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.

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?

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.

Apologies for the confusion in my previous explanation. Let me clarify the process to find the area above the curve.

Given the curve equation y = (1/2*x - 2)^6 + 5 and the line passing through the points (0, y) and (x, y), we need to find the area enclosed between the curve and the line.

To solve this, we first determine the equation of the line. Since the line is horizontal, it will have a constant value of y. Let's suppose the line intersects the curve at y = k. Therefore, the equation of the line can be written as y = k.

Next, we need to find the x-values where the line intersects the curve. We set the curve equation equal to the line equation:

(1/2*x - 2)^6 + 5 = k

Now, we solve this equation to find the x-values. We can do this by taking the sixth root of both sides and then solving for x:

(1/2*x - 2) = (k - 5)^(1/6)

1/2*x = (k - 5)^(1/6) + 2

x = 2 * ((k - 5)^(1/6) + 2)

We now have the x-values where the line intersects the curve. Let's call these values x1 and x2 (x1 < x2).

To find the area enclosed between the curve and the line, we integrate the difference between the curve and the line within the bounds x1 and x2:

Area = ∫[x1, x2] ((1/2*x - 2)^6 + 5 - k) dx

By performing the integration, we obtain the area as a function of k. To calculate the exact area, we need to know the value of k.

Regarding your question about the length becoming 2(x-4) and the unwanted region (∫[2,x] ((t/2-2)^6 + 5) dt)) with bounds 2 and x, it seems there might have been a misunderstanding or an error in the previous explanation. Please disregard that information.

I apologize for any confusion caused by my previous response.