Let $$g\left( x \right) = \int\limits_0^x {f\left( t \right)dt,} $$ where f is such that
$${1 \over 2} \le f\left( t \right) \le 1,$$ for $$t \in \left[ {0,1} \right]$$ and $$\,0 \le f\left( t \right) \le {1 \over 2},$$ for $$t \in \left[ {1,2} \right]$$.
Then $$g(2)$$ satisfies the inequality

If for a real number $$y$$, $$\left[ y \right]$$ is the greatest integer less than or
equal to $$y$$, then the value of the integral $$\int\limits_{\pi /2}^{3\pi /2} {\left[ {2\sin x} \right]dx} $$ is

$$\int\limits_{\pi /4}^{3\pi /4} {{{dx} \over {1 + \cos x}}} $$ is equal to

If $$\int_0^x {f\left( t \right)dt = x + \int_x^1 {t\,\,f\left( t \right)\,\,dt,} } $$ then the value of $$f(1)$$ is