If $f(\pi)=2$ and $\int\limits_0^\pi\left(f(x)+f''(x)\right) \sin x d x=5$ then $f(0)$ is equal to, (it given that $f(x)$ is continuous in $[0, \pi])$

3

$\int\limits_0^\pi\left(f(x)+f''(x)\right) \sin x d x$

$=\int\limits_0^\pi f(x) . \sin x d x+\int\limits_0^\pi f''(x) \sin x d x$

$=f(x) .-\left.\cos x\right|_0 ^\pi+\int\limits_0^\pi \cos x . f'(x) d x+\left.\sin x . f'(x)\right|_0 ^\pi-\int\limits_0^\pi \cos x . f'(x) d x$

$\Rightarrow f(\pi)+f(0)=5$ (given)

$\Rightarrow f(0)=5-f(\pi)=5-2=3$