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 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

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

Consider the polynomial
$$f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}.$$
Let $$s$$ be the sum of all distinct real roots of $$f(x)$$ and let $$t = \left| s \right|.$$

The area bounded by the curve $$y=f(x)$$ and the lines $$x=0,$$ $$y=0$$ and $$x=t,$$ lies in the interval