# Integral problem

1. Mar 24, 2004

### 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?

2. Mar 24, 2004

### Hurkyl

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

3. Mar 28, 2004

### h2

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

4. Mar 28, 2004

### himanshu121

PF is looking better

Ya f(0)=1

But what do u mean by constant function f=1

5. Mar 28, 2004

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

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

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

Last edited: Mar 29, 2004
6. Mar 29, 2004

### h2

(The integral of sine is -cosine)

7. Mar 29, 2004

### HallsofIvy

Staff Emeritus

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

8. Mar 29, 2004

### 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.

9. Mar 29, 2004

### himanshu121

ya i agree thats 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

10. Mar 29, 2004