Area of the region

$$\left\{ {\left( {x,y} \right) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,5y \le x + 9 \le 15} \right\}$$

is equal to

The area enclosed by the curves $$y = \sin x + {\mathop{\rm cosx}\nolimits} $$ and $$y = \left| {\cos x - \sin x} \right|$$ over the interval $$\left[ {0,{\pi \over 2}} \right]$$ is

Let f $$:$$$$\left[ { - 1,2} \right] \to \left[ {0,\infty } \right]$$ be a continuous function such that
$$f\left( x \right) = f\left( {1 - x} \right)$$ for all $$x \in \left[ { - 1,2} \right]$$

Let $${R_1} = \int\limits_{ - 1}^2 {xf\left( x \right)dx,} $$ and $${R_2}$$ be the area of the region bounded by $$y=f(x),$$ $$x=-1,$$ $$x=2,$$ and the $$x$$-axis. Then

Let the straight line $$x=b$$ divide the area enclosed by
$$y = {\left( {1 - x} \right)^2},y = 0,$$ and $$x=0$$ into two parts $${R_1}\left( {0 \le x \le b} \right)$$ and
$${R_2}\left( {b \le x \le 1} \right)$$ such that $${R_1} - {R_2} = {1 \over 4}.$$ Then $$b$$ equals