Given that for each $$a \in \left( {0,1} \right),\,\,\,\mathop {\lim }\limits_{h \to {0^ + }} \,\int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}dt} $$ exists. Let this limit be $$g(a).$$ In addition, it is given that the function $$g(a)$$ is differentiable on $$(0,1).$$

The value of $$g\left( {{1 \over 2}} \right)$$ is

The function $$y=f(x)$$ is the solution of the differential equation
$${{dy} \over {dx}} + {{xy} \over {{x^2} - 1}} = {{{x^4} + 2x} \over {\sqrt {1 - {x^2}} }}\,$$ in $$(-1,1)$$ satisfying $$f(0)=0$$.
Then $$\int\limits_{ - {{\sqrt 3 } \over 2}}^{{{\sqrt 3 } \over 2}} {f\left( x \right)} \,d\left( x \right)$$ is

Let $$f:\left[ {0,2} \right] \to R$$ be a function which is continuous on $$\left[ {0,2} \right]$$ and is differentiable on $$(0,2)$$ with $$f(0)=1$$. Let
$$F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt} $$ for $$x \in \left[ {0,2} \right]$$. If $$F'\left( x \right) = f'\left( x \right)$$ for all $$x \in \left[ {0,2} \right]$$, then $$F(2)$$ equals

Let f₁ : R $$ \to $$ R, f₂ : [0, $$\infty $$) $$ \to $$ R, f₃ : R $$ \to $$ R, and f₄ : R $$ \to $$ [0, $$\infty $$) be defined by

$${f_1}\left( x \right) = \left\{ {\matrix{ {\left| x \right|} & {if\,x < 0,} \cr {{e^x}} & {if\,x \ge 0;} \cr } } \right.$$

f₂(x) = x² ;

$${f_3}\left( x \right) = \left\{ {\matrix{ {\sin x} & {if\,x < 0,} \cr x & {if\,x \ge 0;} \cr } } \right.$$

and

$${f_4}\left( x \right) = \left\{ {\matrix{ {{f_2}\left( {{f_1}\left( x \right)} \right)} & {if\,x < 0,} \cr {{f_2}\left( {{f_1}\left( x \right)} \right) - 1} & {if\,x \ge 0;} \cr } } \right.$$