To evaluate the double integral \[\int_0^8 {\left( {\int_{\frac{{\text{y}}}{2}}^{\frac{{\text{y}}}{2} + 1} {\left( {\frac{{2{\text{x}} - {\text{y}}}}{2}} \right){\text{dx}}} } \right){\text{dy,}}} \]      we make the substitution \[{\text{u}} = \frac{{2{\text{x}} - {\text{y}}}}{2}\]   and \[{\text{v}} = \frac{{\text{y}}}{2}.\]  The integral will reduce to

A. \[\int_0^4 {\left( {\int_0^2 {2{\text{u du}}} } \right){\text{dv}}} \]

B. \[\int_0^4 {\left( {\int_0^1 {2{\text{u du}}} } \right){\text{dv}}} \]

C. \[\int_0^4 {\left( {\int_0^1 {{\text{u du}}} } \right){\text{dv}}} \]

D. \[\int_0^4 {\left( {\int_0^2 {{\text{u du}}} } \right){\text{dv}}} \]

Answer: Option B