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.

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

Dec 22, 2019
Answer is 438+6/7

Dec 22, 2019

Dec 22, 2019
Oops. That's the area under the curve, not the area above the curve. You can adjust it as needed, I guess.

Dec 22, 2019
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 ...

How did length become 2(x-4) and area of unwanted region (∫[2,x] ((t/2-2)^6 + 5) dt)) with bounds 2 and x?

Answer 1

This question is poorly posed. You have x and y being used both as an independent variable and as a function, as well as some unknown point. How about we change things a bit and pick the line through the points (0,k) and (h,k) ? That is, the line y = k.

Now, since the graph of f(x) has a vertex and absolute minimum at (4,5), the line y=k only bounds an area with the curve if k > 5.

So, the area between the line and the curve is
∫k-y dx = ∫k - ((x/2 - 2)^6 + 5) dx

So now we need to determine the interval to integrate on.
x = 2(2±(y-5)^(1/6)) = 4 ± 2(y-5)^(1/6)
So, the line y=k intersects the curve at h = x = 4-2(k-5)^(1/6) and 4+2(k-5)^(1/6)
So, using symmetry, the area is
2∫[4,h] k - ((x/2 - 2)^6 + 5) dx
And you can evaluate that at the integration limits.