Find the number b such that the line y = b divides the region bounded by the curves y = 4x2 and y = 1 into two regions with equal area. (Round your answer to two decimal places.)
To find the number b that divides the region into two equal areas, we need to first find the area of the entire region bounded by the curves y = 4x^2 and y = 1.
Step 1: Setting up the integral
Let's find the x-values where the curves intersect. Equating the two equations gives us:
4x^2 = 1
x^2 = 1/4
Taking the square root of both sides, we find:
x = ±1/2
So the two curves intersect at x = -1/2 and x = 1/2. To find the area between the curves, we integrate from -1/2 to 1/2.
Step 2: Setting up the integral for area between curves
The area between the curves is given by the integral of the difference of the two curves. So we have:
A = ∫[from -1/2 to 1/2] (4x^2 - 1) dx
Step 3: Evaluating the integral
Evaluating the integral will give us the area of the entire region. So let's solve it:
A = ∫[-1/2 to 1/2] (4x^2 - 1) dx
= [4/3 * x^3 - x] evaluated from -1/2 to 1/2
= (4/3 * (1/2)^3 - 1/2) - (4/3 * (-1/2)^3 - (-1/2))
= (4/3 * 1/8 - 1/2) - (4/3 * -1/8 - 1/2)
= (1/6 - 1/2) - (-1/6 - 1/2)
= (1/6 - 3/6) - (-1/6 - 3/6)
= -2/6 + 4/6
= 2/6
= 1/3
So, the area of the entire region bounded by the curves y = 4x^2 and y = 1 is 1/3.
Step 4: Finding the value of b
To divide this region into two equal areas, we need to find the value of b such that the area between the curve y = 4x^2 and the line y = b is equal to 1/3 * 1/2 = 1/6.
We can set up the integral for this area as follows:
1/6 = ∫[-1/2 to a] (4x^2 - b) dx
We need to find the value of a that satisfies this equation. To do that, we integrate and solve for a:
1/6 = [4/3 * x^3 - bx] evaluated from -1/2 to a
1/6 = 4/3 * a^3 - ba - (4/3 * (-1/2)^3 - b(-1/2))
1/6 = 4/3 * a^3 - ba - (4/3 * -1/8 + b/2)
1/6 = 4/3 * a^3 - ba - (-1/6 + b/2)
1/6 = 4/3 * a^3 - ba + 1/6 - b/2
0 = 4/3 * a^3 - ba - b/2
0 = 8/6 * a^3 - 6/6 * ba - 3/6 * b/2
0 = 8/6 * a^3 - 6/6 * ba - 3/12 * b
Simplifying the equation:
0 = 8/6 * a^3 - b * (6a + 3/12)
From this equation, we can solve for b by plugging in the value of a = 1/2 (which gives a/2 for calculating b):
0 = 8/6 * (1/2)^3 - b * (6 * (1/2) + 3/12)
Simplifying further:
0 = 2/6 - b * (3 + 1/4)
0 = 1/3 - b * (3 + 1/4)
Now, we can solve for b:
b * (3 + 1/4) = 1/3
b * (13/4) = 1/3
b = (1/3) / (13/4)
b = 4/39
Therefore, the value of b that divides the region into two equal areas is approximately 0.10 (rounded to two decimal places).
We need the whole area to be twice the
area under the line y=b.
y(1/2) = 1
y(√b/2) = b
So, integrating along x, we need
∫[0,1/2] 1-4x^2 dx = 2∫[0,√b/2] b-4x^2 dx
Now just evaluate the integrals and solve the polynomial for b.