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 :

Suppose we define the definite integral using the following formula $$\int_\limits{a}^{b} f(x) d x=\frac{b-a}{2}(f(a)+f(b))$$, for more accurate result for

$$c \in(a, b) \mathrm{F}(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$$.

When $$c=\frac{a+b}{c}, \int_\limits{a}^{b} f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$$

$$\int_\limits{0}^{\pi / 2} \sin x d x$$ is equal to: