find the area of the region.
2y=4sqrtx , y=3 and 2y+2x=6 ?
To find the area of the region, we need to find the points of intersection between the given curves and calculate the area between them.
First, let's find the points of intersection between the curve 2y = 4√x and the line y = 3:
Substitute y = 3 in the equation 2y = 4√x:
2(3) = 4√x
6 = 4√x
Divide both sides of the equation by 4:
√x = 6/4
√x = 3/2
Square both sides of the equation to eliminate the square root:
x = (3/2)^2
x = 9/4
So, the first point of intersection is (9/4, 3).
Now, let's find the points of intersection between the curve 2y + 2x = 6 and the line y = 3:
Substitute y = 3 in the equation 2y + 2x = 6:
2(3) + 2x = 6
6 + 2x = 6
Subtract 6 from both sides of the equation:
2x = 0
Divide both sides of the equation by 2:
x = 0
So, the second point of intersection is (0, 3).
Now that we have the points of intersection, we can find the area between the curves. Since one of the curves is a line, we'll integrate the other curve with respect to x.
The area between the curves is given by:
A = ∫[x1, x2] (f(x) - g(x)) dx
In this case, f(x) is the curve 2y = 4√x and g(x) is the line y = 3.
Substituting y = 3 in the curve equation, we get:
2(3) = 4√x
6 = 4√x
Divide both sides by 4:
√x = 3/2
Square both sides of the equation to eliminate the square root:
x = (3/2)^2
x = 9/4
So, x1 = 0 and x2 = 9/4.
The area A becomes:
A = ∫[0, 9/4] (4√x - 3) dx
Evaluating the integral:
A = [8/3 * x^(3/2) - 3x] from 0 to 9/4
Plugging in the bounds:
A = (8/3 * (9/4)^(3/2) - 3(9/4)) - (8/3 * 0^(3/2) - 3(0))
A = (8/3 * (9/4)^(1.5) - 27/4) - (0 - 0)
A = (8/3 * (√9/√4)^3 - 27/4) - 0
A = (8/3 * (3/2)^3 - 27/4)
A = (8/3 * 27/8 - 27/4)
A = (3/1 - 27/4)
A = 12/4 - 27/4
A = -15/4
Hence, the area of the region is -15/4 or -3.75 square units.