The value of the integral, $$\int\limits_1^3 {[{x^2} - 2x - 2]dx} $$, where [x] denotes the greatest integer less than or equal to x, is :

Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 $$-$$ x) for all x$$ \in $$ (0, 2), f(0) = 1 and f(2) = e². Then the value of $$\int\limits_0^2 {f(x)} dx$$ is :

Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) $$ \ne $$ 0 for all x $$ \in $$ R. If $$\left| {\matrix{ {f(x)} & {f'(x)} \cr {f'(x)} & {f''(x)} \cr } } \right|$$ = 0, for all x$$ \in $$R, then the value of f(1) lies in the interval :

$$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {\left( {\sin \sqrt t } \right)dt} } \over {{x^3}}}$$ is equal to :