The value of the integral $$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx} $$ is equal to:

Let g(x) = $$\int_0^x {f(t)dt} $$, where f is continuous function in [ 0, 3 ] such that $${1 \over 3}$$ $$ \le $$ f(t) $$ \le $$ 1 for all t$$\in$$ [0, 1] and 0 $$ \le $$ f(t) $$ \le $$ $${1 \over 2}$$ for all t$$\in$$ (1, 3]. The largest possible interval in which g(3) lies is :

Let f : R $$ \to $$ R be defined as f(x) = e^$$-$$xsinx. If F : [0, 1] $$ \to $$ R is a differentiable function with that F(x) = $$\int_0^x {f(t)dt} $$, then the value of $$\int_0^1 {(F'(x) + f(x)){e^x}dx} $$ lies in the interval

If the integral

$$\int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx = \alpha {e^{ - 1}} + \beta {e^{ - {1 \over 2}}} + \gamma $$, where $$\alpha$$, $$\beta$$, $$\gamma$$ are integers and [x] denotes the greatest integer less than or equal to x, then the value of $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to :