If the value of the integral
$$\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$$

is $${k \over 6}$$, then k is equal to :

$$\int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx} $$ is equal to :

Let a function ƒ : [0, 5] $$ \to $$ R be continuous, ƒ(1) = 3 and F be defined as :

$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$

Then for the function F, the point x = 1 is :

If for all real triplets (a, b, c), ƒ(x) = a + bx + cx²; then $$\int\limits_0^1 {f(x)dx} $$ is equal to :