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Indefinite Integrals

  1. Jan 25, 2005 #1
    Hello all

    If you have the funtions: [tex] y = x^2, y = sin x, y = 2x^4 [/tex] and you are asked to set up the corresponding integrations, would it be:

    [tex] \int^b_a 2u du = x^3 [/tex]
    [tex] \int^b_a -\cos u du = \sin x [/tex]
    [tex] \int^b_a 8u^3 du = 2x^4 [/tex]

    Thanks
     
  2. jcsd
  3. Jan 26, 2005 #2

    dextercioby

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    Explain why you think your 3 formulas are correct.Personally,i find them wrong...

    Daniel.
     
  4. Jan 26, 2005 #3
    I think they are correct because the derivative of the function placed inside the integral, and the original function is on the right side of the equal sign.

    Isnt't it true that the indefinite integral [tex] \Phi (x) = \int^x_a f(u) du [/tex] of a continuous function [tex] f(x) [/tex] always possesses a derivative [tex] \Phi'(x) [/tex] and [tex] \Phi' (x) = f(x)[/tex]?

    Thanks
     
    Last edited: Jan 26, 2005
  5. Jan 26, 2005 #4

    dextercioby

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    Explain to me the first formula.Why doesn't it include "a" & "b"?

    Daniel.
     
  6. Jan 26, 2005 #5
    Oh I see

    This is an indefinite integral so it should be:

    [tex] \int^x_a 2u \ du = x^2 [/tex]

    Is this right?
     
  7. Jan 26, 2005 #6

    jtbell

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    Indefinite integrals don't have limits. Definite integrals are the ones that have limits. What you have is a definite integral.

    No, you still haven't taken account of the lower limit, a, of your integral.
     
  8. Jan 26, 2005 #7
    ok so [tex] \int^x_a 2u \ du = x^2 [/tex] or [tex] \int^b_x 2u \ du [/tex] and we ahve [tex] \int^x_a 2u \ du +C = x^2 [/tex] [tex] \int^b_x 2u \ du +C [/tex] . Is this correct?
     
  9. Jan 26, 2005 #8

    dextercioby

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    Do you know the Fundamental Theorem of Calculus...??

    What u have written is total nonsense...Definite integrals do not have integration constants...

    Daniel.
     
  10. Jan 26, 2005 #9
    What are you talking about?? Isn't [tex] F(x) = c + \int^x_a 2u \ du = \int 2x \ dx [/tex] an indefinite integral? Let's say we know the the indefinite integral is [tex] \Phi (x) = \int^x_a f(u) \ du [/tex] and we know a primitive function [tex] F(x) = \int f(x) \ dx [/tex]. Then [tex] \Phy (x) = F(x) + c [/tex]. But [tex] \int ^a_a f(u) \ du = 0 [/tex] and [tex] 0 = \Phi (a) = F(a) + c [/tex] and solving for c we get [tex] c = -F(a) [/tex] and [tex] \Phi (x) = F(x) - F(a) [/tex] So to evaluate the definite integral [tex] \int^b_a f(u) \ du = F(b) - F(a) [/tex]
     
    Last edited: Jan 26, 2005
  11. Jan 26, 2005 #10
    I think I got it: IT would simply be : [tex] \int 2x \ dx = x^2 [/tex] or [tex] \int 2x \ dx = -x^2 [/tex] because they differ only by an arbitrary constant.
     
  12. Jan 26, 2005 #11

    dextercioby

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    No.A definite integral is
    [tex] \int_{a}^{b} f(x) \ dx [/tex]

    and an indefinite one is
    [tex] \int f(x) \ dx [/tex]

    Daniel.
     
  13. Jan 26, 2005 #12

    dextercioby

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    No,what u've written is wrong...

    [tex] \int 2x \ dx =x^{2}+C [/tex]

    Find the constant C as to obtain your second formula...

    Daniel.
     
  14. Jan 26, 2005 #13
    [tex] C = -2u^2 [/tex] Is this right?
     
  15. Jan 26, 2005 #14

    dextercioby

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    What has "u" gotta do with your integral,which is done wrt "x"?Better said,who is "u"??

    Daniel.
     
  16. Jan 26, 2005 #15
    oh ok so it would be [tex] C = -2x^2 [/tex] but this isn't a constant is it?
     
  17. Jan 26, 2005 #16

    dextercioby

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    Obviously no.Good thing that u eventually realized it by your own.

    Daniel.
     
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