Given that for each $$a \in \left( {0,1} \right),\,\,\,\mathop {\lim }\limits_{h \to {0^ + }} \,\int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}dt} $$ exists. Let this limit be $$g(a).$$ In addition, it is given that the function $$g(a)$$ is differentiable on $$(0,1).$$

The value of $$g\left( {{1 \over 2}} \right)$$ is

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 $$f$$ $$:\,\,\left[ {{1 \over 2},1} \right] \to R$$ (the set of all real number) be a positive,
non-constant and differentiable function such that
$$f'\left( x \right) < 2f\left( x \right)$$ and $$f\left( {{1 \over 2}} \right) = 1.$$ Then the value of $$\int\limits_{1/2}^1 {f\left( x \right)} \,dx$$ lies in the interval

The value of the integral $$\int\limits_{ - \pi /2}^{\pi /2} {\left( {{x^2} + 1n{{\pi + x} \over {\pi - x}}} \right)\cos xdx} $$ is