# Integral problem

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??

## Answers and Replies

Staff Emeritus
Science Advisor
Gold Member
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
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

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

Last edited:
h2
(The integral of sine is -cosine)

Science Advisor
Homework Helper
cookiemonster said:
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.

Science Advisor
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.

himanshu121
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 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