$$\int_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $$,

where [t] denotes greatest integer less than or equal to t, is equal to:

Let f : R $$\to$$ R be a differentiable function such that $$f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0$$ and $$f'\left( {{\pi \over 2}} \right) = 1$$ and

let $$g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt} $$ for $$x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)$$. Then $$\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$$ is equal to :

Let f : R $$\to$$ R be a continuous function satisfying f(x) + f(x + k) = n, for all x $$\in$$ R where k > 0 and n is a positive integer. If $${I_1} = \int\limits_0^{4nk} {f(x)dx} $$ and $${I_2} = \int\limits_{ - k}^{3k} {f(x)dx} $$, then :

Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $$\int\limits_0^1 {[ - 8{x^2} + 6x - 1]dx} $$ is equal to :