Early uni/college Calculus (one question)

I cant read the value of the integral. It looks either like $4$ or $4\pi$. I assume its $4$, otherwise you need to correct the very first and the very last step. We split the integral into two sums:
$$4=\int_0^\pi(f(x) + f''(x))\sin(x)\,dx=\int_0^\pi f(x)\sin(x)\,dx + \int_0^\pi f''(x)\sin(x)\,dx$$
now we do partial integration with the first sum to get (taking $u' =\sin(x)$, $v=f(x)$ so that $u = -\cos(x)$):
$$\int_0^\pi f(x)\sin(x)\,dx = [f(x)\cdot(-\cos(x))]_{x=0}^{x=\pi}-\int_0^\pi f'(x)(-\cos(x))\,dx$$
which becomes:
$$=f(\pi)\cdot(- 1)-f(0)\cdot (-1) +\int_0^\pi f'(x)\cos(x)\,dx = f(0)-f(\pi)+\int_0^{\pi}f'(x)\cos(x)\,dx$$

We do partial integration with the second term (here we take $u'=f''(x)$ and $v=\sin(x)$ so that $u=f'(x)$ and $v=\cos(x)$):

$$\int_0^\pi  f''(x)\sin(x) = [f'(x)\cdot \sin(x)]_{x=0}^{x=\pi} -\int_0^\pi f'(x)\cos(x)\,dx$$
which becomes:
$$f'(\pi)\cdot 0-f''(0)\cdot 0-\int_0^\pi f'(x)\cos(x)\,dx= \int_0^\pi f'(x)\cos(x)\,dx$$

adding both terms together again yields

$$4 = f(0)-f(\pi) +\int_0^\pi f'(x)\cos(x)\,dx - \int_0^\pi f'(x)\cos(x)\,dx = f(0)-f(\pi) = f(0)-9$$
turn this around to get:
$$f(0)= 4+9 = 13$$