The region R is the region in the first quadrant bounded by the curves y = x^2-4x+4, x=0, and x=2.
Find a value h such that the vertical line x = h divided the region R into two regions of equal area. You get h = ?
I solved for the total area under the curve (8/3) and the area of each half (4/3). How do I get the h value from here?
.4126
You are correct in the total area to be 8/3 and half of that would be 4/3
You also must have had the correct integral to obtain the 8/3
so just evaluate it from x = 0 to h instead of 0 to 2, then set that equal to 4/3
that is,
(1/3)(h^3) -2(h^2) + 4h - 0 = 4/3
This does not solve easily, unless I made an arithmetic error
I think you forgot to include y=0 as part of the boundary,
and the line x=2 plays no role, since the vertex of the curve is at (2,0)
So, I'll go on assuming the triangular region bounded by the curve and the x- and y-axes.
As you say, ∫[0,2] (x^2-4x+4) dx = 8/3
So now you want
∫[0,h] (x^2-4x+4) dx = 4/3
That means that
1/3 h^3 - 2h^2 + 4h = 4/3
h^3-6h^2+12h-4 = 0
h^3-6h^2+12h-8 = -4
(h-2)^3 = -4
h-2 = -∛4
h = 2-∛4
Thanks oobleck, my weakness is not reading questions
completely.
Well, finding the value of h is like finding the perfect slice in a pizza with equal toppings on both sides. In this case, we want to find the perfect vertical line that divides our region R into two equal halves.
Now that you know each half should have an area of 4/3, let's use our clown math skills to solve this. We can set up an integral to find the area of one half of the region:
∫[0 to h] (x^2 - 4x + 4) dx = 4/3
Integrating that will give us:
[(x^3/3) - 2x^2 + 4x] from 0 to h = 4/3
Now, let's plug in the values for the upper and lower limit:
(h^3/3) - 2h^2 + 4h - 0 = 4/3
Simplifying this equation will give us the value of h that divides the region equally. Solve for h, and voila! You'll have the value you're looking for.
To find the value of h, we need to set up an integral that represents the area between the curve and the dividing line x = h.
Since the region R is bounded by the curves y = x^2 - 4x + 4, x = 0, and x = 2, we can set up the integral as follows:
A(h) = ∫[0 to h] (x^2 - 4x + 4) dx
To find the value of h that divides the region into two equal areas, we need to solve the equation:
∫[0 to h] (x^2 - 4x + 4) dx = (1/2) * ∫[0 to 2] (x^2 - 4x + 4) dx
Simplifying this equation gives:
2 * ∫[0 to h] (x^2 - 4x + 4) dx = ∫[0 to 2] (x^2 - 4x + 4) dx
Now, let's integrate both sides:
[2 * (1/3)x^3 - 2x^2 + 4x] from x = 0 to h = [(1/3)h^3 - 2h^2 + 4h] - [(1/3)0^3 - 2(0)^2 + 4(0)]
Simplifying further, we get:
(1/3)h^3 - 2h^2 + 4h = (1/3) * 2^3 - 2(2)^2 + 4(2)
(1/3)h^3 - 2h^2 + 4h = 8/3 - 8 + 8
(1/3)h^3 - 2h^2 + 4h = 8/3
Now, to solve for h, we can move all terms to one side:
(1/3)h^3 - 2h^2 + 4h - 8/3 = 0
Multiply through by 3 to get rid of the fraction:
h^3 - 6h^2 + 12h - 8 = 0
At this point, we can use numerical methods (such as Newton's method or a graphing calculator) to solve for h.
Alternatively, we can simplify the equation to:
(h - 2)^3 = 0
This implies h = 2.
Therefore, the value of h that divides the region R into two regions of equal area is h = 2.