The minimum value of the function $$f(x) = \int\limits_0^2 {{e^{|x - t|}}dt} $$ is :

$$\int_{{{3\sqrt 2 } \over 4}}^{{{3\sqrt 3 } \over 4}} {{{48} \over {\sqrt {9 - 4{x^2}} }}dx} $$ is equal to

The area enclosed by the curves $${y^2} + 4x = 4$$ and $$y - 2x = 2$$ is :

If $$[t]$$ denotes the greatest integer $$\leq t$$, then the value of $$\int_{0}^{1}\left[2 x-\left|3 x^{2}-5 x+2\right|+1\right] \mathrm{d} x$$ is :