The minimum value of the twice differentiable function $$f(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$$, $$x \in \mathbf{R}$$, is :

Let $$f(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}$$.

Consider

$$(\mathrm{S} 1): f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$$

$$(\mathrm{S} 2): \int\limits_{-2}^{2} f(x) \mathrm{d} x=12$$

Then,

$$\int\limits_{0}^{2}\left(\left|2 x^{2}-3 x\right|+\left[x-\frac{1}{2}\right]\right) \mathrm{d} x$$, where [t] is the greatest integer function, is equal to :

Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a function defined as

$$f(x)=a \sin \left(\frac{\pi[x]}{2}\right)+[2-x], a \in \mathbb{R}$$ where $$[t]$$ is the greatest integer less than or equal to $$t$$. If $$\mathop {\lim }\limits_{x \to -1 } f(x)$$ exists, then the value of $$\int\limits_{0}^{4} f(x) d x$$ is equal to