Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y = 4(x^(1/2)), y=4, and 2y +2x = 6
I keep getting an area around 21.3 but it is incorrect. Am I close?
Thank you!
odd that one function is written as
2y = 4√x
why not just y = 2√x ?
Anyway, assuming the functions are correctly written, the area is roughly triangular, with vertices at (-1,4),(1,2),(4,4)
Integrating along x, we need to divide it up into two regions, over [-1,1] and [1,4]
a = ∫[-1,1] 4 - (3-x) dx + ∫[1,4] 4 - 2√x dx
= (x + 1/2 x^2)[-1,1] + (4x - 4/3 x^(3/2))[1,4]
= 2 + 8/3
= 14/3
Integrating along y is easier, since we just subtract the left from the right:
a = ∫[2,4] y^2/4 - (3-y) dy
= y^3/12 + y^2/2 - 3y [2,4]
= 14/3
To find the area of the region enclosed by the given curves, we need to first sketch the region and decide whether to integrate with respect to x or y.
Let's start by sketching the curves. The first equation, 2y = 4(x^(1/2)), represents a curve that is a square root function. To sketch it, we can rewrite the equation as y = 2(x^(1/2)). The resulting curve is a right-ward opening parabola that passes through the origin (0,0).
The second equation, y = 4, represents a horizontal line parallel to the x-axis and passing through y = 4.
The third equation, 2y + 2x = 6, can be rewritten as y = 3 - x. This equation represents a straight line with a negative slope passing through the points (0,3) and (3,0).
Now, let's determine the region enclosed by the curves. The region is enclosed between the square root function curve y = 2(x^(1/2)), the horizontal line y = 4, and the line y = 3 - x.
To find the area of this region, we need to integrate with respect to x or y. In this case, it is simpler to integrate with respect to x.
To find the interval for integration, we need to find the x-values where the curves intersect.
The square root function curve and the horizontal line intersect when 2(x^(1/2)) = 4. Solving for x, we get x = 4.
The square root function curve and the straight line intersect when 2(x^(1/2)) = 3 - x. Solving for x, we get x = 1.
Therefore, the interval for integration is from x = 1 to x = 4.
To find the area, we need to integrate the difference between the top curve and the bottom curve within this interval. The top curve is y = 4, and the bottom curve is y = 2(x^(1/2)).
Since we are integrating with respect to x, the integral for the area is given by:
A = ∫[1 to 4] [4 - 2(x^(1/2))] dx
Evaluating this integral, we get:
A = [4x - 4(x^(3/2))/3] [1 to 4]
= [16 - 4(8/3)/3] - [4 - 4(1^(3/2))/3]
= (16 - 32/3) - (4 - 4/3)
= (48/3 - 32/3) - (12/3 - 4/3)
= (16/3) - (8/3)
= 8/3
Therefore, the area of the region enclosed by the given curves is 8/3 or approximately 2.67 units squared.
Your calculated area is 21.3, which is significantly larger than the correct answer. It seems like a mistake was made in the calculation. Double-check the integration and arithmetic steps to ensure accuracy.