# Integral problem

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?

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

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

h2 said:
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

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:
(The integral of sine is -cosine)

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

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

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

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

matt grime
Homework Helper
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.

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
$$\int_o^{\pi} f(x) sinx dx + \int_o^{\pi} f^{''}{(x) sinx dx$$
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

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