Let $f:\mathrm{ℝ}\to \mathrm{ℝ}$ be a differentiable function such that its derivative $f^{\text{'}}$ is continuous and $f\left(\pi \right)=-6$. If $F:\left[0,\pi \right]\to \mathrm{ℝ}$ is defined by $F\left(x\right)={\int }_0^{x}f\left(t\right)dt,$ and if

${\int }_0^{\pi }\left(f^{\text{'}}\left(x\right)+F\left(x\right)\right)\cos x dx=2$, then the value of $f\left(0\right)$ is_______

$I={\int }_0^{\pi }\left(f^{\text{'}}\left(x\right)·\cos x+F\left(x\right)·\cos x\right)dx=2$

$={\int }_0^{\pi }{f}^{\text{'}}\left(x\right)·\cos x·dx+{\int }_0^{\pi }F\left(x\right)\cos x·dx=2$

$={\left(\cos x·f\left(x\right)\right)}_0^{\pi }-{\int }_0^{\pi }\left(-\sin x\right)·f\left(x\right)dx+{\int }_0^{\pi }F\left(x\right)·\cos x·dx=2$

$\Rightarrow \left(\cos \pi .f\left(\pi \right)-\cos 0.f\left(0\right)\right)+{\int }_0^{\pi }\sin x·f\left(x\right)dx+{\int }_0^{\pi }F\left(x\right)·\cos xdx=2$

$\Rightarrow \left(\left(-1\right)·\left(-6\right)-\mathrm{f}\left(0\right)\right)+{\left(\sin x.F\left(x\right)\right)}_0^{\pi }-{\int }_0^{\pi }\cos x·F\left(x\right)dx+{\int }_0^{\pi }F\left(x\right)·\cos x·dx=2$

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

$\Rightarrow f\left(0\right)=4$.