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Something interesting i realized. (more than one set of limits, change of variables)

  1. Dec 16, 2009 #1
    suppose i want to find the following integral:
    7
    [tex]\int[/tex]x dx
    3

    now suppose for some demented reason i decided not to do it straightforward and get (49-9)/2=20

    instead i use the substitution x=u2+4u+5
    giving
    u1
    [tex]\int[/tex](u2+4u+5)(2u+4)du
    u0
    u1
    [tex]\int[/tex]2u3+12u2+26u+20 du
    u0
    the indefinite integral is .5u4+4u3+13u2+20u

    now its time to find the limits of integration,
    by the quadratic formula,
    u1=(-4[tex]\pm[/tex][tex]\sqrt{24}[/tex])/2=0.44948974278318,-4.44948974278318

    u0=(-4[tex]\pm[/tex][tex]\sqrt{8}[/tex])/2=-0.5857864376269, -3.4142135623731



    what happens when you enter these limits of integration?
    Here is whats fascinating:
    if i use both values of u1 as the limits of integration, i get 0 (the case is the same with using both values of u0).
    if i use either value of u1 as the upper limit and either value of u0 as the lower limit, i get the correct answer of 20. there is a total of 4 ways to correctly get the answer. it doesnt matter which of the values i use as long as i use a value of u1 as the upper limit and a value of u0 as the lower limit.

    you can go ahead and test it to see what i mean.
     
    Last edited: Dec 16, 2009
  2. jcsd
  3. Dec 17, 2009 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Re: something interesting i realized. (more than one set of limits, change of variabl

    Yes, of course. The two values of u1 both correspond to x= 7 and the integral of any function from "7" to "7" is 0.

    Somewhere in your calculation you are doing the equivalent putting your values of u0 and u1 into the formula u2+ 4u+ 5 and getting values of 7 for both values of u1 and 3 for both values of u0. This behavior is eactly what you should expect.

     
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