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

Question

Note.It is a straight line.

Answer 1

Huh? It seems that you want the area under some arbitrary horizontal line.
So, let's say the line is through (0,6) and (2,6)
The area would then be (using symmetry of the region)
A = 2∫[0,2] 6 - ((1/2*x-2)^6+5) dx
That's just a simple substitution, letting
u = x/2 - 2
2du = dx
A = 2∫[-2,-1] 1-u^6 du

Answer 2

Answer is 438+6/7

Answer 3

I guess if you want an arbitrary x value, you could use this.

https://www.wolframalpha.com/input/?i=2%E2%88%AB%5B0%2Cx%5D+%28%28t%2F2-2%29%5E6+%2B+5%29+dt

Answer 4

Oops. That's the area under the curve, not the area above the curve. You can adjust it as needed, I guess.

Answer 5

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 ...