If $$f\left( y \right) = {e^y},$$ $$g\left( y \right) = y;y > 0$$ and

$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} $$ then :

Let $$f(x)$$ be a function satisfying $$f'(x)=f(x)$$ with $$f(0)=1$$ and $$g(x)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}$$. Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,} $$ is

If $$f\left( {a + b - x} \right) = f\left( x \right)$$ then $$\int\limits_a^b {xf\left( x \right)dx} $$ is equal to

The value of the integral $$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $$ is