Integral Problem: Compute f(0) Given f'', f(\pi)=1

[Jupiter]

Suppose f'' is continuous and
\int_0^{\pi}[f(x)+f''(x)]\sin xdx=2. Given that f(\pi)=1, 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??

 

[Hurkyl]

What about a substitution? x \leftarrow \pi - x looks particularly tempting. Integration by parts could be helpful too.

 

[h2]

f(0)=1.
The constant function f=1 is a solution... (and f'' is continuous)

 

[himanshu121]

f(0)=1.
The constant function f=1 is a solution... (and f'' is continuous)

PF is looking better

Ya f(0)=1

But what do u mean by constant function f=1

 

[cookiemonster]

The function f(x) = 1 satisfies the integral.

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

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

cookiemonster

 

[h2]

(The integral of sine is -cosine)

 

[HallsofIvy]

The function f(x) = 1 satisfies the integral.

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

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

cookiemonster

Yes, that's true but the original problem was to show that that is the ONLY function that satisfies the equation.

 

[matt grime]

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.

 

[himanshu121]

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 that's why i wanted out to point that f(x) is not a constant function

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

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

with the given conditions u get f(0)=1 not f(x)=1

 

[cookiemonster]

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

cookiemonster