Let $$f\left( x \right) = \int {{e^x}\left( {x - 1} \right)\left( {x - 2} \right)dx.} $$ Then $$f$$ decreases in the interval

If $$f\left( x \right) = \left\{ {\matrix{ {{e^{\cos x}}\sin x,} & {for\,\,\left| x \right| \le 2} \cr {2,} & {otherwise,} \cr } } \right.$$ then $$\int\limits_{ - 2}^3 {f\left( x \right)dx = } $$

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

The value of the integral $$\int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {{{{{\log }_e}x} \over x}} \right|dx} $$ is :