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Integral problem

  1. Mar 24, 2004 #1
    Suppose [tex]f''[/tex] is continuous and
    [tex]\int_0^{\pi}[f(x)+f''(x)]\sin xdx=2[/tex]. Given that [tex]f(\pi)=1[/tex], compute f(0).
    I'm stuck on this, and I'm not sure where to start. The problem seems like a quickie, and the assumption that f'' is continuous seems curious. Do I have to use the FTC?
    A hint, please??
     
  2. jcsd
  3. Mar 24, 2004 #2

    Hurkyl

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    What about a substitution? [itex]x \leftarrow \pi - x[/itex] looks particularly tempting. Integration by parts could be helpful too.
     
  4. Mar 28, 2004 #3

    h2

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    f(0)=1.
    The constant function f=1 is a solution.... (and f'' is continuous)
     
  5. Mar 28, 2004 #4
    PF is looking better :biggrin:

    Ya f(0)=1

    But what do u mean by constant function f=1
     
  6. Mar 28, 2004 #5
    The function f(x) = 1 satisfies the integral.

    f(x) = 1
    f''(x) = 0

    [tex]\int_0^\pi [f(x) + f''(x)]\sin{x} \,dx = \int_0^\pi [1 + 0]\sin{x} \,dx = - \cos{x}\Big|_0^\pi = 2[/tex]

    cookiemonster
     
    Last edited: Mar 29, 2004
  7. Mar 29, 2004 #6

    h2

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    (The integral of sine is -cosine)
     
  8. Mar 29, 2004 #7

    HallsofIvy

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    Yes, that's true but the original problem was to show that that is the ONLY function that satisfies the equation.
     
  9. Mar 29, 2004 #8

    matt grime

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    Well, the orignal question was to determine what f(0) is. f itself cannot be determined. Since f(x)=1 works, and yields f(0)=1 we can say that if a unique answer to what f(0) is exists it must be f(0)=1. This is a useful observation, but the solution is found be integration by parts.
     
  10. Mar 29, 2004 #9
    ya i agree thats why i wanted out to point that f(x) is not a constant function

    u have
    [tex] \int_o^{\pi} f(x) sinx dx + \int_o^{\pi} f^{''}{(x) sinx dx [/tex]
    u will have

    [tex] - cosx * f(x)|_0^{\pi} + sinx * f^{'} (x)|_0^{\pi} [/tex]

    with the given conditions u get f(0)=1 not f(x)=1
     
  11. Mar 29, 2004 #10
    I was just expanding on h2's post! Don't kill the messenger... =\

    cookiemonster
     
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