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 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

If $$(x, y)$$ is continuous function defined over $$\left( {x,y} \right) \in \left[ {0,1} \right] \times \left[ {0,1} \right].\,\,\,$$ Given two constants, $$\,x > {y^2}$$ and $$\,y > {x^2},$$ the volume under $$f(x, y)$$ is