Question 763107
Find the area of the region in the {{{xy}}} - plane under the line {{{y = x/2}}}  above the {{{x-axis}}}, and between the lines {{{x=8}}} and {{{x=24}}} 


{{{drawing(600,600,   -30, 30, -10,15,blue(line(24,15,24,-15)), blue(line(8,15,8,-15)), grid(0),
graph( 600,600,   -30, 30, -10, 15,x/2)) }}}

 the region in the xy - plane is right-angular trapezoid and its area is:

{{{A=(1/2)(h[1]+h[2])*a}}}

in this case {{{a=24-8=16}}}, {{{h[1]=4}}}, and {{{h[2]=12}}} because if {{{x=24}}} then {{{y = 24/2=12}}}

{{{A=(1/2)(4+12)*16}}}

{{{A=(1/cross(2))(16)*cross(16)8}}}

{{{A=16*8}}}

{{{A=128}}}