Value of function defined using its integral

Question

Suppose we have a function $f$. Knowing its second derivative is continuous, $f(\pi)=1$, and

$$\int_0^\pi{(f(x)+f''(x))\sin{x}dx}=2$$

What is the value of $f(0)$?

Answer 1

HINT:

Write $$\int_0^\pi(f(x)+f''(x))\sin{x}\,dx=\int_0^\pi f(x)\sin{x}\,dx+\int_0^\pi f''(x)\sin{x}\,dx.$$ For the first integral use integration by parts twice, differentiating $f(x)$ twice and integrating $\sin x$ twice. This gives us $$\begin{align}\int_0^\pi f(x)\sin{x}\,dx&=[-f(x)\cos x]_0^\pi+\int_0^\pi f'(x)\cos x\,dx\\&=f(\pi)+f(0)+\left([f'(x)\sin x]_0^\pi-\int_0^\pi f''(x)\sin x\,dx\right)\end{align}$$

For the second integral, use integration by parts twice, differentiating $\sin x$ twice and integrating $f''(x)$ twice. $$\begin{align}\int_0^\pi f''(x)\sin{x}\,dx&=[f'(x)\sin x]_0^\pi-\int_0^\pi f'(x)\cos x\,dx\\&=-\left([f(x)\cos x]_0^\pi+\int_0^\pi f(x)\sin x\,dx\right)\end{align}$$

Now add them and equate to $2$ to solve for $f(0)$.