To evaluate the double integral $$\int\limits_0^8 {\left( {\int\limits_{y/2}^{\left( {y/2} \right) + 1} {\left( {{{2x - y} \over 2}} \right)dx} } \right)dy,\,\,} $$ we make the substitution $$u = \left( {{{2x - y} \over 2}} \right)$$ and $$v = {y \over 2}.$$ The integral will reduce to

The minimum value of the function $$f\left( x \right) = {x^3} - 3{x^2} - 24x + 100$$ in the interval $$\left[ { - 3,3} \right]$$ is

The maximum value of $$f\left( x \right) = {x^3} - 9{x^2} + 24x + 5$$ in the interval $$\left[ {1,6} \right]$$ is

The value of the quantity, where $$P = \int\limits_0^1 {x{e^x}\,dx\,\,\,} $$ is