Find the number b such that the line y = b divides the region bounded by the curves y = 4x2 and y = 16 into two regions with equal area. (Round your answer to two decimal places.)
Because of summetry, we only need to look at the first quadrant
Area of whole thing in quadrant I
= ∫(16 - 4x^2) dx from 0 to 2
= [16x - (4/3)x^3] from 0 to 2
= 32 - (4/3)(8) - 0
= 64/3
So the area created by y = b must be 32/3
the intersection of y = b and y = 4x^2:
4x^2 = b
x = √b/2
so we would have:
area = ∫( b - 4x^2) dx from 0 to √b/2
= [bx - (4/3)x^3] from 0 to √b
= b√b/2 - (4/3)b√b/8 - 0
= (1/3)b√b
(1/3)b√b = 32/3
b√b = 32
square both sides
b^3 = 1024
b = 10.079
b = appr 10.08
Proof
http://www.wolframalpha.com/input/?i=area+between+y+%3D+16+and+y+%3D+4x%5E2
http://www.wolframalpha.com/input/?i=area+between+y+%3D+10.079+and+y+%3D+4x%5E2
notice in my graph I did the whole thing, but
21.3322 = appr (1/2)(128/3)
To find the value of b that divides the region into two equal areas, we need to find the x-values where the two curves intersect.
First, let's find the intersection points.
Setting the equations equal to each other:
4x^2 = 16
Dividing both sides by 4:
x^2 = 4
Taking the square root of both sides:
x = ±2
So, the intersection points are x = -2 and x = 2.
To find the area between the curves, we need to integrate the positive difference between the two functions from x = -2 to x = 2.
The area between the curves is given by the integral of (16 - 4x^2) with respect to x, from -2 to 2.
A = ∫(16 - 4x^2) dx evaluated from -2 to 2
Integrating:
A = [16x - (4/3)x^3] from -2 to 2
Evaluating the integral:
A = [16(2) - (4/3)(2)^3] - [16(-2) - (4/3)(-2)^3]
A = [32 - (4/3)(8)] - [-32 - (4/3)(-8)]
A = [32 - 32/3] - [-32 + 32/3]
A = 96/3 - 32/3 - 96/3 + 32/3
A = 0
This tells us that the area between the two curves is zero, so any value of b will divide it into two regions with equal area.
Therefore, there is no specific value of b that satisfies the condition of dividing the region into two equal areas.
To find the number b such that the line y = b divides the region bounded by the curves y = 4x^2 and y = 16 into two regions with equal area, we need to set up an integral expression and solve for b.
First, let's find the x-values where the two curves intersect. Set the two equations equal to each other:
4x^2 = 16
Divide both sides by 4:
x^2 = 4
Taking the square root of both sides, we have:
x = ±2
So, the two curves intersect at x = -2 and x = 2.
Next, to calculate the area between the curves, integrate the difference between the two equations with respect to x, from -2 to 2:
A = ∫[from -2 to 2] (16 - 4x^2) dx
Let's calculate this integral:
A = ∫[from -2 to 2] (16 - 4x^2) dx
= ∫[from -2 to 2] 16 dx - ∫[from -2 to 2] 4x^2 dx
= [16x] [from -2 to 2] - [4 * (1/3) * x^3] [from -2 to 2]
Evaluating these integrals at the limits, we get:
A = [16(2) - 16(-2)] - [4 * (1/3) * (2^3) - 4 * (1/3) * (-2^3)]
= [32 - (-32)] - [(32/3) - (-32/3)]
= 64 - (64/3)
= (192/3) - (64/3)
= 128/3
Now, since we want the line y = b to divide this area into two equal parts, the area below the line y = b will be (128/3) / 2 = 64/3.
We can now set up and solve the following equation to find b:
64/3 = ∫[from -2 to 2] (b - 4x^2) dx
Let's calculate this integral:
64/3 = ∫[from -2 to 2] (b - 4x^2) dx
= [bx] [from -2 to 2] - [4 * (1/3) * x^3] [from -2 to 2]
Evaluating these integrals at the limits, we get:
64/3 = [b(2) - b(-2)] - [4 * (1/3) * (2^3) - 4 * (1/3) * (-2^3)]
= [2b - (-2b)] - [(32/3) - (-32/3)]
= 4b - (64/3)
To solve for b, rearrange the equation:
4b - (64/3) = 64/3
4b = 64/3 + 64/3
4b = 128/3
b = (128/3) / 4
b = 32/3
Therefore, the number b such that the line y = b divides the region into two equal parts is b = 32/3, rounded to two decimal places.