be as efficient. Once he computes that integral Cyosis got him, I think he will be able to see in future why it may not be wise to compute an integral as such.
     
  13. Aug 5, 2009 #12

    Cyosis

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    You raise a very good point. I guess it's all too easy to see if something is going to be inefficient or not when you have some experience with the topic at hand.

    Note: I didn't take the limits into account so you still have to do that yourself.
     
  14. Aug 5, 2009 #13
    You are too harsh. Sure, for this function there are easy ways to do the integral. But for other functions f maybe you really should do the substitution. And why not learn to do the substitution on simple problems that can be evaluated in other ways? Then, for example, you can check your answers and see when you make a mistake.
     
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