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.
To find the area above the curve y = (1/2x - 2)^6 + 5, enclosed by a line that cuts at (0, y) and (x, y), we need to calculate the area using integration.
First, note that the axis of symmetry is x = 4, not x = 0. This means that we will consider x values greater than or equal to 4.
The desired area is a rectangle with base (x - 4) and height y. The formula for the area of a rectangle is given by A = base * height. So, the area we are looking for can be expressed as:
A = 2(x - 4)y
Now, we need to find the value of y to substitute into the formula. We can do this by evaluating the equation of the curve at x.
Substituting x gives:
y = (1/2x - 2)^6 + 5
Next, we need to subtract the area under the curve from the rectangle's area. To find the area under the curve, we need to evaluate the integral of the curve function from x = 2 to x.
The integral of y = (1/2x - 2)^6 + 5 with respect to x over the interval [2, x] gives us the area under the curve.
Integrating this equation with respect to x gives:
∫[(1/2t - 2)^6 + 5]dt
Now, we can substitute this integral into the formula for the area above the curve:
A = 2[(x - 4) * ((1/2x - 2)^6 + 5)] - ∫[(1/2t - 2)^6 + 5]dt
Simplifying this equation will give us the final expression for the area.