Integral problem

  • Thread starter Jupiter
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  • #1
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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??
 

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  • #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.
 
  • #3
h2
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f(0)=1.
The constant function f=1 is a solution.... (and f'' is continuous)
 
  • #4
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h2 said:
f(0)=1.
The constant function f=1 is a solution.... (and f'' is continuous)

PF is looking better :biggrin:

Ya f(0)=1

But what do u mean by constant function f=1
 
  • #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:
  • #6
h2
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(The integral of sine is -cosine)
 
  • #7
HallsofIvy
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cookiemonster said:
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


Yes, that's true but the original problem was to show that that is the ONLY function that satisfies the equation.
 
  • #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.
 
  • #9
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matt grime said:
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.

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
 
  • #10
I was just expanding on h2's post! Don't kill the messenger... =\

cookiemonster
 

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