Find the area above curve y=(1/2*x-2)^6+5 enclosed by a line cutting at (0,y) and (x,y).

Note.It is a straight line

Oops. I forgot to note that the axis of symmetry is x=4, not x=0. So the area would be, for some arbitrary x>=4, since the area desired is a rectangle of area 2(x-4)y, you'd get
2((x-4)*((x/2-2)^6+5)-(∫[2,x] ((t/2-2)^6 + 5) dt))
= 53/1952 (x-4)^7 - 72/7

Better double-check my math ...
Please explain this.