Let $$T>0$$ be a fixed real number . Suppose $$f$$ is a continuous
function such that for all $$x \in R$$, $$f\left( {x + T} \right) = f\left( x \right)$$.

If $$I = \int\limits_0^T {f\left( x \right)dx} $$ then the value of $$\int\limits_3^{3 + 3T} {f\left( {2x} \right)dx} $$ is

The value of $$\int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}dx,\,a > 0,} $$ is

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 :