If $$f\left( x \right) = \int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}} dt,$$ then $$f(x)$$ increases in

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 integral $$\int\limits_{ - 1/2}^{1/2} {\left( {\left[ x \right] + \ell n\left( {{{1 + x} \over {1 - x}}} \right)} \right)dx} $$ equal to

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