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=4x^1/2,y=5,2y+1x=5
Do you really mean "2y+1x=5" ?
It is not customary to use the coefficient 1 in front of a variable. It is unnecessary.
Since y=5 is one boundary, it would be easiest to integrate from the lowest y value of the region to y=5.
The equation "2y=4x^1/2"
would be better written as y = 2x^1/2
To sketch the region, let's start by plotting the given curves on a graph:
1. Plot the line y = 5. This is a horizontal line that intersects the y-axis at y = 5 and is parallel to the x-axis.
2. Plot the curve y = 2x^(1/2) (or y = 2√x). This is a square root function that starts at the origin (0, 0) and increases as x increases.
Now, let's look at the third equation. Assuming you meant 2y + x = 5 (without the coefficient of 1), we can rewrite it as x = 5 - 2y. This is a linear equation in the form of y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope is -2, and the y-intercept is 5.
To find the region enclosed by the given curves, we need to determine the limits of integration. Since the line y = 5 is the upper boundary, we will integrate from the lowest y-value of the region to y = 5. To find the lowest y-value, we can set y = 2√x (the curve) equal to y = 5 (the line) and solve for x:
2√x = 5
√x = 5/2
x = (5/2)^2
x = 25/4
So, the lowest y-value occurs at x = 25/4.
To find the area of the region, we need to set up the integral. Since we integrated with respect to y, the integral will be:
A = ∫[from y = 2√x to y = 5] (5 - 2y) dy
Simplifying the expression inside the integral:
A = ∫[from y = 2√x to y = 5] (5dy - 2ydy)
A = ∫[from y = 2√x to y = 5] (5 - 2y) dy
Evaluating the integral:
A = [5y - y^2] [from y = 2√x to y = 5]
Now, substitute the limits of integration:
A = [5(5) - (5)^2] - [5(2√x) - (2√x)^2]
A = [25 - 25] - [10√x - 4x]
A = -10√x + 4x
This is the equation for the area of the region enclosed by the given curves.