Let f: ℝ → ℝ be a differentiable function such that its derivative...

$I = {∫}_{0}^{\pi}\left({f}^{'}\left(x\right) \cdot \cos x + F\left(x\right) \cdot \cos x\right)dx = 2$

$= {∫}_{0}^{\pi}{f}^{'}\left(x\right) \cdot \cos x \cdot dx + {∫}_{0}^{\pi}F\left(x\right)\cos x \cdot dx = 2$

$= {\left(\cos x \cdot f ( x )\right)}_{0}^{\pi} - {∫}_{0}^{\pi}\left(- \sin x\right) \cdot f\left(x\right)dx + {∫}_{0}^{\pi}F\left(x\right) \cdot \cos x \cdot dx = 2$

$⇒ \left(cos\pi . f\left(\pi\right) - \cos 0 . f\left(0\right)\right) + {∫}_{0}^{\pi}\sin x \cdot f\left(x\right)dx + {∫}_{0}^{\pi}F\left(x\right) \cdot cosxdx = 2$

$⇒ \left(\left(- 1\right) \cdot \left(- 6\right) - f\left(0\right)\right) + {\left(\sin x . F ( x )\right)}_{0}^{\pi} - {∫}_{0}^{\pi}\cos x \cdot F\left(x\right)dx + {∫}_{0}^{\pi}F\left(x\right) \cdot \cos x \cdot dx = 2$

$⇒ 6 - f\left(0\right) + \left(sin\pi \cdot F\left(\pi\right) - \sin 0 . F\left(0\right)\right) = 2$

$⇒ f\left(0\right) = 4$.