If $$f\left( x \right) = {x^a}\log x$$ and $$f\left( 0 \right) = 0,$$ then the value of $$\alpha $$ for which Rolle's theorem can be applied in $$\left[ {0,1} \right]$$ is

The value of the integral $$\int\limits_0^1 {\sqrt {{{1 - x} \over {1 + x}}} dx} $$ is

The area enclosed between the curves $$y = a{x^2}$$ and
$$x = a{y^2}\left( {a > 0} \right)$$ is $$1$$ sq. unit, then the value of $$a$$ is

If $$f(x)$$ is differentiable and $$\int\limits_0^{{t^2}} {xf\left( x \right)dx = {2 \over 5}{t^5},} $$ then $$f\left( {{4 \over {25}}} \right)$$ equals