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=3sqrt(x) , y=3 and 2y+3x=6.

I cant sketch it for you. I think I would integrate dy

You have posted this same question 6 times.

PLease check back to see if it has been answered.

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To sketch the region enclosed by the curves, let's start by graphing each of the curves separately.

1. The equation 2y = 3√x can be rewritten as y = (3/2)√x.
This curve is a square root function that starts at the origin and increases as x increases.

2. The equation y = 3 represents a horizontal line at y = 3.

3. The equation 2y + 3x = 6 can be rewritten as y = (6-3x)/2.
This equation represents a line with negative slope and y-intercept at (0, 3) or (0, 3/2).

Now, let's find the points of intersection to determine the boundaries of the enclosed region:

1. The first curve is y = (3/2)√x.
2. The second curve is y = 3.
3. The third curve is y = (6-3x)/2.

To find the points of intersection, equate the first curve with the second curve and the third curve.

1. (3/2)√x = 3
Divide both sides by (3/2): √x = 2
Square both sides: x = 4
Now, substitute x = 4 into the first curve: y = (3/2)√4 = (3/2) * 2 = 3
So, the first curve intersects with the second curve at the point (4, 3).

2. Substitute y = (6-3x)/2 into the second curve:
(6-3x)/2 = 3
Multiply both sides by 2: 6-3x = 6
Subtract 6 from both sides: -3x = 0
Divide both sides by -3: x = 0
Now, substitute x = 0 into the third curve: y = (6-3*0)/2 = 6/2 = 3
So, the second curve intersects with the third curve at the point (0, 3).

Now, we can sketch the region enclosed by the curves:

1. Start by graphing the curves individually.
- The first curve is a square root function that starts at the origin and increases as x increases.
- The second curve is a horizontal line at y = 3.
- The third curve is a line with negative slope and y-intercept at (0, 3) or (0, 3/2).

2. The points of intersection are at (4, 3) and (0, 3).
Draw vertical lines at x = 0 and x = 4 to mark the boundaries of the region.

3. Shade the region enclosed by the curves.

To find the area of the region, we need to determine whether to integrate with respect to x or y.

Looking at the sketch, we can see that the region lies between two vertical lines. Therefore, we need to integrate with respect to x.

To find the area, integrate the difference in y-values with respect to x between the boundaries, which are x = 0 and x = 4. So, we need to calculate ∫[0, 4] (3 - (6-3x)/2) dx.

Evaluating the integral, we get the area of the region enclosed by the curves.