Find the area above curve y=(1/2*x-2)^6+5 enclosed by a line cutting at (0,y) and (x,y).
Ans.438.85
To find the area enclosed by the curve y=(1/2*x-2)^6+5 and the line segment (0,y) to (x,y), we can use integration.
First, let's find the x-values where the curve intersects the line segment. Since the line is a horizontal line at y, the equation of the line can be written as y = y.
Set y = (1/2*x-2)^6+5 equal to y, and solve for x:
y = (1/2*x-2)^6+5
y - 5 = (1/2*x-2)^6
√(y - 5) = |1/2*x-2|
√(y - 5) = 1/2*x-2 (taking the positive value)
√(y - 5) + 2 = 1/2*x
2*(√(y - 5) + 2) = x
Now we have the x-coordinate of the points where the curve intersects the line segment.
To find the area enclosed by the curve, we can set up the integral. The area can be expressed as the definite integral from 0 to x:
Area = ∫[0 to x] [(1/2*t-2)^6+5] dt
Integrating this expression will give us the area.
Evaluate the integral, and the resulting value will be the area enclosed by the curve and the line segment.
Using proper integration techniques, you should be able to compute the integral and find the area.
If you have the mathematical tools or software, you can simply input the equation and limits into an appropriate integral calculator to get the numerical result.