Let $$f$$ be a non-negative function defined on the interval $$[0,1]$$.

If $$\int\limits_0^x {\sqrt {1 - {{(f'(t))}^2}dt} = \int\limits_0^x {f(t)dt,0 \le x \le 1} } $$, and $$f(0) = 0$$, then

Consider the functions defined implicitly by the equation $$y^3-3y+x=0$$ on various intervals in the real line. If $$x\in(-\infty,-2)\cup(2,\infty)$$, the equation implicitly defines a unique real valued differentiable function $$y=f(x)$$. If $$x\in(-2,2)$$, the equation implicitly defines a unique real valued differentiable function $$y=g(x)$$ satisfying $$g(0)=0$$

The area of the region bounded by the curve $$y=f(x),$$ the
$$x$$-axis, and the lines $$x=a$$ and $$x=b$$, where $$ - \infty < a < b < - 2,$$ is :