If the area of the region bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$ is equal to $\alpha$, then $6 \alpha$. equals

Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that

$f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$ for all $x \in[0, \infty)$.

Then the area of the region bounded by $y=f(x)$ and the coordinate axes is

Let the area enclosed between the curves $|y| = 1 - x^2$ and $x^2 + y^2 = 1$ be $\alpha$. If $9\alpha = \beta \pi + \gamma; \beta, \gamma$ are integers, then the value of $|\beta - \gamma|$ equals: